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  • How do I do a mail merge that includes images?

    - by Ian Ringrose
    I am trying to find out the practicalities of doing a mail merge when each “record” to be merged on includes some images. I need to: print letters And envelopes Both the letters and the envelopes have: Fixed text Fixed images Text that come from the mail merge record Images that come from the mail merge record I don’t know if all images will be the same size for every record, so a bit of simple “on the fly” automatic formatting may be needed . I need to be able to repeat a single item if I get a problem (e.g when folding the letter). What problems am I likely to have? Is Word 2007 up to this sort of mail merging, or should I be looking at a report writing tool? How do I restart a print run after a printer jam etc? What format should I store the “records” and there images in? E.g Can standard software cope with images that are stored in separate files named after the “CustomerId” that is in the “record” (I can write custom software if needed, but would rather use standard “of-the-shelf” software for the printing, I am planning on custom software for the data creation, so can output in whatever format is needed)

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  • Why does HP Update at remote system trigger RDP printing at local system?

    - by lcbrevard
    This is obscure. When connected with RDP to another system that has HP Update installed on it, either directly running the HP Update or having the notification pop up to ask if you want to run HP Update causes the local system to try to print something to peculiarly-chosen-local-printer. Case 1: Desktop Win 7 Ult system RDP connected to HP Laptop Win 7 Ult system. When HP Update runs on the laptop a dialog for XPS Writer Save As... appears on Desktop system. Even if you put in a name, nothing gets generated and the dialog repeats. And repeats. Until you (a) close the RDP connection and (b) clean out the queued entries. If the HP Update pops up the request to run the update and you are not at the desk when this happens, there can be dozens of queued requests for this bogus printing. NOTE: the XPS Writer is not selected as a default printer on either system. Case 2: (Different) HP Laptop Win 7 Ult system RDP connected to XP Pro "brand X" desktop system but with HP printer drivers installed. If the request to run HP Update notification pops on the XP system, dozens of attempts to print, in this case to a Versa Check Printer driver, are queued. Dismissing the HP request, closing RDP, and cleaning out the queue are required to stop this. NOTE: the Versa Check Writer is not selected as a default printer on either system. THE QUESTION: What the heck is going on here? Some kind of scripting or COM activity that is misdirected?

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  • How do I get a network printer installed in ubuntu 9.04?

    - by SoaperGEM
    My girlfriend's work computer now has Linux on one of the partitions, and for the most part it's running fine--except that I can't seem to get the network printers configured right. There are two of them: a Lanier MP 7500/LD275 and a Lanier MP C3000/LD430c, and Linux seems to have found them both automatically. I'll go through the steps of what I did, and what exactly went wrong. I went to Administration Printing, and clicked the new printer button. It searched for printers and found them both, listed under "Network Printers." I added them as new printers in succession. However, when I clicked "Print a Test Page," it failed saying there was a broken pipe. The device URIs were saved as socket://[ip address]:9100. I changed these to lpd://[ip address] per some online tutorial, which at first looked like it might have worked (but didn't). Then when I tried to print a test page, it first said Processing (and sometimes even Processing - printing test page, 4%, but always subsequently displays Idle - /usr/lib/cups/backend/lpd failed. Help! What do I do? It seems like Linux can find these printers just fine, and the drivers seem to be in place, so what's going wrong?

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  • Can't open shared drive after disconnecting vpn

    - by Matt McMinn
    I use a VPN to connect to my office network. On my local network I have another WinXP machine that shares a printer and a few shared folders. While I'm connected to my work VPN, I can access the shared printer and folders on the other machine just fine, and vice versa. Once I disconnect the VPN, I can't access the local machine any more, and the other machine can't access my machine. The network itself seems ok - I can ping the other machine, get to the internet, and get on to a web server shared by the other machine, but I can't get to the shared folders or printer. If I reconnect to the VPN, my access is restored. I'm guessing this is some sort of authentication thing, but I don't know what. Any ideas? Update This problem is bothering me again, so here's an update. Depending on when I first access the WinXP machine, I either have this problem, or the opposite problem. After a reboot, if I (for example) print, then connect to the VPN, I can't access the machine while on the VPN. If after a reboot I connect to the VPN, then print, I can't access the machine off the VPN. In both cases, if I enable/disable the VPN again, I can access the machine again. Thanks

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  • How can I totally flatten a PDF in Mac OS on the command line?

    - by Matthew Leingang
    I use Mac OS X Snow Leopard. I have a PDF with form fields, annotations, and stamps on it. I would like to freeze (or "flatten") that PDF so that the form fields can't be changed and the annotations/stamps are no longer editable. Since I actually have many of these PDFs, I want to do this automatically on the command line. Some things I've tried/considered, with their degree of success: Open in Preview and Print to File. This creates a totally flat PDF without changing the file size. The only way to automate seems to be to write a kludgy UI-based AppleScript, though, which I've been trying to avoid. Open in Acrobat Pro and use a JavaScript function to flatten. Again, not sure how to automate this on the command line. Use pdftk with the flatten option. But this only flattens form fields, not stamps and other annotations. Use cupsfilter which can create PDF from many file formats. Like pdftk this flattened only the form fields. Use cups-pdf to hook into the Mac's printserver and save a PDF file instead of print. I used the macports version. The resulting file is flat but huge. I tried this on an 8MB file; the flattened PDF was 358MB! Perhaps this can be combined with a ghostscript call as in Ubuntu Tip:Howto reduce PDF file size from command line. Any other suggestions would be appreciated.

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  • Excel equivilant of java's String.contains(String otherString)

    - by corsiKa
    I have a cell that has a fairly archaic String. (It's the mana cost of a Magic: the Gathering spell.) Examples are 3g, 2gg, 3ur, and bg. There are 5 possible letters (g w u b r). I have 5 columns and would like to count at the bottom how many of each it contains. So my spreadsheet might look like this A B C D E F G +-------------------------------------------- 1|Name Cost G W U B R 2|Centaur Healer 1gw 1 1 0 0 0 3|Sunspire Griffin 1ww 0 1 0 0 0 // just 1, even though 1ww 4|Rakdos Shred-Freak {br}{br} 0 0 0 1 1 Basically, I want something that looks like =if(contains($A2,C$1),1,0) and I can drag it across all 5 columns and down all 270 some cards. (Those are actual data, by the way. It's not mocked :-) .) In Java I would do this: String[] colors = { "B", "G", "R", "W", "U" }; for(String color : colors) { System.out.print(cost.toUpperCase().contains(color) ? 1 : 0); System.out.print("\t"); } Is there something like this in using Excel 2010. I tried using find() and search() and they work great if the color exists. But if the color doesn't exist, it returns #value - so I get 1 1 #value #value #value instead of 1 1 0 0 0 for, example, Centaur Healer (row 2). The formula used was if(find($A2,C$1) > 0, 1, 0).

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  • Firefox won't start

    - by Daniel R Hicks
    OK, I've got this problem again, only this time the problem only seems to affect Firefox and Thunderbird. Rebooted several times. Tried resetting to the last restore point, but that didn't work. Tried setting a new Firefox profile, and that didn't work either. The symptom is that you click on the Firefox or Thunderbird icon, the process appears in the Process Explorer list, but the window never opens. Curiously, if Firefox has been "started" this way, Internet Explorer hangs starting until I kill the Firefox process. Any ideas? I suppose the next thing to try is uninstalling and reinstalling Firefox/Thunderbird, but this whole thing is getting old. The box is a Sony Vaio running Windows Vista. It was completely restored from scratch less than two weeks ago, after the last fiasco. (I'm suspecting that my aborted install of Acronis True Image may have mucked things up this time.) Sigh! Another symptom: It occurred to me to try printing something, but if I open "Printers" it just sits there "searching". So something is rotten in the bowels of Windows. Minor update: It occurred to me to kill Internet Explorer (where I'd attempted printing). Then Printers comes up fairly quickly -- with no printers defined. Clicking "Add a printer" does nothing. Update: Well, following this suggestion to stop and restart the print spooler brought the printers back. And, wonder of wonders, Firefox now starts OK. Stopping and restarting the print spooler!!

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  • sharing a USB printer in SOHO environment [migrated]

    - by Registered User
    Here is a situation I am facing, there is USB printer which works only on a Windows XP machine, there are other devices in LAN it is a Small Office Home Office environment. How can this USB printer attached to Windows XP machine be shared so that other laptops or users in Network who have Windows 7 or Linux on their laptops can use this printer. The printer model number is Canon Laser Shot LBP-1210 http://www.canon-europe.com/For_Home/Product_Finder/Printers/Laser/LaserShot_LBP1210/index.asp a print server is not available to me I need to make it work in this situation only.What can I do? the clients are unable to connect to this.It is not a network or TCP/IP printer If a from Windows 7 machine some one wants to use this printer so that he can take a print he gets an error while adding the printer to his machine which is a Windows 7 machine (where as the printer is USB printer on Windows XP machine) Start--->Devices and Printers---> Add Printer---> Find Printer by name or IP address--->Selected a shared printer by name-->\\PC-Name-printer3 and select browse it gives a message Windows can not find a driver for Canon LASER SHOT LBP-1210 on the network what does this mean do I need to install some kind of software at client machine or on the machine where printer is present?

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  • Run Python script at startup using upstart

    - by MarcusMaximus
    I'm trying to create an upstart script to run a python script on startup. In theory it looks simple enough but I just can't seem to get it to work. I'm using a skeleton script I found here and altered. description "Used to start python script as a service" author "Me <[email protected]>" # Stanzas # # Stanzas control when and how a process is started and stopped # See a list of stanzas here: http://upstart.ubuntu.com/wiki/Stanzas#respawn # When to start the service start on runlevel [2345] # When to stop the service stop on runlevel [016] # Automatically restart process if crashed respawn # Essentially lets upstart know the process will detach itself to the background expect fork # Start the process script exec python /usr/local/scripts/script.py end script The test script I want it to run is currently a simple python script that runs without any issue when run from a terminal. #!/usr/bin/python2 import os, sys, time if __name__ == "__main__": for i in range (10000): message = "UpstartTest " , i , time.asctime() , " - Username: " , os.getenv("USERNAME") #print message time.sleep(60) out = open("/var/log/scripts/scriptlogfile", "a") print >> out, message out.close() The location/var/log/scripts has permissions 777 The file /usr/local/scripts/script.py has permissions 775 The upstart script /etc/init.d/pythonupstart.conf has permissions 755

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  • The script not working as expected files dump path

    - by user3319390
    I have a script needs to be dump matching cname from my file contains and then matching scode to dump file to $cname/$year/$month/$day/ into files like access and error logs #!/bin/sh #base_dir="/home/vizion/Desktop" path="/home/vizion/Desktop/adn_DF9D_20140515_0005.log" name=$(basename "$path" ".log") for x in *.log; do year=${x:9:4}; month=${x:13:2}; day=${x:15:2}; done while read -r line do cname=$(echo ${line} | awk '{split($7,c,"/"); print c[3]}') scode=$(echo ${line} | awk -F"[ ]" '{print $9}') [[ ! -d "$cname/$year/$month/$day" ]] && mkdir -p "$cname/$year/$month/$day/" [[ ( ${scode} -ge 200 ) && ( ${scode} -le 399 ) ]] && { # [[ ! -d "$cname/$year/$month/$day" ]] && mkdir -p "$cname/$year/$month/$day/" echo ${line} >> /home/vizion/Desktop/$cname/$year/$month/$day/${cname}_${name}_access.log } [[ ( ${scode} -ge 400 ) && ( ${scode} -le 599 ) ]] && { [[ ! -d "$cname/$year/$month/$day" ]] && mkdir -p "$cname/$year/$month/$day" echo ${line} >> ${cname}_${name}_error.log } done < $path i am able to filter logs but not not dumping the exact location It's going other locations suggest to me correction in script

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  • How do I do a mail merge that includes images? (Maybe in Word 2007)

    - by Ian Ringrose
    I am trying to find out the practicalities of doing a mail merge when each “record” to be merged on includes some images. I need to: print letters And envelopes Both the letters and the envelopes have: Fixed text Fixed images Text that come from the mail merge record Images that come from the mail merge record I don’t know if all images will be the same size for every record, so a bit of simple “on the fly” automatic formatting may be needed . I need to be able to repeat a single item if I get a problem (e.g when folding the letter). What problems am I likely to have? Is Word 2007 up to this sort of mail merging, or should I be looking at a report writing tool? How do I restart a print run after a printer jam etc? What format should I store the “records” and there images in? E.g Can standard software cope with images that are stored in separate files named after the “CustomerId” that is in the “record” (I can write custom software if needed, but would rather use standard “of-the-shelf” software for the printing, I am planning on custom software for the data creation, so can output in whatever format is needed)

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  • Printer deployment via Group Policy not working on a single system

    - by Aron Rotteveel
    One of my coworkers just got a new laptop running Windows 7 Pro x64. We use a GPO to deploy the printers to every system, but for some reason it is not working on this system. I have been breaking my head over this for the past 3 hours now without any result. The strange thing is that gpresult /H seems to indicate that the GPO did run. The hardware: Laptop: Windows 7 Professional x64 Print server: Windows Server 2008 x64 R1 HP Color LaserJet 2605dn HP LaserJet P2015 Driver packages on server: HP universal printer driver PCL5, both X86 as X64 Oddities and other info: GPO working flawlessly on every other system, including my own Windows 7 Ultimate X64 laptop gpresult /H shows the GPO being ran Windows Firewall completely disabled on the new laptop Below is the output for gpresult /H (in Dutch sadly, but I think you'll recognize it): Beleidsregels Windows-instellingen Printerverbindingen Pad Dominerend groepsbeleidsobject \\Server2008\HP Color LaserJet 2605dn Printers \\Server2008\HP LaserJet P2015 Printers Beheersjablonen Beleidsdefinities (ADMX-bestanden) opgehaald van de lokale computer. Configuratiescherm/Printers Beleid Instelling Dominerend groepsbeleidsobject Beperkingen van point-and-print Uitgeschakeld Printers Like I said, I have been trying to figure this out for the past few hours or so without any result, so you are my last hope. Any help is appreciated.

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  • Same script, different behavior [migrated]

    - by Antoine_935
    I just stumbled upon an interesting bug... Still trying to figure out what is exactly happening. Maybe you can help. First, the context. I'm currently building yet another man to html converter (for some reasons I won't motivate here, but I need it). So, have a look at the screenshot below (see the link), more precisely at the outlined spots. See? On the upper shell, I have &lt ; and &gt ;, that is, escaped html. While on the shell below I have < and directly. But as you can see (or do I seriously need looking glass ?), the command man 2 semget | webmanneris the same on both sides, as is the which webmanner. The two are executed roughly at the same moment, with no modification made to the script between. [Oops, cannot post pictures just yet... Here comes the link] http://aspyct.org/media/webmanner-bug.png But the shell below is older (open about 1 hour ago). Newer shells all print out &lt ;. So my first guess was that it somehow had a cached reference to the old inode of the file, or old blocks or whatever. So I modified parts of the script, at the start and then at the end, to print different messages. And, surprise, the message shown up on both terminals. But still, same difference between &lt ; and <. I'm confused... How to explain that behavior? I'm working on a OSX 10.8 (Mountain Lion) EDIT: OK, there is one big difference: the shell below uses ruby 1.9.3, while above is 1.8.7. Is there any known difference in string handling between the two versions ?

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  • SSH & SFTP: Should I assign one port to each user to facilitate bandwidth monitoring?

    - by BertS
    There is no easy way to track real-time per-user bandwidth usage for SSH and SFTP. I think assigning one port to each user may help. Idea of implementation Use case Bob, with UID 1001, shall connect on port 31001. Alice, with UID 1002, shall connect on port 31002. John, with UID 1003, shall connect on port 31003. (I do not want to lauch several sshd instances as proposed in question 247291.) 1. Setup for SFTP: In /etc/ssh/sshd_config: Port 31001 Port 31002 Port 31003 Subsystem sftp /usr/bin/sftp-wrapper.sh The file sftp-wrapper.sh starts the sftp server only if the port is the correct one: #!/bin/sh mandatory_port=3`id -u` current_port=`echo $SSH_CONNECTION | awk '{print $4}'` if [ $mandatory_port -eq $current_port ] then exec /usr/lib/openssh/sftp-server fi 2. Additional setup for SSH: A few lines in /etc/profile prevents the user from connecting on the wrong port: if [ -n "$SSH_CONNECTION" ] then mandatory_port=3`id -u` current_port=`echo $SSH_CONNECTION | awk '{print $4}'` if [ $mandatory_port -ne $current_port ] then echo "Please connect on port $mandatory_port." exit 1 fi fi Benefits Now it should be easy to monitor per-user bandwidth usage. A Rrdtool-based application could produce charts like this: I know this won't be a perfect calculation of the bandwidth usage: for example, if somebody launches a bruteforce attack on port 31001, there will be a lot of traffic on this port although not from Bob. But this is not a problem to me: I do not need an exact computation of per-user bandwidth usage, but an indicator that is approximately correct in standard situations. Questions Is the idea of assigning one port for each user is a good one? Is the proposed setup an reliable one? If I have to open dozens of ports for many users, should I expect a performance drawback? Do you know a rrdtool-based application which could make the chart above?

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  • HP LaserJet 1515: Disable "refill" warning

    - by Pekka
    I have a HP LaserJet 1515 connected to a Windows 7 PC. The Magenta cartridge is empty; the printer shows a warning to that effect, and won't let me print even black-and-white documents any more. I can't turn the warning off manually using the printer's small console: When I try to enter any menu, the display says "Menu access disabled". I have no idea why. There is a setting to override the warning, but it can't be changed using the Network interface in the browser (Although it is there on the status page) According to the manual,the HP printing tool is supposed to offer a switch for this, but it won't install on my Windows 7. It just rumbles about for half an hour, to magnificently exit with an "unknown error" requiring a reboot. On second look, the problem seems to be that Windows 7 just isn't supported. There is no download link for the tool when you specify Windows 7 as your OS. I just want to print a black-and-white-document on a printer whose black cartridge is still 65% full. Is this indeed impossible? On second thought, I'm cross-posting this on the HP support forum. I'll update here if anything comes up.

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  • Sudo won't execute command as another user

    - by TOdorus
    I'm trying to get a unicorn server to start when the server boots. I've created a shell script which works if I log as the ubuntu user and run /etc/init.d/unicorn start Shell script #!/bin/sh case "$1" in start) cd /home/ubuntu/projects/asbest/current/ unicorn_rails -c /home/ubuntu/projects/asbest/current/config/unicorn.rb -D -E production ;; stop) if ps aux | awk '{print $2 }' | grep `cat ~/projects/asbest/current/tmp/pids/unicorn.pid`> /dev/null; then kill `cat ~/projects/asbest/current/tmp/pids/uni$ ;; restart) $0 stop $0 start ;; esac When I rebooted the server I noticed that the unicorn server wasn't listening to a socket. Since I ran the code succesfully as the ubuntu user I modified the script to let it always use the ubuntu user via sudo. #!/bin/sh case "$1" in start) cd /home/ubuntu/projects/asbest/current/ sudo -u ubuntu unicorn_rails -c /home/ubuntu/projects/asbest/current/config/unicorn.rb -D -E production ;; stop) if ps aux | awk '{print $2 }' | grep `cat ~/projects/asbest/current/tmp/pids/unicorn.pid`> /dev/null; then sudo -u ubuntu kill `cat ~/projects/asbest/current/tmp/pids/uni$ ;; restart) $0 stop $0 start ;; esac After rebooting unicorn still wouldn't start, so I tried running the script from the command line. Now I get the following error sudo: unicorn_rails: command not found I've searched high and low to what could cause this, but I'm afraid I've tapped my limited understanding of Linux. From what I can understand is that although sudo should use the ubuntu user to execute the commands, it still uses the environment of the root user, which isn't configured to run ruby or unicorn. Does anybody have any experience with this?

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  • Finding Font Issues in Acrobat

    - by Jayme
    Ok let me try this again, so sorry for not be clear. We create our PDFs through Quark, then send to print. I usually create outlines on my EPS files before I load in Quark but forgot this time. We bypassed the font error that Quark gave us by accident and found out our PDF was bad too late and it cost a lot of money to fix. We are trying to find a way to check our PDF for font problems before we send it to print, in case this problem happens again. We just want to be extra sure that we have tried everything. What I see in Quark is what the font is supposed to look like. When I view my PDF, the text is mixed up. Its readable but doesn't look like its supposed to and the spacing is all off within the text. My boss told me about the preflight in Quark and the Internal Structure for the fonts. She was asking me if this would help and what the lingo all meant. (which is where my first question started) The image on the left is my EPS that is correct, the image on the right is from the PDF. The white text in the top right and the website at the bottom left is what is messed up. I am running Mac 10.5.8, Quark 7.5 and Acrobat 8.3.1. Thanks, Jayme

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  • WINDOWS 7: Make the contents of two folders appear in one

    - by big_smile
    In Windows 7, I have three folders: "Images", "assets" and "all". I want the contents of "Images " and "assets" to appear in "all" automatically without copying those files into that folder (e.g. I don't want to duplicate the files). I also only want the contents to be copied over and not the folders themslves (The reason for this is that if the folders are copied over, they will become sub-directories. I am using a printing hot folder that access "all" but it can't see any subdirectories in "all"). When Images and Assets are updated (e.g. with files being added or deleted), "all" should automatically update as well. How can I do this? This is what I have tried: Libraries: This is a feature built into Windows. It works exactly as I want. However, the print hot folder cannot recognise the library as a folder. Sym Link Extension: I can use this to make the "images" and "assets" folders appear as a sub directories of "all". However, I want the contents of "images"/"assets" to appear in the "all" folder (I don't want the directories to appear as sub directories, because as stated, the print hot folder cannot access sub directories).

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  • Convert Java program to C

    - by imicrothinking
    I need a bit of guidance with writing a C program...a bit of quick background as to my level, I've programmed in Java previously, but this is my first time programming in C, and we've been tasked to translate a word count program from Java to C that consists of the following: Read a file from memory Count the words in the file For each occurrence of a unique word, keep a word counter variable Print out the top ten most frequent words and their corresponding occurrences Here's the source program in Java: package lab0; import java.io.File; import java.io.FileReader; import java.util.ArrayList; import java.util.Calendar; import java.util.Collections; public class WordCount { private ArrayList<WordCountNode> outputlist = null; public WordCount(){ this.outputlist = new ArrayList<WordCountNode>(); } /** * Read the file into memory. * * @param filename name of the file. * @return content of the file. * @throws Exception if the file is too large or other file related exception. */ public char[] readFile(String filename) throws Exception{ char [] result = null; File file = new File(filename); long size = file.length(); if (size > Integer.MAX_VALUE){ throw new Exception("File is too large"); } result = new char[(int)size]; FileReader reader = new FileReader(file); int len, offset = 0, size2read = (int)size; while(size2read > 0){ len = reader.read(result, offset, size2read); if(len == -1) break; size2read -= len; offset += len; } return result; } /** * Make article word by word. * * @param article the content of file to be counted. * @return string contains only letters and "'". */ private enum SPLIT_STATE {IN_WORD, NOT_IN_WORD}; /** * Go through article, find all the words and add to output list * with their count. * * @param article the content of the file to be counted. * @return words in the file and their counts. */ public ArrayList<WordCountNode> countWords(char[] article){ SPLIT_STATE state = SPLIT_STATE.NOT_IN_WORD; if(null == article) return null; char curr_ltr; int curr_start = 0; for(int i = 0; i < article.length; i++){ curr_ltr = Character.toUpperCase( article[i]); if(state == SPLIT_STATE.IN_WORD){ article[i] = curr_ltr; if ((curr_ltr < 'A' || curr_ltr > 'Z') && curr_ltr != '\'') { article[i] = ' '; //printf("\nthe word is %s\n\n",curr_start); if(i - curr_start < 0){ System.out.println("i = " + i + " curr_start = " + curr_start); } addWord(new String(article, curr_start, i-curr_start)); state = SPLIT_STATE.NOT_IN_WORD; } }else{ if (curr_ltr >= 'A' && curr_ltr <= 'Z') { curr_start = i; article[i] = curr_ltr; state = SPLIT_STATE.IN_WORD; } } } return outputlist; } /** * Add the word to output list. */ public void addWord(String word){ int pos = dobsearch(word); if(pos >= outputlist.size()){ outputlist.add(new WordCountNode(1L, word)); }else{ WordCountNode tmp = outputlist.get(pos); if(tmp.getWord().compareTo(word) == 0){ tmp.setCount(tmp.getCount() + 1); }else{ outputlist.add(pos, new WordCountNode(1L, word)); } } } /** * Search the output list and return the position to put word. * @param word is the word to be put into output list. * @return position in the output list to insert the word. */ public int dobsearch(String word){ int cmp, high = outputlist.size(), low = -1, next; // Binary search the array to find the key while (high - low > 1) { next = (high + low) / 2; // all in upper case cmp = word.compareTo((outputlist.get(next)).getWord()); if (cmp == 0) return next; else if (cmp < 0) high = next; else low = next; } return high; } public static void main(String args[]){ // handle input if (args.length == 0){ System.out.println("USAGE: WordCount <filename> [Top # of results to display]\n"); System.exit(1); } String filename = args[0]; int dispnum; try{ dispnum = Integer.parseInt(args[1]); }catch(Exception e){ dispnum = 10; } long start_time = Calendar.getInstance().getTimeInMillis(); WordCount wordcount = new WordCount(); System.out.println("Wordcount: Running..."); // read file char[] input = null; try { input = wordcount.readFile(filename); } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); System.exit(1); } // count all word ArrayList<WordCountNode> result = wordcount.countWords(input); long end_time = Calendar.getInstance().getTimeInMillis(); System.out.println("wordcount: completed " + (end_time - start_time)/1000000 + "." + (end_time - start_time)%1000000 + "(s)"); System.out.println("wordsort: running ..."); start_time = Calendar.getInstance().getTimeInMillis(); Collections.sort(result); end_time = Calendar.getInstance().getTimeInMillis(); System.out.println("wordsort: completed " + (end_time - start_time)/1000000 + "." + (end_time - start_time)%1000000 + "(s)"); Collections.reverse(result); System.out.println("\nresults (TOP "+ dispnum +" from "+ result.size() +"):\n" ); // print out result String str ; for (int i = 0; i < result.size() && i < dispnum; i++){ if(result.get(i).getWord().length() > 15) str = result.get(i).getWord().substring(0, 14); else str = result.get(i).getWord(); System.out.println(str + " - " + result.get(i).getCount()); } } public class WordCountNode implements Comparable{ private String word; private long count; public WordCountNode(long count, String word){ this.count = count; this.word = word; } public String getWord() { return word; } public void setWord(String word) { this.word = word; } public long getCount() { return count; } public void setCount(long count) { this.count = count; } public int compareTo(Object arg0) { // TODO Auto-generated method stub WordCountNode obj = (WordCountNode)arg0; if( count - obj.getCount() < 0) return -1; else if( count - obj.getCount() == 0) return 0; else return 1; } } } Here's my attempt (so far) in C: #include <stdio.h> #include <stdlib.h> #include <stdbool.h> #include <string.h> // Read in a file FILE *readFile (char filename[]) { FILE *inputFile; inputFile = fopen (filename, "r"); if (inputFile == NULL) { printf ("File could not be opened.\n"); exit (EXIT_FAILURE); } return inputFile; } // Return number of words in an array int wordCount (FILE *filePointer, char filename[]) {//, char *words[]) { // count words int count = 0; char temp; while ((temp = getc(filePointer)) != EOF) { //printf ("%c", temp); if ((temp == ' ' || temp == '\n') && (temp != '\'')) count++; } count += 1; // counting method uses space AFTER last character in word - the last space // of the last character isn't counted - off by one error // close file fclose (filePointer); return count; } // Print out the frequencies of the 10 most frequent words in the console int main (int argc, char *argv[]) { /* Step 1: Read in file and check for errors */ FILE *filePointer; filePointer = readFile (argv[1]); /* Step 2: Do a word count to prep for array size */ int count = wordCount (filePointer, argv[1]); printf ("Number of words is: %i\n", count); /* Step 3: Create a 2D array to store words in the file */ // open file to reset marker to beginning of file filePointer = fopen (argv[1], "r"); // store words in character array (each element in array = consecutive word) char allWords[count][100]; // 100 is an arbitrary size - max length of word int i,j; char temp; for (i = 0; i < count; i++) { for (j = 0; j < 100; j++) { // labels are used with goto statements, not loops in C temp = getc(filePointer); if ((temp == ' ' || temp == '\n' || temp == EOF) && (temp != '\'') ) { allWords[i][j] = '\0'; break; } else { allWords[i][j] = temp; } printf ("%c", allWords[i][j]); } printf ("\n"); } // close file fclose (filePointer); /* Step 4: Use a simple selection sort algorithm to sort 2D char array */ // PStep 1: Compare two char arrays, and if // (a) c1 > c2, return 2 // (b) c1 == c2, return 1 // (c) c1 < c2, return 0 qsort(allWords, count, sizeof(char[][]), pstrcmp); /* int k = 0, l = 0, m = 0; char currentMax, comparedElement; int max; // the largest element in the current 2D array int elementToSort = 0; // elementToSort determines the element to swap with starting from the left // Outer a iterates through number of swaps needed for (k = 0; k < count - 1; k++) { // times of swaps max = k; // max element set to k // Inner b iterates through successive elements to fish out the largest element for (m = k + 1; m < count - k; m++) { currentMax = allWords[k][l]; comparedElement = allWords[m][l]; // Inner c iterates through successive chars to set the max vars to the largest for (l = 0; (currentMax != '\0' || comparedElement != '\0'); l++) { if (currentMax > comparedElement) break; else if (currentMax < comparedElement) { max = m; currentMax = allWords[m][l]; break; } else if (currentMax == comparedElement) continue; } } // After max (count and string) is determined, perform swap with temp variable char swapTemp[1][20]; int y = 0; do { swapTemp[0][y] = allWords[elementToSort][y]; allWords[elementToSort][y] = allWords[max][y]; allWords[max][y] = swapTemp[0][y]; } while (swapTemp[0][y++] != '\0'); elementToSort++; } */ int a, b; for (a = 0; a < count; a++) { for (b = 0; (temp = allWords[a][b]) != '\0'; b++) { printf ("%c", temp); } printf ("\n"); } // Copy rows to different array and print results /* char arrayCopy [count][20]; int ac, ad; char tempa; for (ac = 0; ac < count; ac++) { for (ad = 0; (tempa = allWords[ac][ad]) != '\0'; ad++) { arrayCopy[ac][ad] = tempa; printf("%c", arrayCopy[ac][ad]); } printf("\n"); } */ /* Step 5: Create two additional arrays: (a) One in which each element contains unique words from char array (b) One which holds the count for the corresponding word in the other array */ /* Step 6: Sort the count array in decreasing order, and print the corresponding array element as well as word count in the console */ return 0; } // Perform housekeeping tasks like freeing up memory and closing file I'm really stuck on the selection sort algorithm. I'm currently using 2D arrays to represent strings, and that worked out fine, but when it came to sorting, using three level nested loops didn't seem to work, I tried to use qsort instead, but I don't fully understand that function as well. Constructive feedback and criticism greatly welcome (...and needed)!

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  • Why is my simple recusive method for this game always off by 1?

    - by FrankTheTank
    I'm attempting to create a text-based version of this game: http://www.cse.nd.edu/java/SameGame.html Here is the code I have so far: #include <iostream> #include <vector> #include <ctime> class Clickomania { public: Clickomania(); std::vector<std::vector<int> > board; int move(int, int); bool isSolved(); void print(); void pushDown(); bool isValid(); }; Clickomania::Clickomania() : board(12, std::vector<int>(8,0)) { srand((unsigned)time(0)); for(int i = 0; i < 12; i++) { for(int j = 0; j < 8; j++) { int color = (rand() % 3) + 1; board[i][j] = color; } } } void Clickomania::pushDown() { for(int i = 0; i < 8; i++) { for(int j = 0; j < 12; j++) { if (board[j][i] == 0) { for(int k = j; k > 0; k--) { board[k][i] = board[k-1][i]; } board[0][i] = 0; } } } } int Clickomania::move(int row, int col) { bool match = false; int totalMatches = 0; if (row > 12 || row < 0 || col > 8 || col < 0) { return 0; } int currentColor = board[row][col]; board[row][col] = 0; if ((row + 1) < 12) { if (board[row+1][col] == currentColor) { match = true; totalMatches++; totalMatches += move(row+1, col); } } if ((row - 1) >= 0) { if (board[row-1][col] == currentColor) { match = true; totalMatches++; totalMatches += move(row-1, col); } } if ((col + 1) < 8) { if (board[row][col+1] == currentColor) { match = true; totalMatches++; totalMatches += move(row, col+1); } } if ((col - 1) >= 0) { if (board[row][col-1] == currentColor) { match = true; totalMatches++; totalMatches += move(row, col-1); } } return totalMatches; } void Clickomania::print() { for(int i = 0; i < 12; i++) { for(int j = 0; j < 8; j++) { std::cout << board[i][j]; } std::cout << "\n"; } } int main() { Clickomania game; game.print(); int row; int col; std::cout << "Enter row: "; std::cin >> row; std::cout << "Enter col: "; std::cin >> col; int numDestroyed = game.move(row,col); game.print(); std::cout << "Destroyed: " << numDestroyed << "\n"; } The method that is giving me trouble is my "move" method. This method, given a pair of coordinates, should delete all the squares at that coordinate with the same number and likewise with all the squares with the same number connected to it. If you play the link I gave above you'll see how the deletion works on a click. int Clickomania::move(int row, int col) { bool match = false; int totalMatches = 0; if (row > 12 || row < 0 || col > 8 || col < 0) { return 0; } int currentColor = board[row][col]; board[row][col] = 0; if ((row + 1) < 12) { if (board[row+1][col] == currentColor) { match = true; totalMatches++; totalMatches += move(row+1, col); } } if ((row - 1) >= 0) { if (board[row-1][col] == currentColor) { match = true; totalMatches++; totalMatches += move(row-1, col); } } if ((col + 1) < 8) { if (board[row][col+1] == currentColor) { match = true; totalMatches++; totalMatches += move(row, col+1); } } if ((col - 1) >= 0) { if (board[row][col-1] == currentColor) { match = true; totalMatches++; totalMatches += move(row, col-1); } } return totalMatches; } My move() method above works fine, as in, it will delete the appropriate "blocks" and replace them with zeros. However, the number of destroyed (value returned) is always one off (too small). I believe this is because the first call of move() isn't being counted but I don't know how to differentiate between the first call or subsequent calls in that recursive method. How can I modify my move() method so it returns the correct number of destroyed blocks?

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  • Help with Java Program for Prime numbers

    - by Ben
    Hello everyone, I was wondering if you can help me with this program. I have been struggling with it for hours and have just trashed my code because the TA doesn't like how I executed it. I am completely hopeless and if anyone can help me out step by step, I would greatly appreciate it. In this project you will write a Java program that reads a positive integer n from standard input, then prints out the first n prime numbers. We say that an integer m is divisible by a non-zero integer d if there exists an integer k such that m = k d , i.e. if d divides evenly into m. Equivalently, m is divisible by d if the remainder of m upon (integer) division by d is zero. We would also express this by saying that d is a divisor of m. A positive integer p is called prime if its only positive divisors are 1 and p. The one exception to this rule is the number 1 itself, which is considered to be non-prime. A positive integer that is not prime is called composite. Euclid showed that there are infinitely many prime numbers. The prime and composite sequences begin as follows: Primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, … Composites: 1, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, … There are many ways to test a number for primality, but perhaps the simplest is to simply do trial divisions. Begin by dividing m by 2, and if it divides evenly, then m is not prime. Otherwise, divide by 3, then 4, then 5, etc. If at any point m is found to be divisible by a number d in the range 2 d m-1, then halt, and conclude that m is composite. Otherwise, conclude that m is prime. A moment’s thought shows that one need not do any trial divisions by numbers d which are themselves composite. For instance, if a trial division by 2 fails (i.e. has non-zero remainder, so m is odd), then a trial division by 4, 6, or 8, or any even number, must also fail. Thus to test a number m for primality, one need only do trial divisions by prime numbers less than m. Furthermore, it is not necessary to go all the way up to m-1. One need only do trial divisions of m by primes p in the range 2 p m . To see this, suppose m 1 is composite. Then there exist positive integers a and b such that 1 < a < m, 1 < b < m, and m = ab . But if both a m and b m , then ab m, contradicting that m = ab . Hence one of a or b must be less than or equal to m . To implement this process in java you will write a function called isPrime() with the following signature: static boolean isPrime(int m, int[] P) This function will return true or false according to whether m is prime or composite. The array argument P will contain a sufficient number of primes to do the testing. Specifically, at the time isPrime() is called, array P must contain (at least) all primes p in the range 2 p m . For instance, to test m = 53 for primality, one must do successive trial divisions by 2, 3, 5, and 7. We go no further since 11 53 . Thus a precondition for the function call isPrime(53, P) is that P[0] = 2 , P[1] = 3 , P[2] = 5, and P[3] = 7 . The return value in this case would be true since all these divisions fail. Similarly to test m =143 , one must do trial divisions by 2, 3, 5, 7, and 11 (since 13 143 ). The precondition for the function call isPrime(143, P) is therefore P[0] = 2 , P[1] = 3 , P[2] = 5, P[3] = 7 , and P[4] =11. The return value in this case would be false since 11 divides 143. Function isPrime() should contain a loop that steps through array P, doing trial divisions. This loop should terminate when 2 either a trial division succeeds, in which case false is returned, or until the next prime in P is greater than m , in which case true is returned. Function main() in this project will read the command line argument n, allocate an int array of length n, fill the array with primes, then print the contents of the array to stdout according to the format described below. In the context of function main(), we will refer to this array as Primes[]. Thus array Primes[] plays a dual role in this project. On the one hand, it is used to collect, store, and print the output data. On the other hand, it is passed to function isPrime() to test new integers for primality. Whenever isPrime() returns true, the newly discovered prime will be placed at the appropriate position in array Primes[]. This process works since, as explained above, the primes needed to test an integer m range only up to m , and all of these primes (and more) will already be stored in array Primes[] when m is tested. Of course it will be necessary to initialize Primes[0] = 2 manually, then proceed to test 3, 4, … for primality using function isPrime(). The following is an outline of the steps to be performed in function main(). • Check that the user supplied exactly one command line argument which can be interpreted as a positive integer n. If the command line argument is not a single positive integer, your program will print a usage message as specified in the examples below, then exit. • Allocate array Primes[] of length n and initialize Primes[0] = 2 . • Enter a loop which will discover subsequent primes and store them as Primes[1] , Primes[2], Primes[3] , ……, Primes[n -1] . This loop should contain an inner loop which walks through successive integers and tests them for primality by calling function isPrime() with appropriate arguments. • Print the contents of array Primes[] to stdout, 10 to a line separated by single spaces. In other words Primes[0] through Primes[9] will go on line 1, Primes[10] though Primes[19] will go on line 2, and so on. Note that if n is not a multiple of 10, then the last line of output will contain fewer than 10 primes. Your program, which will be called Prime.java, will produce output identical to that of the sample runs below. (As usual % signifies the unix prompt.) % java Prime Usage: java Prime [PositiveInteger] % java Prime xyz Usage: java Prime [PositiveInteger] % java Prime 10 20 Usage: java Prime [PositiveInteger] % java Prime 75 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 % 3 As you can see, inappropriate command line argument(s) generate a usage message which is similar to that of many unix commands. (Try doing the more command with no arguments to see such a message.) Your program will include a function called Usage() having signature static void Usage() that prints this message to stderr, then exits. Thus your program will contain three functions in all: main(), isPrime(), and Usage(). Each should be preceded by a comment block giving it’s name, a short description of it’s operation, and any necessary preconditions (such as those for isPrime().) See examples on the webpage.

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  • Why is my simple recusive method's final return value always off by 1?

    - by FrankTheTank
    I'm attempting to create a text-based version of this game: http://www.cse.nd.edu/java/SameGame.html Here is the code I have so far: #include <iostream> #include <vector> #include <ctime> class Clickomania { public: Clickomania(); std::vector<std::vector<int> > board; int move(int, int); bool isSolved(); void print(); void pushDown(); bool isValid(); }; Clickomania::Clickomania() : board(12, std::vector<int>(8,0)) { srand((unsigned)time(0)); for(int i = 0; i < 12; i++) { for(int j = 0; j < 8; j++) { int color = (rand() % 3) + 1; board[i][j] = color; } } } void Clickomania::pushDown() { for(int i = 0; i < 8; i++) { for(int j = 0; j < 12; j++) { if (board[j][i] == 0) { for(int k = j; k > 0; k--) { board[k][i] = board[k-1][i]; } board[0][i] = 0; } } } } int Clickomania::move(int row, int col) { bool match = false; int totalMatches = 0; if (row > 12 || row < 0 || col > 8 || col < 0) { return 0; } int currentColor = board[row][col]; board[row][col] = 0; if ((row + 1) < 12) { if (board[row+1][col] == currentColor) { match = true; totalMatches++; totalMatches += move(row+1, col); } } if ((row - 1) >= 0) { if (board[row-1][col] == currentColor) { match = true; totalMatches++; totalMatches += move(row-1, col); } } if ((col + 1) < 8) { if (board[row][col+1] == currentColor) { match = true; totalMatches++; totalMatches += move(row, col+1); } } if ((col - 1) >= 0) { if (board[row][col-1] == currentColor) { match = true; totalMatches++; totalMatches += move(row, col-1); } } return totalMatches; } void Clickomania::print() { for(int i = 0; i < 12; i++) { for(int j = 0; j < 8; j++) { std::cout << board[i][j]; } std::cout << "\n"; } } int main() { Clickomania game; game.print(); int row; int col; std::cout << "Enter row: "; std::cin >> row; std::cout << "Enter col: "; std::cin >> col; int numDestroyed = game.move(row,col); game.print(); std::cout << "Destroyed: " << numDestroyed << "\n"; } The method that is giving me trouble is my "move" method. This method, given a pair of coordinates, should delete all the squares at that coordinate with the same number and likewise with all the squares with the same number connected to it. If you play the link I gave above you'll see how the deletion works on a click. int Clickomania::move(int row, int col) { bool match = false; int totalMatches = 0; if (row > 12 || row < 0 || col > 8 || col < 0) { return 0; } int currentColor = board[row][col]; board[row][col] = 0; if ((row + 1) < 12) { if (board[row+1][col] == currentColor) { match = true; totalMatches++; totalMatches += move(row+1, col); } } if ((row - 1) >= 0) { if (board[row-1][col] == currentColor) { match = true; totalMatches++; totalMatches += move(row-1, col); } } if ((col + 1) < 8) { if (board[row][col+1] == currentColor) { match = true; totalMatches++; totalMatches += move(row, col+1); } } if ((col - 1) >= 0) { if (board[row][col-1] == currentColor) { match = true; totalMatches++; totalMatches += move(row, col-1); } } return totalMatches; } My move() method above works fine, as in, it will delete the appropriate "blocks" and replace them with zeros. However, the number of destroyed (value returned) is always one off (too small). I believe this is because the first call of move() isn't being counted but I don't know how to differentiate between the first call or subsequent calls in that recursive method. How can I modify my move() method so it returns the correct number of destroyed blocks?

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  • Why does C qicksort function implementation works much slower (tape comparations, tape swapping) than bobble sort function?

    - by Artur Mustafin
    I'm going to implement a toy tape "mainframe" for a students, showing the quickness of "quicksort" class functions (recursive or not, does not really matters, due to the slow hardware, and well known stack reversal techniques) comparatively to the "bubblesort" function class, so, while I'm clear about the hardware implementation ans controllers, i guessed that quicksort function is much faster that other ones in terms of sequence, order and comparation distance (it is much faster to rewind the tape from the middle than from the very end, because of different speed of rewind). Unfortunately, this is not the true, this simple "bubble" code shows great improvements comparatively to the "quicksort" functions in terms of comparison distances, direction and number of comparisons and writes. So I have 3 questions: Does I have mistaken in my implememtation of quicksort function? Does I have mistaken in my implememtation of bubblesoft function? If not, why the "bubblesort" function is works much faster in (comparison and write operations) than "quicksort" function? I already have a "quicksort" function: void quicksort(float *a, long l, long r, const compare_function& compare) { long i=l, j=r, temp, m=(l+r)/2; if (l == r) return; if (l == r-1) { if (compare(a, l, r)) { swap(a, l, r); } return; } if (l < r-1) { while (1) { i = l; j = r; while (i < m && !compare(a, i, m)) i++; while (m < j && !compare(a, m, j)) j--; if (i >= j) { break; } swap(a, i, j); } if (l < m) quicksort(a, l, m, compare); if (m < r) quicksort(a, m, r, compare); return; } } and the kind of my own implememtation of the "bubblesort" function: void bubblesort(float *a, long l, long r, const compare_function& compare) { long i, j, k; if (l == r) { return; } if (l == r-1) { if (compare(a, l, r)) { swap(a, l, r); } return; } if (l < r-1) { while(l < r) { i = l; j = l; while (i < r) { i++; if (!compare(a, j, i)) { continue; } j = i; } if (l < j) { swap(a, l, j); } l++; i = r; k = r; while(l < i) { i--; if (!compare(a, i, k)) { continue; } k = i; } if (k < r) { swap(a, k, r); } r--; } return; } } I have used this sort functions in a test sample code, like this: #include <stdio.h> #include <stdlib.h> #include <math.h> #include <conio.h> long swap_count; long compare_count; typedef long (*compare_function)(float *, long, long ); typedef void (*sort_function)(float *, long , long , const compare_function& ); void init(float *, long ); void print(float *, long ); void sort(float *, long, const sort_function& ); void swap(float *a, long l, long r); long less(float *a, long l, long r); long greater(float *a, long l, long r); void bubblesort(float *, long , long , const compare_function& ); void quicksort(float *, long , long , const compare_function& ); void main() { int n; printf("n="); scanf("%d",&n); printf("\r\n"); long i; float *a = (float *)malloc(n*n*sizeof(float)); sort(a, n, &bubblesort); print(a, n); sort(a, n, &quicksort); print(a, n); free(a); } long less(float *a, long l, long r) { compare_count++; return *(a+l) < *(a+r) ? 1 : 0; } long greater(float *a, long l, long r) { compare_count++; return *(a+l) > *(a+r) ? 1 : 0; } void swap(float *a, long l, long r) { swap_count++; float temp; temp = *(a+l); *(a+l) = *(a+r); *(a+r) = temp; } float tg(float x) { return tan(x); } float ctg(float x) { return 1.0/tan(x); } void init(float *m,long n) { long i,j; for (i = 0; i < n; i++) { for (j=0; j< n; j++) { m[i + j*n] = tg(0.2*(i+1)) + ctg(0.3*(j+1)); } } } void print(float *m, long n) { long i, j; for(i = 0; i < n; i++) { for(j = 0; j < n; j++) { printf(" %5.1f", m[i + j*n]); } printf("\r\n"); } printf("\r\n"); } void sort(float *a, long n, const sort_function& sort) { long i, sort_compare = 0, sort_swap = 0; init(a,n); for(i = 0; i < n*n; i+=n) { if (fmod (i / n, 2) == 0) { compare_count = 0; swap_count = 0; sort(a, i, i+n-1, &less); if (swap_count == 0) { compare_count = 0; sort(a, i, i+n-1, &greater); } sort_compare += compare_count; sort_swap += swap_count; } } printf("compare=%ld\r\n", sort_compare); printf("swap=%ld\r\n", sort_swap); printf("\r\n"); }

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  • Unnecessary Error Message Being Displayed

    - by ThatMacLad
    I've set up a form to update my blog and it was working fine up until about this morning. It keeps on turning up with an Invalid Entry ID error on the edit post page when I click the update button despite the fact that it updates the homepage. All help is seriously appreciated. <html> <head> <title>Ultan's Blog | New Post</title> <link rel="stylesheet" href="css/editpost.css" type="text/css" /> </head> <body> <div class="new-form"> <div class="header"> </div> <div class="form-bg"> <?php mysql_connect ('localhost', 'root', 'root') ; mysql_select_db ('tmlblog'); if (isset($_POST['update'])) { $id = htmlspecialchars(strip_tags($_POST['id'])); $month = htmlspecialchars(strip_tags($_POST['month'])); $date = htmlspecialchars(strip_tags($_POST['date'])); $year = htmlspecialchars(strip_tags($_POST['year'])); $time = htmlspecialchars(strip_tags($_POST['time'])); $entry = $_POST['entry']; $title = htmlspecialchars(strip_tags($_POST['title'])); if (isset($_POST['password'])) $password = htmlspecialchars(strip_tags($_POST['password'])); else $password = ""; $entry = nl2br($entry); if (!get_magic_quotes_gpc()) { $title = addslashes($title); $entry = addslashes($entry); } $timestamp = strtotime ($month . " " . $date . " " . $year . " " . $time); $result = mysql_query("UPDATE php_blog SET timestamp='$timestamp', title='$title', entry='$entry', password='$password' WHERE id='$id' LIMIT 1") or print ("Can't update entry.<br />" . mysql_error()); header("Location: post.php?id=" . $id); } if (isset($_POST['delete'])) { $id = (int)$_POST['id']; $result = mysql_query("DELETE FROM php_blog WHERE id='$id'") or print ("Can't delete entry.<br />" . mysql_error()); if ($result != false) { print "The entry has been successfully deleted from the database."; exit; } } if (!isset($_GET['id']) || empty($_GET['id']) || !is_numeric($_GET['id'])) { die("Invalid entry ID."); } else { $id = (int)$_GET['id']; } $result = mysql_query ("SELECT * FROM php_blog WHERE id='$id'") or print ("Can't select entry.<br />" . $sql . "<br />" . mysql_error()); while ($row = mysql_fetch_array($result)) { $old_timestamp = $row['timestamp']; $old_title = stripslashes($row['title']); $old_entry = stripslashes($row['entry']); $old_password = $row['password']; $old_title = str_replace('"','\'',$old_title); $old_entry = str_replace('<br />', '', $old_entry); $old_month = date("F",$old_timestamp); $old_date = date("d",$old_timestamp); $old_year = date("Y",$old_timestamp); $old_time = date("H:i",$old_timestamp); } ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <p><input type="hidden" name="id" value="<?php echo $id; ?>" /> <strong><label for="month">Date (month, day, year):</label></strong> <select name="month" id="month"> <option value="<?php echo $old_month; ?>"><?php echo $old_month; ?></option> <option value="January">January</option> <option value="February">February</option> <option value="March">March</option> <option value="April">April</option> <option value="May">May</option> <option value="June">June</option> <option value="July">July</option> <option value="August">August</option> <option value="September">September</option> <option value="October">October</option> <option value="November">November</option> <option value="December">December</option> </select> <input type="text" name="date" id="date" size="2" value="<?php echo $old_date; ?>" /> <select name="year" id="year"> <option value="<?php echo $old_year; ?>"><?php echo $old_year; ?></option> <option value="2004">2004</option> <option value="2005">2005</option> <option value="2006">2006</option> <option value="2007">2007</option> <option value="2008">2008</option> <option value="2009">2009</option> <option value="2010">2010</option> </select> <strong><label for="time">Time:</label></strong> <input type="text" name="time" id="time" size="5" value="<?php echo $old_time; ?>" /></p> <p><strong><label for="title">Title:</label></strong> <input type="text" name="title" id="title" value="<?php echo $old_title; ?>" size="40" /> </p> <p><strong><label for="password">Password protect?</label></strong> <input type="checkbox" name="password" id="password" value="1"<?php if($old_password == 1) echo " checked=\"checked\""; ?> /></p> <p><textarea cols="80" rows="20" name="entry" id="entry"><?php echo $old_entry; ?></textarea></p> <p><input type="submit" name="update" id="update" value="Update"></p> </form> <p><strong>Be absolutely sure that this is the post that you wish to remove from the blog!</strong><br /> </p> <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <input type="hidden" name="id" id="id" value="<?php echo $id; ?>" /> <input type="submit" name="delete" id="delete" value="Delete" /> </form> </div> </div> </div> <div class="bottom"></div> </body> </html>

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  • what "Debug Assertion Failed" mean and how to fix it, c++?

    - by nonon
    hi, why this program gives me a "Debug Assertion Failed" Error Message while running #include "stdafx.h" #include "iostream" #include "fstream" #include "string" using namespace std; int conv_ch(char b) { int f; f=b; b=b+0; switch(b) { case 48: f=0; break; case 49: f=1; break; case 50: f=2; break; case 51: f=3; break; case 52: f=4; break; case 53: f=5; break; case 54: f=6; break; case 55: f=7; break; case 56: f=8; break; case 57: f=9; break; default: f=0; } return f; } class Student { public: string id; size_t id_len; string first_name; size_t first_len; string last_name; size_t last_len; string phone; size_t phone_len; string grade; size_t grade_len; void print(); void clean(); }; void Student::clean() { id.erase (id.begin()+6, id.end()); first_name.erase (first_name.begin()+15, first_name.end()); last_name.erase (last_name.begin()+15, last_name.end()); phone.erase (phone.begin()+10, phone.end()); grade.erase (grade.begin()+2, grade.end()); } void Student::print() { int i; for(i=0;i<6;i++) { cout<<id[i]; } cout<<endl; for(i=0;i<15;i++) { cout<<first_name[i]; } cout<<endl; for(i=0;i<15;i++) { cout<<last_name[i]; } cout<<endl; for(i=0;i<10;i++) { cout<<phone[i]; } cout<<endl; for(i=0;i<2;i++) { cout<<grade[i]; } cout<<endl; } int main() { Student k[80]; char data[1200]; int length,i,recn=0; int rec_length; int counter = 0; fstream myfile; char x1,x2; char y1,y2; char zz; int ad=0; int ser,j; myfile.open ("example.txt",ios::in); int right; int left; int middle; string key; while(!myfile.eof()){ myfile.get(data,1200); char * pch; pch = strtok (data, "#"); printf ("%s\n", pch); j=0; for(i=0;i<6;i++) { k[recn].id[i]=data[j]; j++; } for(i=0;i<15;i++) { k[recn].first_name[i]=data[j]; j++; } for(i=0;i<15;i++) { k[recn].last_name[i]=data[j]; j++; } for(i=0;i<10;i++) { k[recn].phone[i]=data[j]; j++; } for(i=0;i<2;i++) { k[recn].grade[i]=data[j]; j++; } recn++; j=0; } //cout<<recn; string temp1; size_t temp2; int temp3; for(i=0;i<recn-1;i++) { for(j=0;j<recn-1;j++) { if(k[i].id.compare(k[j].id)<0) { temp1 = k[i].first_name; k[i].first_name = k[j].first_name; k[j].first_name = temp1; temp2 = k[i].first_len; k[i].first_len = k[j].first_len; k[j].first_len = temp2; temp1 = k[i].last_name; k[i].last_name = k[j].last_name; k[j].last_name = temp1; temp2 = k[i].last_len; k[i].last_len = k[j].last_len; k[j].last_len = temp2; temp1 = k[i].grade; k[i].grade = k[j].grade; k[j].grade = temp1; temp2 = k[i].grade_len; k[i].grade_len = k[j].grade_len; k[j].grade_len = temp2; temp1 = k[i].id; k[i].id = k[j].id; k[j].id = temp1; temp2 = k[i].id_len; k[i].id_len = k[j].id_len; k[j].id_len = temp2; temp1 = k[i].phone; k[i].phone = k[j].phone; k[j].phone = temp1; temp2 = k[i].phone_len; k[i].phone_len = k[j].phone_len; k[j].phone_len = temp2; } } } for(i=0;i<recn-1;i++) { k[i].clean(); } char z; string id_sear; cout<<"Enter 1 to display , 2 to search , 3 to exit:"; cin>>z; while(1){ switch(z) { case '1': for(i=0;i<recn-1;i++) { k[i].print(); } break; case '2': cin>>key; right=0; left=recn-2; while(right<=left) { middle=((right+left)/2); if(key.compare(k[middle].id)==0){ cout<<"Founded"<<endl; k[middle].print(); break; } else if(key.compare(k[middle].id)<0) { left=middle-1; } else { right=middle+1; } } break; case '3': exit(0); break; } cout<<"Enter 1 to display , 2 to search , 3 to exit:"; cin>>z; } return 0; } the program reads from a file example.txt 313121crewwe matt 0114323111A # 433444cristinaee john 0113344325A+# 324311matte richee 3040554032B # the idea is to read fixed size field structure with a text seprator record strucutre

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