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  • C/C++ Headers in NetBeans

    - by sterh
    Hi to all, I installed IDE NetBeans for C/C++ development. I created C project with main.c and makefile generated netbeans. When i add test.h file in my project and try to compile it, i see that netbeans does not see this header file, and that I would not write it, even if there are errors anyway compilation is successful. How to connect a header file in netbeans? Thanks.

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  • penalty for "inlined" classes

    - by 2di
    Hi All Visual studio allow you to create "inlined" classes (if I am not mistaken with the name). So class header and implementation all in one file. H. file contain definitions and declarations of the class and functions, there is no .cpp file at all. So I was wondering if there is any penalty for doing it that way? any disadvantages ? Thanks a lot

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  • how do i avoid this linking error ?

    - by Yogesh
    if i have defined a global variable(with initialization) in header file, and included this file in two file and try to compile and link, compiler gives linking error ----------------- >>headers.h #ifndef __HEADERS #define __HEADERS int x = 10; #endif >>1.c #include "headers.h" main () { } --------------------- >>2.c #include "headers.h" fun () {}

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  • open url through header in php

    - by sumit
    I have a .php file in which i have a added a simple code: <?php header("Location:http//www.google.com"); ?> when i run this code, then instead of opening google.com it opens a download file, which is the same as my php file.

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  • c++ class in cocoa

    - by joels
    Any ideas why Xcode wont let me define a c++ class in my cocoa project? I am trying to use a C++ class in my cocoa project but I am getting build errors just creating the c++ header file. class SomeClass{ public: int count; } Expected '=', ',', ';', 'asm' or 'attribute' before 'SomeClass' in ..... If I remove all code from the header file, ?the cpp file builds without any errors and is included in the list of compiled sources...

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  • How can i get the flash player working

    - by Selase
    Am only looking to play some flash content on my page. found this code on the web and just copied it into my codes and changed the file name. However the intellisense underlines part of the code in green and says "File 'swfobject.js' was not found" this is the code i copied below. <asp:Content ID="HeaderContent" runat="server" ContentPlaceHolderID="HeadContent"> <script type='text/javascript' src='swfobject.js'></script> <script type='text/javascript'> var s1 = new SWFObject('cmsflash.swf','player', '400', '300', '9'); s1.addParam('allowfullscreen', 'false'); s1.addParam('allowscriptaccess', 'always'); s1.addParam('flashvars', 'file=video.flv'); s1.write('preview'); </script> <style type="text/css"> and it underlines only (swfobject.js) saying file not found. the file i wish to display is named cmsflash and is in the root directory probably ~/cmsflash.swf where and what could be going wrong? the block where the flash should display is declared as follows <table class="style2" style="height: 309px; width: 76%"> <tr> <td> <p id='preview'>The player will show in this paragraph</p> </td> </tr> </table> <span class="style6"> <br /> Please Help

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  • Get the top nth values from a rectangular array

    - by user355925
    I am reading a txt file for strings that represent integers. The file is space delimited. I have created an array[10,2]. Everytime the strings 1~10 is found in the file I increment array[n,0] by 1. I also feed array[n,1] with numbers 1~10. i.e. txt file contents: 1/1/1 10/1/2001 1 1 10 2 2 3 1 5 10 word word 3 3 etc.. streamreader reads 1/1/1 and determines that is is not 1~10 streamreader reads 10/1/2001 and determines that it is not 1~10 streamreader reads 1 and ++array[0,0] streamreader reads 1 and ++array[0,0] streamreader reads 10 and ++array[9,0] etc.. The result will be: '1' was found 3 times '2' was found 2 times '3' was found 3 times '5' was found 1 time '10' was found 2 times My problem is that I need this array placed in order(sorted) by value of column 0 so that it would be: 1 3 2 10 5

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  • Objects in Java ArrayList don't get updated.

    - by Sbm007
    This is going to be a very long post, hopefully you can understand what I'm talking about and I appreciate any help. Thanks Basically, I've created a personal, non-commercial project (which I don't plan to release) that can read ZIP and RAR files. It can only read the contents in the archive, the folders inside, the files inside the folders and its properties (such as last modified date, last modified time, CRC checksum, uncompressed size, compressed size and file name). It can't extract files either, so it's really a ZIP/RAR viewer if you may. Anyway that's slightly irrelevant to my problem but I thought I'd give you some background info. Now for my problem: I can successfully list all the folders and files inside a ZIP archive, so now I want to take that raw input and link it together in some useful way. I made 2 classes: ArchiveFile (represents a file inside a ZIP) and ArchiveFolder (represents a folder inside a ZIP). They both have some useful methods such as getLastModifiedDate, getName, getPath and so on. But the difference is that ArchiveFolder can hold an ArrayList of ArchiveFile's and additional ArchiveFolder's (think of this as files and folders inside a folder). Now I want to populate my raw input into one root ArchiveFolder, which will have all the files in the root dir of the ZIP in the ArchiveFile's ArrayList and any additional folders in the root dir of the ZIP in the ArchiveFolder's ArrayList (and this process can continue on like this like a chain reaction (more files/folders in that ArchiveFolder etc etc). So I came up with the following code: while (archive.hasMore()) { String path = ""; ArchiveFolder current = root; String[] contents = archive.getName().split("/"); for (int x = 0; x < contents.length; ++x) { if (x == (contents.length - 1) && !archive.getName().endsWith("/")) { // If on last item and item is a file path += contents[x]; // Update final path ArchiveFile file = new ArchiveFile(path, contents[x], archive.getUncompressedSize(), archive.getCompressedSize(), archive.getModifiedTime(), archive.getModifiedDate(), archive.getCRC()); current.addFile(file); // Create and add the file to the current ArchiveFolder } else if (x == (contents.length - 1)) { // Else if we are on last item and it is a folder path += contents[x] + "/"; // Update final path ArchiveFolder folder = new ArchiveFolder(path, contents[x], archive.getModifiedTime(), archive.getModifiedDate()); current.addFolder(folder); // Create and add this folder to the current ArchiveFile } else { // Else if we are still traversing through the path path += contents[x] + "/"; // Update path ArchiveFolder folder = new ArchiveFolder(path, contents[x]); current.addFolder(folder); // Create and add folder (remember we do not know the modified date/time as all we know is the path, so we can deduce the name only) current = folder; // Update current ArchiveFolder to the newly created one for the next iteration of the for loop } } archive.getNext(); } Assume that root is the root ArchiveFolder (initially empty). And that archive.getName() returns the name of the current file OR folder in the following fashion: file.txt or folder1/file2.txt or folder4/folder2/ (this is a empty folder) etc. So basically the relative path from the root of the ZIP archive. Please read through the comments in the above code to familiarize yourself with it. Also assume that the addFolder method in an ArchiveFile, only adds the folder if it doesn't exist already (so there are no multiple folders) and it also updates the time and date of an existing folder if it is blank (ie it was a intermediate folder we only knew the name of, but now we know its details). The code for addFolder is (pretty self-explanitory): public void addFolder(ArchiveFolder folder) { int loc = folders.indexOf(folder); // folders is the ArrayList containing ArchiveFolder's if (loc == -1) { folders.add(folder); } else { ArchiveFolder real = folders.get(loc); if (real.time == null) { real.setTime(folder.getTime()); real.setDate(folder.getDate()); } } } So I can't see anything wrong with the code, it works and after finishing, the root ArchiveFolder contains all the files in the root of the ZIP as I want it to, and it contains all the direcories in the root folder as I want it to. So you'd think it works as expected, but no the ArchiveFolder's in the root folder don't contain the data inside those 'child' folders, it's just a blank folder with no additional files and folders (while it does really contain some more files/folders when viewed in WinZip). After debugging using Eclipse, the for loop does iterate through all the files (even those not included above), so this led me to believe that there is a problem with this line of the code: current = folder; What it does is, it updates the current folder (used as an intermediate by the loop) to the newly added folder. I thought Java passed by reference and thus all new operations and new additions in future ArchiveFile's and ArchiveFolder's are automatically updated, and parent ArchiveFolder's will be updated accordingly. But that does not appear to be the case? I know this is a long ass post and I really hope anyone can help me out with this. Thanks in advance.

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  • returning values from function or method multiple times by only calling the class once

    - by Sokhrat Sikar
    I have a members.php file that shows my websites members. I echo members name by using foreach method. A method of Members class returns an array, then I use foreach loop in the members.php file to echo the members. I am trying to aovid writing php code in my members.php file. Is there a way to avoid using foreach inside members.php file? For example, is it possible to return value from a method couple of times? (by only calling the object once). Just like how we normally call the functions? This question doesn't make sense, but I am just trying to see if there is a away around this issue?

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  • Changed the AllowOverride to All and still nothing

    - by Asaf
    I tried to write a .htaccess file on my local pc's website, I've realized I need to set AllowOverride All instead of None searched, found the file /etc/apache2/conf.d/security in the file I found #<Directory /> #AllowOverride None #Order Deny,Allow #Deny from all #</Directory> changed it to <Directory /> AllowOverride All Order Deny,Allow Deny from all </Directory> typed service apache2 restart and... .htaccess still didn't work :I the file by the way, holds one line, deny from all.

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  • Is it possible to make this Flex/Flash application safe?

    - by Frank
    I'm back with another Flex/Flash security question. I've already received some help from the community on this topic, but I'm still not quite sure this is the best way to do. Here's the thing. A flex web app, a lot of users (1000+), custom configuration of the application depending of the user group. Can I make this thing safe... or safer. For the moment, when a user comes to the application, there is only one configuration possible, but for the next version we've implented a multi-configuration protocol, this way : 1. The user connect to Default.aspx, server code process the windows credentials (whe are on intranet) and give the correct xml configuration file. 2. The flex app loads with the xml conf file as a flashvar and then the app 'builds' itself with the content of the xml file. As we know, since this is a flex application the swf is downloaded on the client computer and the xml file too. If more than one user connects to the app, from the same computer, the can possibly see the other xml file in the windows temp folder. The current directory of the application looks that way : Web site |-> default.aspx |-> index.swf |-> configAdmin.xml |-> configUserType1.xml |-> configUserType2.xml |-> com |-> a lot of swf and xml files I was first thinking making another directory (without read access for the client) containing all the configurations xml files, picking the right one, copying it to the client and deleting it afterwards. But it seems like I must let know the user know when downloading/deleting content on it's computer... I'm running out of ideas, so I hope you have some great ones. It's there are some design flaws (in the way the app is build, not in Flash :p) please share. I'm always looking forward to improve. Thanks Update : In browser Flash/Flex (without AIR that is) doesn't allow deleting file localy silently (on the client computer, where the application is). It's also not yet possible to get session data.

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  • Need help making this code more efficient

    - by Rendicahya
    I always use this method to easily read the content of a file. Is it efficient enough? Is 1024 good for the buffer size? public static String read(File file) { FileInputStream stream = null; StringBuilder str = new StringBuilder(); try { stream = new FileInputStream(file); } catch (FileNotFoundException e) { } FileChannel channel = stream.getChannel(); ByteBuffer buffer = ByteBuffer.allocate(1024); try { while (channel.read(buffer) != -1) { buffer.flip(); while (buffer.hasRemaining()) { str.append((char) buffer.get()); } buffer.rewind(); } } catch (IOException e) { } finally { try { channel.close(); stream.close(); } catch (IOException e) { } } return str.toString(); }

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  • How can I check if a value is in a list in Perl?

    - by ablimit
    I have a file in which every line is an integer which represents an id. What I want to do is just check whether some specific ids are in this list. But the code didn't work. It never tells me it exists even if 123 is a line in that file. I don't know why? Help appreciated. open (FILE, "list.txt") or die ("unable to open !"); my @data=<FILE>; my %lookup =map {chop($_) => undef} @data; my $element= '123'; if (exists $lookup{$element}) { print "Exists"; } Thanks in advance.

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  • how to get around the command line length limit

    - by Crash893
    I've been working on a small and simple program that i drop files onto and based on certian rules they are moved to diffrent places. the program works fine unless i drop more than a few files on the program then it kicks back an error (that appears to be more windows than anything) that the start up command c:\myapp.exe \file \file \file is to long I realize i could set up a background proccess but i really would prefer that this program not run in the backround (where it would be idle 99.999999999% of the time) Can anyone think of any way around this limitation?

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  • undefined method `output_data' for #<EventManager:0x007fa4220320c8> (NoMethodError)

    - by Roger Camps
    I keep getting this error: event_manager.rb:83:in': undefined method output_data' for #<EventManager:0x007fc5018320c0> (NoMethodError) I am following the exercise on this website: Here is my code (My error comes towards the end with DEF OUTPUT_DATA ...): # Dependencies require "csv" # Class Definition class EventManager INVALID_PHONE_NUMBER = "0000000000" INVALID_ZIPCODE = "00000" def initialize puts "EventManager Initialized." filename = "event_attendees.csv" @file = CSV.open(filename, {:headers => true, :header_converters => :symbol}) end def print_names @file.each do |line| puts line.inspect puts line[2] + " " + line[3] end end #printing home phone number method def print_numbers @file.each do |line| number = clean_number(line[:homephone]) puts number end end #cleaning numbers method def clean_number(number) cleaner= number.delete('.' + ')' + '(' + '-') if cleaner.length == 10 # Do Nothing elsif cleaner.length == 11 if cleaner.start_with?("1") cleaner = cleaner[1..-1] else cleaner = INVALID_PHONE_NUMBER end else cleaner = INVALID_PHONE_NUMBER end return cleaner end def clean_zipcode(original) if original.nil? zipcode = INVALID_ZIPCODE elsif original.length < 5 while original.length < 5 original = original.insert(0, "0") end else return original end return zipcode end def print_zipcodes @file.each do |line| zipcode = clean_zipcode(line[:zipcode]) puts zipcode end def output_data output = CSV.open("event_attendees_clean.csv", "w") @file.each do |line| output << line end end end end # Script manager = EventManager.new #manager.print_numbers #manager.print_zipcodes manager.output_data I've tried several things, checked all through the internet and I just can't figure it out myself. I will really appreciate any help. Thank you in advance!

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  • Windows [C++] - Creating new folders if they don't exist for fopen

    - by Gbps
    I have a C++ program that takes user input for fopen in order to initiate a file write. Could someone help me find a function which will return a FILE* and use the Windows specific version of mkdir in order to create the folder structure for fopen to never fail to open a new file in the specified location because one of the folders does not exist. Thanks a bunch!

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  • KindError: Property r must be an instance of SecondModel, why ?

    - by zjm1126
    class FirstModel(db.Model): p = db.StringProperty() r=db.ReferenceProperty(SecondModel) class SecondModel(db.Model): r = db.ReferenceProperty(FirstModel) class sss(webapp.RequestHandler): def get(self): a=FirstModel() a.p='sss' a.put() b=SecondModel() b.r=a b.put() a.r=b a.put() self.response.out.write(str(b.r.p)) the error is : Traceback (most recent call last): File "D:\Program Files\Google\google_appengine\google\appengine\ext\webapp\__init__.py", line 511, in __call__ handler.get(*groups) File "D:\zjm_code\helloworld\a.py", line 158, in get a.r=b File "D:\Program Files\Google\google_appengine\google\appengine\ext\db\__init__.py", line 3009, in __set__ value = self.validate(value) File "D:\Program Files\Google\google_appengine\google\appengine\ext\db\__init__.py", line 3048, in validate (self.name, self.reference_class.kind())) KindError: Property r must be an instance of SecondModel thanks

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  • how to return value from groovy to java

    - by Sid
    Hi I am very new to groovy and having trouble with some parts. I have a jsp page tied to a servlet that runs groovy scripts. I am able to get to the groovy script from the servlet. But after the script runs how do I return the response from the groovy script back to the servlet to be displayed in the jsp page? My java servlet code is as follows: File file = new File("TestScript.groovy"); ClassLoader parent = getClass().getClassLoader(); GroovyClassLoader loader = new GroovyClassLoader(parent); Class groovyClass = loader.parseClass(file); Object[] args = {}; GroovyObject groovyObject = (GroovyObject) groovyClass.newInstance(); groovyObject.invokeMethod("runTest", args);

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  • How to optimize this script

    - by marks34
    I have written the following script. It opens a file, reads each line from it splitting by new line character and deleting first character in line. If line exists it's being added to array. Next each element of array is splitted by whitespace, sorted alphabetically and joined again. Every line is printed because script is fired from console and writes everything to file using standard output. I'd like to optimize this code to be more pythonic. Any ideas ? import sys def main(): filename = sys.argv[1] file = open(filename) arr = [] for line in file: line = line[1:].replace("\n", "") if line: arr.append(line) for line in arr: lines = line.split(" ") lines.sort(key=str.lower) line = ''.join(lines) print line if __name__ == '__main__': main()

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  • Upload files from website

    - by Andrew
    All I want to do is allow the user to browse through their folders to look for a file, select it and then press submit which saves the file to my server as well as the path to the saved file. How would someone do this? (Some sort of tutorial website would help a lot)

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