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  • Warning: expects resource but string given

    - by Damien
    I get: "Resource id #8 Warning: mysql_fetch_array() expects parameter 1 to be resource, string given" Heres the code: $sql="SELECT password FROM user WHERE userid=$userid"; echo $password=mysql_query($sql); while($row = mysql_fetch_array($password)) { $password = $row['password']; } Any ideas?

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  • Using @Resource to load environment entries

    - by a1ex07
    Hi, I'm trying to load bean runtime configuration. @Stateless public class MyBean implements MyLocal{ @Resource String runtimeSetting1="default_value"; //.... } I cannot find out how to create custom resource on app server side (Glassfish) - I have no idea what I should enter in "Factory Class" field. Maybe there is a better way of loading configuration... Thanks.

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  • ASP.NET: Resource strings

    - by JamesBrownIsDead
    I have an .ascx file. This file has an associated .ascx.resx in multiple languages. I want to get a resource string for a specific CultureInfo. Ordinarily I'd use this.GetLocalResourceObject in the codebehind, but I don't want a local resource string, I want one for a specific language.

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  • a good resource or book for architecting object-oriented software

    - by Ygam
    I have looked at a couple of books and all I have looked at were just discussing the technicalities of OOP. By technicalities I mean, here's a concept, here's some code, now get working. I have yet to see a book that discusses the architectural process, what are the ways of doing this, why doing this is bad, how to actually incorporate design patterns in a real-world project, etc. Can you recommend a good resource or book? I am mainly programming with PHP but a language-agnostic book/resource would do :)

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  • rspec nested ( has many resource ) no route found error

    - by Surya
    My calendar resource is a nested resource under profile map.resources :profiles, :has_many=>[:calendar] I am trying to write a rspec spec for calendarcontroller it "should use supplied date" do get :show , :month = '09' , :year = '2010' end But i get an error stating No route matches {:month="09", :year="2010", :controller="calendar", :action="show"} Any idea how i could get around this ?

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  • no resource found that matches the given name Android 1.6

    - by xaero212
    I've got 'no resource found that matches the given name' but everything is set up correctly. error: Error: No resource found that matches the given name (at 'text' with value '@string/labReminderClear'). In AndroidManifest.xml: <application android:label="MyName" ..... In Strings.xml: <string name="app_name">MyName</string> ... <string name="labReminderClear">Clear</string> What could be wrong?

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  • SharePoint OCR image files indexing

    Introduction This article describes how to setup indexing of the image files (including TIFF, PDF, JPEG, BMP...) using OCR technology. The indexing described below utilizes Microsoft IFilter technology and as such is not specific to SharePoint, but can be used with any product that uses Microsoft indexing: Microsoft Search, Desktop search, SQL Server search, and through the plug-ins with Google desktop search. I however use it with Microsoft Windows SharePoint Services 2003. For those other products, the registration may need to be slightly different. Background  One of the projects I was working on required a storage of old documents scanned into PDF files. Then there was a separate team of people responsible for providing a tags for a search engine so those image documents could be found. The whole process was clumsy, labor intensive, and error prone. That was what started me on my exploration path. OCR The first search I fired was for the Open Source OCR products. Pretty quickly, I narrowed it down to TESSERACT (http://code.google.com/p/tesseract-ocr/). Tesseract is an orphaned brain child of HP that worked on it from 1985 to 1995. Then it was moved to the Open Source, and now if I understand it correctly, Google is working on it. With credentials like that, it's no wonder that Tesseract scores one of the highest marks on OCR recognition and accuracy. After downloading and struggling just a bit, I got Tesseract to work. The struggling part was that the home page claims that its base input format is a TIFF file. May be my TIFFs were bad, but I was able to get it to work only for BMP files. Image files conversion So now that I have an OCR that can convert BMP files into text, how do I get text out of the image PDF files? One more search, and I settled down on ImageMagic (http://www.imagemagick.org/). This is another wonderful Open Source utility that can convert any file into image. It did work out of the box, converting any TIFF files into bitmaps, but to get PDF files converted, it requires a GhostScript (http://mirror.cs.wisc.edu/pub/mirrors/ghost/GPL/gs864/gs864w32.exe). Dealing with text PDFs With that utility installed, I was cooking - I can convert any file (in particular PDF and TIFF) into bitmap, and then I can extract the text out of the bitmap. The only consideration was to somehow treat PDF files containing text differently - after all, OCR is very computation intensive and somewhat error prone even with perfect image quality and resolution. So another quick search, and I have a PDFTOTEXT (ftp://ftp.foolabs.com/pub/xpdf/xpdf-3.02pl4-win32.zip) - thank God for Open Source! With these guys, I can pull text out of PDF in an eye blink. However, I would get nothing for pure image PDFs, but I already have a solution for that! Batch process It took another 15 minutes to setup a batch script to automate the process: Check the file extension If file is a PDF file try to extract text out of it if there is more than certain amount of text in the file - done! if there is no text, convert first page into bitmap run OCR on the bitmap For any other file type, convert file into bitmap Run OCR on the bitmap Once you unzip the attached project, check out the bin\OCR.BAT file. It will create a temporary file in the directory where your source file is with the same name + the '.txt' extension.Continue span.fullpost {display:none;}

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  • Convert .3GP and .3G2 Files to AVI / MPEG for Free

    - by DigitalGeekery
    3GP and .3G2 are common video capture formats used on many mobile phones, but they may not be supported by your favorite media player. Today we’ll show you a quick and easy way to convert those files to AVI or MPG format with the free Windows application, Pazera Free 3GP to AVI Converter. Download the Pazera Free 3GP to AVI Converter. You’ll have to unzip the download folder, but there is no need to install the application. Just double-click the 3gptoavi.exe file to run the application. To add your 3GP or 3G2 files to the queue to be converted, click on the Add files  button at the top left. Browse for your file, and click Open.   Your video will be added to the Queue. You can add multiple files to the queue and convert them all at one time.   Most users will find it preferable to use one of the pre-configured profiles for their conversion settings. To load a profile, choose one from the Profile drop down list and then click the Load button. You will see the profile update the settings in the panels at the bottom of the application. We tested Pazera Free 3GP to AVI Converter with 3GP files recorded on a Motorola Droid, and found the AVI H.264 Very High Q. profile to return the best results for AVI output, and the MPG – DVD NTSC: MPEG-2 the best results for MPG output. Other profiles produced smaller file sizes, but at a cost of reduced quality video output.   More advanced users may tweak video and audio settings to their liking in the lower panels. Click on the AVI button under Output file format / Video settings to adjust settings AVI… Or the MPG button to adjust the settings for MPG output. By default, the converted file will be output to the same location as the input directory. You can change it by clicking the text box input radio button and browsing for a different folder. When you’ve chosen your settings, click Convert to begin the conversion process.   A conversion output box will open and display the progress. When finished, click Close. Now you’re ready to enjoy your video in your favorite media player. Pazera Free 3GP to AVI Converter isn’t the most robust media conversion tool, but it does what it is intended to do. It handles the task of 3GP to AVI / MPG conversion very well. It’s easy enough for the beginner to manage without much trouble, but also has enough options to please more experienced users. Download Pazera Free 3GP to AVI Converter Similar Articles Productive Geek Tips How To Convert Video Files to MP3 with VLCEasily Change Audio File Formats with XRECODEConvert PDF Files to Word Documents and Other FormatsConvert Video and Remove Commercials in Windows 7 Media Center with MCEBuddy 1.1Compress Large Video Files with DivX / Xvid and AutoGK TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips DVDFab 6 Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 Install, Remove and HIDE Fonts in Windows 7 Need Help with Your Home Network? Awesome Lyrics Finder for Winamp & Windows Media Player Download Videos from Hulu Pixels invade Manhattan Convert PDF files to ePub to read on your iPad

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  • Copy New Files Only in .NET

    - by psheriff
    Recently I had a client that had a need to copy files from one folder to another. However, there was a process that was running that would dump new files into the original folder every minute or so. So, we needed to be able to copy over all the files one time, then also be able to go back a little later and grab just the new files. After looking into the System.IO namespace, none of the classes within here met my needs exactly. Of course I could build it out of the various File and Directory classes, but then I remembered back to my old DOS days (yes, I am that old!). The XCopy command in DOS (or the command prompt for you pure Windows people) is very powerful. One of the options you can pass to this command is to grab only newer files when copying from one folder to another. So instead of writing a ton of code I decided to simply call the XCopy command using the Process class in .NET. The command I needed to run at the command prompt looked like this: XCopy C:\Original\*.* D:\Backup\*.* /q /d /y What this command does is to copy all files from the Original folder on the C drive to the Backup folder on the D drive. The /q option says to do it quitely without repeating all the file names as it copies them. The /d option says to get any newer files it finds in the Original folder that are not in the Backup folder, or any files that have a newer date/time stamp. The /y option will automatically overwrite any existing files without prompting the user to press the "Y" key to overwrite the file. To translate this into code that we can call from our .NET programs, you can write the CopyFiles method presented below. C# using System.Diagnostics public void CopyFiles(string source, string destination){  ProcessStartInfo si = new ProcessStartInfo();  string args = @"{0}\*.* {1}\*.* /q /d /y";   args = string.Format(args, source, destination);   si.FileName = "xcopy";  si.Arguments = args;  Process.Start(si);} VB.NET Imports System.Diagnostics Public Sub CopyFiles(source As String, destination As String)  Dim si As New ProcessStartInfo()  Dim args As String = "{0}\*.* {1}\*.* /q /d /y"   args = String.Format(args, source, destination)   si.FileName = "xcopy"  si.Arguments = args  Process.Start(si)End Sub The CopyFiles method first creates a ProcessStartInfo object. This object is where you fill in name of the command you wish to run and also the arguments that you wish to pass to the command. I created a string with the arguments then filled in the source and destination folders using the string.Format() method. Finally you call the Start method of the Process class passing in the ProcessStartInfo object. That's all there is to calling any command in the operating system. Very simple, and much less code than it would have taken had I coded it using the various File and Directory classes. Good Luck with your Coding,Paul Sheriff ** SPECIAL OFFER FOR MY BLOG READERS **Visit http://www.pdsa.com/Event/Blog for a free video on Silverlight entitled Silverlight XAML for the Complete Novice - Part 1.  

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  • Kill all the project files!

    - by jamiet
    Like many folks I’m a keen podcast listener and yesterday my commute was filled by listening to Scott Hunter being interviewed on .Net Rocks about the next version of ASP.Net. One thing Scott said really struck a chord with me. I don’t remember the full quote but he was talking about how the ASP.Net project file (i.e. the .csproj file) is going away. The rationale being that the main purpose of that file is to list all the other files in the project, and that’s something that the file system is pretty good at. In Scott’s own words (that someone helpfully put in the comments): A file that lists files is really redundant when the OS already does this Romeliz Valenciano correctly pointed out on Twitter that there will still be a project.json file however no longer will there be a need to keep a list of files in a project file. I suspect project.json will simply contain a list of exclusions where necessary rather than the current approach where the project file is a list of inclusions. On the face of it this seems like a pretty good idea. I’ve long been a fan of convention over configuration and this is a great example of that. Instead of listing all the files in a separate file, just treat all the files in the directory as being part of the project. Ostensibly the approach is if its in the directory, its part of the project. Simple. Now I’m not an ASP.net developer, far from it, but it did occur to me that the same approach could be applied to the two Visual Studio project types that I am most familiar with, SSIS & SSDT. Like many people I’ve long been irritated by SSIS projects that display a faux file system inside Solution Explorer. As you can see in the screenshot below the project has Miscellaneous and Connection Managers folders but no such folders exist on the file system: This may seem like a minor thing but it means useful Solution Explorer features like Show All Files and Open Folder in Windows Explorer don’t work and quite frankly it makes me feel like a second class citizen in the Microsoft ecosystem. I’m a developer, treat me like one. Don’t try and hide the detail of how a project works under the covers, show it to me. I’m a big boy, I can handle it! Would it not be preferable to simply treat all the .dtsx files in a directory as being part of a project? I think it would, that’s pretty much all the .dtproj file does anyway (that, and present things in a non-alphabetic order – something else that wildly irritates me), so why not just get rid of the .dtproj file? In the case of SSDT the .sqlproj actually does a whole lot more than simply list files because it also states the BuildAction of each file (Build, NotInBuild, Post-Deployment, etc…) but I see no reason why the convention over configuration approach can’t help us there either. Want to know which is the Post-deployment script? Well, its the one called Post-DeploymentScript.sql! Simple! So that’s my new crusade. Let’s kill all the project files (well, the .dtproj & .sqlproj ones anyway). Are you with me? @Jamiet

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  • What are the valid DepthBuffer Texture formats in DirectX 11? And which are also valid for a staging resource?

    - by sebf
    I am trying to read the contents of the depth buffer into main memory so that my CPU side code can do Some Stuff™ with it. I am attempting to do this by creating a staging resource which can be read by the CPU, which I will copy the contents of the depth buffer into before reading it. I keep encountering errors however, because of, I believe, incompatibilities between the resource format and the view formats. Threads like these lead me to believe it is possible in DX11 to access the depth buffer as a resource, and that I can create a resource with a typeless format and have it interpreted in the view as another, but I cannot get it to work. What are the valid formats for the resource to be used as the depth buffer? Which of these are also valid for a CPU accessible staging resource?

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  • Cannot delete old NFS directory: Device or resource busy

    - by Jakobud
    On server1, we had an NFS share mounted from server 2 like this: /nfs/server2/share Recently, we took down server2 to install a new OS on it. Now we can't get NFS setup the way it was. When I do this: ls -l /nfs I get this: drwxr-xr-x 2 root root 0 2010-03-15 09:59 server2 Notice how the directory size is 0 instead of 4096 like usual? Anyways I go into server2 expecting to see a share directory, but I don't. It's empty. So therefore I cannot mount my share at /nfs/server2/share. When I try to create /nfs/server2/share directory, I get mkdir: cannot create directory `share': No such file or directory I think this is because it doesn't really think the /nfs/server2 directory really exists. Even if I use the -p option with mkdir, it doesn't work. Next I tried to remove /nfs/server2 so I could just recreate it. I try to rm -r /nfs/server2 but I get rm: cannot remove directory `/nfs/server2': Device or resource busy So now I'm at a loss. I need to mount this NFS share in the same exact place on server1 (at /nfs/server2/share) because other software on server1 depend on this. But if I can't create that share directory and I can't remove that directory, what do I do? Also, just for testing, I attempted to mount the share at /nfs/testing/share and it mounted just fine. But like I said, I need to mount it back in the same location.

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  • .htaccess authorization requiring username/password for every resource

    - by webworm
    I am using Apache2 on Ubuntu and I have having some "weird" user authorization issues. I am using .htaccess to control access to my directories. I have many users and have grouped them into user groups which are defined in a "group" file. I then use .htaccess within each directory to define which users have access to the directory and which do not. Here is an example .htaccess file. AuthUserFile /var/local/.htpasswd AuthGroupFile /var/local/groups AuthName "Username and Password Required" AuthType Basic require group design admin Everything is working with one exception. I added a new user to one of my groups and though they can gain access to the directory they are prompted for a username and password for every resource (i.e. image, CSS). After a while I can just keep selecting "cancel" and I will get a page with just html with no images or CSS. I would think the browser would just cache the username/password. It seems to be working well for other users. Any thoughts?

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  • .htaccess authorization requiring username/password for every resource

    - by webworm
    I am using Apache2 on Ubuntu and I have having some "weird" user authorization issues. I am using .htaccess to control access to my directories. I have many users and have grouped them into user groups which are defined in a "group" file. I then use .htaccess within each directory to define which users have access to the directory and which do not. Here is an example .htaccess file. AuthUserFile /var/local/.htpasswd AuthGroupFile /var/local/groups AuthName "Username and Password Required" AuthType Basic require group design admin Everything is working with one exception. I added a new user to one of my groups and though they can gain access to the directory they are prompted for a username and password for every resource (i.e. image, CSS). After a while I can just keep selecting "cancel" and I will get a page with just html with no images or CSS. I would think the browser would just cache the username/password. It seems to be working well for other users. Any thoughts?

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  • Directory tree in a Resource without extraction...

    - by Corelgott
    Hi all, i am looking for a way to store a complete directory including sub directories in an application's resource and not have to extract it to use it. Details: We would like to use GeckoFx (Gecko as C# Component) in one of our applications. GeckoFX needs the XUL-Runner and needs to find it's folder structure We have some other data which I would not prefer to extracted to the customer's pc; At least not onto something persistent like a hdd... Getting the complete directory into the resources is not that kind of a big deal. Compress to one file and done. But not writing it to the disk to use it is something else. I have a strong dislike against temp folders and such things. Would anything like a RAM drive be possible? Some part of the RAM beeing mounted? Does something like this even exist as a lib, or would this only be possible by a device driver? Any thoughts on this? Thanks in advance! Corelgott

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  • Homebrew large data cluster access for 2 user levels?

    - by Yegor
    The title probably makes little sense, so here is an example. I have a file hosting site, that serves a large amount of semi-randomly accessed files. The setup is as follows: High horsepower front-end +DB server that also does encoding for files that need encoding Fresh file server, which stores newly uploaded content, thats probably (and usually) rapidly accessible, which has 500GB of raided SSD storage, that can push over 3GBit of traffic. 3 cheap node servers, containing 2 x 750GB SATA drives in raid1, where files older than 2 weeks are archived, from the SSD server (mentioned above). Files on each server are accessed via subdomains (via modsec) in a straight forward fashion (server1.domain.com, server2.domain.com, etc) Where I have the problem is this. I introduced a "premium" service where people pay a small fee every month, and get ad-free, quick accesses to stuff on the site. Once they are logged in, they access same files via premium.server1.domain.com via a different modsec script, with a different pass phrase. That all works fine and dandy.... except the cheap node servers are all IO bound, so accessing the files on them via a different, unsaturated network makes no difference, since it cannot read off the drive fast enough. What would be a good way to make files on the site be accessible via 2 different network routes, 1 of which will be saturated (the "free network") while all other files are on an un-saturated "premium" network?

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  • Chef bash resource not executing as specified user

    - by Arthur Maltson
    I'm writing a Chef cookbook to install Hubot. In the recipe, I do the following: bash "install hubot" do user hubot_user group hubot_group cwd install_dir code <<-EOH wget https://github.com/downloads/github/hubot/hubot-#{node['hubot']['version']}.tar.gz && \ tar xzvf hubot-#{node['hubot']['version']}.tar.gz && \ cd hubot && \ npm install EOH end However, when I try to run chef-client on the server installing the cookbook, I'm getting a permission denied writing to the directory of the user that runs chef-client, not the hubot user. For some reason, npm is trying to run under the wrong user, not the user specified in the bash resource. I am able to run sudo su - hubot -c "npm install /usr/local/hubot/hubot" manually, and this gets the result I want (installs hubot as the hubot user). However, it seems chef-client isn't executing the command as the hubot user. Below you'll find the chef-client execution. Thank you in advance. Saving to: `hubot-2.1.0.tar.gz' 0K ...... 100% 563K=0.01s 2012-01-23 12:32:55 (563 KB/s) - `hubot-2.1.0.tar.gz' saved [7115/7115] npm ERR! Could not create /home/<user-chef-client-uses>/.npm/log/1.2.0/package.tgz npm ERR! Failed creating the tarball. npm ERR! couldn't pack /tmp/npm-1327339976597/1327339976597-0.13104878342710435/contents/package to /home/<user-chef-client-uses>/.npm/log/1.2.0/package.tgz npm ERR! error installing [email protected] Error: EACCES, permission denied '/home/<user-chef-client-uses>/.npm/log' ... npm not ok ---- End output of "bash" "/tmp/chef-script20120123-25024-u9nps2-0" ---- Ran "bash" "/tmp/chef-script20120123-25024-u9nps2-0" returned 1

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  • How to simulate inner join on very large files in java (without running out of memory)

    - by Constantin
    I am trying to simulate SQL joins using java and very large text files (INNER, RIGHT OUTER and LEFT OUTER). The files have already been sorted using an external sort routine. The issue I have is I am trying to find the most efficient way to deal with the INNER join part of the algorithm. Right now I am using two Lists to store the lines that have the same key and iterate through the set of lines in the right file once for every line in the left file (provided the keys still match). In other words, the join key is not unique in each file so would need to account for the Cartesian product situations ... left_01, 1 left_02, 1 right_01, 1 right_02, 1 right_03, 1 left_01 joins to right_01 using key 1 left_01 joins to right_02 using key 1 left_01 joins to right_03 using key 1 left_02 joins to right_01 using key 1 left_02 joins to right_02 using key 1 left_02 joins to right_03 using key 1 My concern is one of memory. I will run out of memory if i use the approach below but still want the inner join part to work fairly quickly. What is the best approach to deal with the INNER join part keeping in mind that these files may potentially be huge public class Joiner { private void join(BufferedReader left, BufferedReader right, BufferedWriter output) throws Throwable { BufferedReader _left = left; BufferedReader _right = right; BufferedWriter _output = output; Record _leftRecord; Record _rightRecord; _leftRecord = read(_left); _rightRecord = read(_right); while( _leftRecord != null && _rightRecord != null ) { if( _leftRecord.getKey() < _rightRecord.getKey() ) { write(_output, _leftRecord, null); _leftRecord = read(_left); } else if( _leftRecord.getKey() > _rightRecord.getKey() ) { write(_output, null, _rightRecord); _rightRecord = read(_right); } else { List<Record> leftList = new ArrayList<Record>(); List<Record> rightList = new ArrayList<Record>(); _leftRecord = readRecords(leftList, _leftRecord, _left); _rightRecord = readRecords(rightList, _rightRecord, _right); for( Record equalKeyLeftRecord : leftList ){ for( Record equalKeyRightRecord : rightList ){ write(_output, equalKeyLeftRecord, equalKeyRightRecord); } } } } if( _leftRecord != null ) { write(_output, _leftRecord, null); _leftRecord = read(_left); while(_leftRecord != null) { write(_output, _leftRecord, null); _leftRecord = read(_left); } } else { if( _rightRecord != null ) { write(_output, null, _rightRecord); _rightRecord = read(_right); while(_rightRecord != null) { write(_output, null, _rightRecord); _rightRecord = read(_right); } } } _left.close(); _right.close(); _output.flush(); _output.close(); } private Record read(BufferedReader reader) throws Throwable { Record record = null; String data = reader.readLine(); if( data != null ) { record = new Record(data.split("\t")); } return record; } private Record readRecords(List<Record> list, Record record, BufferedReader reader) throws Throwable { int key = record.getKey(); list.add(record); record = read(reader); while( record != null && record.getKey() == key) { list.add(record); record = read(reader); } return record; } private void write(BufferedWriter writer, Record left, Record right) throws Throwable { String leftKey = (left == null ? "null" : Integer.toString(left.getKey())); String leftData = (left == null ? "null" : left.getData()); String rightKey = (right == null ? "null" : Integer.toString(right.getKey())); String rightData = (right == null ? "null" : right.getData()); writer.write("[" + leftKey + "][" + leftData + "][" + rightKey + "][" + rightData + "]\n"); } public static void main(String[] args) { try { BufferedReader leftReader = new BufferedReader(new FileReader("LEFT.DAT")); BufferedReader rightReader = new BufferedReader(new FileReader("RIGHT.DAT")); BufferedWriter output = new BufferedWriter(new FileWriter("OUTPUT.DAT")); Joiner joiner = new Joiner(); joiner.join(leftReader, rightReader, output); } catch (Throwable e) { e.printStackTrace(); } } } After applying the ideas from the proposed answer, I changed the loop to this private void join(RandomAccessFile left, RandomAccessFile right, BufferedWriter output) throws Throwable { long _pointer = 0; RandomAccessFile _left = left; RandomAccessFile _right = right; BufferedWriter _output = output; Record _leftRecord; Record _rightRecord; _leftRecord = read(_left); _rightRecord = read(_right); while( _leftRecord != null && _rightRecord != null ) { if( _leftRecord.getKey() < _rightRecord.getKey() ) { write(_output, _leftRecord, null); _leftRecord = read(_left); } else if( _leftRecord.getKey() > _rightRecord.getKey() ) { write(_output, null, _rightRecord); _pointer = _right.getFilePointer(); _rightRecord = read(_right); } else { long _tempPointer = 0; int key = _leftRecord.getKey(); while( _leftRecord != null && _leftRecord.getKey() == key ) { _right.seek(_pointer); _rightRecord = read(_right); while( _rightRecord != null && _rightRecord.getKey() == key ) { write(_output, _leftRecord, _rightRecord ); _tempPointer = _right.getFilePointer(); _rightRecord = read(_right); } _leftRecord = read(_left); } _pointer = _tempPointer; } } if( _leftRecord != null ) { write(_output, _leftRecord, null); _leftRecord = read(_left); while(_leftRecord != null) { write(_output, _leftRecord, null); _leftRecord = read(_left); } } else { if( _rightRecord != null ) { write(_output, null, _rightRecord); _rightRecord = read(_right); while(_rightRecord != null) { write(_output, null, _rightRecord); _rightRecord = read(_right); } } } _left.close(); _right.close(); _output.flush(); _output.close(); } UPDATE While this approach worked, it was terribly slow and so I have modified this to create files as buffers and this works very well. Here is the update ... private long getMaxBufferedLines(File file) throws Throwable { long freeBytes = Runtime.getRuntime().freeMemory() / 2; return (freeBytes / (file.length() / getLineCount(file))); } private void join(File left, File right, File output, JoinType joinType) throws Throwable { BufferedReader leftFile = new BufferedReader(new FileReader(left)); BufferedReader rightFile = new BufferedReader(new FileReader(right)); BufferedWriter outputFile = new BufferedWriter(new FileWriter(output)); long maxBufferedLines = getMaxBufferedLines(right); Record leftRecord; Record rightRecord; leftRecord = read(leftFile); rightRecord = read(rightFile); while( leftRecord != null && rightRecord != null ) { if( leftRecord.getKey().compareTo(rightRecord.getKey()) < 0) { if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.LeftExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, leftRecord, null); } leftRecord = read(leftFile); } else if( leftRecord.getKey().compareTo(rightRecord.getKey()) > 0 ) { if( joinType == JoinType.RightOuterJoin || joinType == JoinType.RightExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, null, rightRecord); } rightRecord = read(rightFile); } else if( leftRecord.getKey().compareTo(rightRecord.getKey()) == 0 ) { String key = leftRecord.getKey(); List<File> rightRecordFileList = new ArrayList<File>(); List<Record> rightRecordList = new ArrayList<Record>(); rightRecordList.add(rightRecord); rightRecord = consume(key, rightFile, rightRecordList, rightRecordFileList, maxBufferedLines); while( leftRecord != null && leftRecord.getKey().compareTo(key) == 0 ) { processRightRecords(outputFile, leftRecord, rightRecordFileList, rightRecordList, joinType); leftRecord = read(leftFile); } // need a dispose for deleting files in list } else { throw new Exception("DATA IS NOT SORTED"); } } if( leftRecord != null ) { if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.LeftExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, leftRecord, null); } leftRecord = read(leftFile); while(leftRecord != null) { if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.LeftExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, leftRecord, null); } leftRecord = read(leftFile); } } else { if( rightRecord != null ) { if( joinType == JoinType.RightOuterJoin || joinType == JoinType.RightExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, null, rightRecord); } rightRecord = read(rightFile); while(rightRecord != null) { if( joinType == JoinType.RightOuterJoin || joinType == JoinType.RightExclusiveJoin || joinType == JoinType.FullExclusiveJoin || joinType == JoinType.FullOuterJoin ) { write(outputFile, null, rightRecord); } rightRecord = read(rightFile); } } } leftFile.close(); rightFile.close(); outputFile.flush(); outputFile.close(); } public void processRightRecords(BufferedWriter outputFile, Record leftRecord, List<File> rightFiles, List<Record> rightRecords, JoinType joinType) throws Throwable { for(File rightFile : rightFiles) { BufferedReader rightReader = new BufferedReader(new FileReader(rightFile)); Record rightRecord = read(rightReader); while(rightRecord != null){ if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.RightOuterJoin || joinType == JoinType.FullOuterJoin || joinType == JoinType.InnerJoin ) { write(outputFile, leftRecord, rightRecord); } rightRecord = read(rightReader); } rightReader.close(); } for(Record rightRecord : rightRecords) { if( joinType == JoinType.LeftOuterJoin || joinType == JoinType.RightOuterJoin || joinType == JoinType.FullOuterJoin || joinType == JoinType.InnerJoin ) { write(outputFile, leftRecord, rightRecord); } } } /** * consume all records having key (either to a single list or multiple files) each file will * store a buffer full of data. The right record returned represents the outside flow (key is * already positioned to next one or null) so we can't use this record in below while loop or * within this block in general when comparing current key. The trick is to keep consuming * from a List. When it becomes empty, re-fill it from the next file until all files have * been consumed (and the last node in the list is read). The next outside iteration will be * ready to be processed (either it will be null or it points to the next biggest key * @throws Throwable * */ private Record consume(String key, BufferedReader reader, List<Record> records, List<File> files, long bufferMaxRecordLines ) throws Throwable { boolean processComplete = false; Record record = records.get(records.size() - 1); while(!processComplete){ long recordCount = records.size(); if( record.getKey().compareTo(key) == 0 ){ record = read(reader); while( record != null && record.getKey().compareTo(key) == 0 && recordCount < bufferMaxRecordLines ) { records.add(record); recordCount++; record = read(reader); } } processComplete = true; // if record is null, we are done if( record != null ) { // if the key has changed, we are done if( record.getKey().compareTo(key) == 0 ) { // Same key means we have exhausted the buffer. // Dump entire buffer into a file. The list of file // pointers will keep track of the files ... processComplete = false; dumpBufferToFile(records, files); records.clear(); records.add(record); } } } return record; } /** * Dump all records in List of Record objects to a file. Then, add that * file to List of File objects * * NEED TO PLACE A LIMIT ON NUMBER OF FILE POINTERS (check size of file list) * * @param records * @param files * @throws Throwable */ private void dumpBufferToFile(List<Record> records, List<File> files) throws Throwable { String prefix = "joiner_" + files.size() + 1; String suffix = ".dat"; File file = File.createTempFile(prefix, suffix, new File("cache")); BufferedWriter writer = new BufferedWriter(new FileWriter(file)); for( Record record : records ) { writer.write( record.dump() ); } files.add(file); writer.flush(); writer.close(); }

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  • Creating Visual Studio projects that only contain static files

    - by Eilon
    Have you ever wanted to create a Visual Studio project that only contained static files and didn’t contain any code? While working on ASP.NET MVC we had a need for exactly this type of project. Most of the projects in the ASP.NET MVC solution contain code, such as managed code (C#), unit test libraries (C#), and Script# code for generating our JavaScript code. However, one of the projects, MvcFuturesFiles, contains no code at all. It only contains static files that get copied to the build output folder: As you may well know, adding static files to an existing Visual Studio project is easy. Just add the file to the project and in the property grid set its Build Action to “Content” and the Copy to Output Directory to “Copy if newer.” This works great if you have just a few static files that go along with other code that gets compiled into an executable (EXE, DLL, etc.). But this solution does not work well if the projects only contains static files and has no compiled code. If you create a new project in Visual Studio and add static files to it you’ll still get an EXE or DLL copied to the output folder, despite not having any actual code. We wanted to avoid having a teeny little DLL generated in the output folder. In ASP.NET MVC 2 we came up with a simple solution to this problem. We started out with a regular C# Class Library project but then edited the project file to alter how it gets built. The critical part to get this to work is to define the MSBuild targets for Build, Clean, and Rebuild to perform custom tasks instead of running the compiler. The Build, Clean, and Rebuild targets are the three main targets that Visual Studio requires in every project so that the normal UI functions properly. If they are not defined then running certain commands in Visual Studio’s Build menu will cause errors. Once you create the class library projects there are a few easy steps to change it into a static file project: The first step in editing the csproj file is to remove the reference to the Microsoft.CSharp.targets file because the project doesn’t contain any C# code: <Import Project="$(MSBuildToolsPath)\Microsoft.CSharp.targets" /> .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; } The second step is to define the new Build, Clean, and Rebuild targets to delete and then copy the content files: <Target Name="Build"> <Copy SourceFiles="@(Content)" DestinationFiles="@(Content->'$(OutputPath)%(RelativeDir)%(Filename)%(Extension)')" /> </Target> <Target Name="Clean"> <Exec Command="rd /s /q $(OutputPath)" Condition="Exists($(OutputPath))" /> </Target> <Target Name="Rebuild" DependsOnTargets="Clean;Build"> </Target> .csharpcode, .csharpcode pre { font-size: small; color: black; font-family: consolas, "Courier New", courier, monospace; background-color: #ffffff; /*white-space: pre;*/ } .csharpcode pre { margin: 0em; } .csharpcode .rem { color: #008000; } .csharpcode .kwrd { color: #0000ff; } .csharpcode .str { color: #006080; } .csharpcode .op { color: #0000c0; } .csharpcode .preproc { color: #cc6633; } .csharpcode .asp { background-color: #ffff00; } .csharpcode .html { color: #800000; } .csharpcode .attr { color: #ff0000; } .csharpcode .alt { background-color: #f4f4f4; width: 100%; margin: 0em; } .csharpcode .lnum { color: #606060; } The third and last step is to add all the files to the project as normal Content files (as you would do in any project type). To see how we did this in the ASP.NET MVC 2 project you can download the source code and inspect the MvcFutureFules.csproj project file. If you’re working on a project that contains many static files I hope this solution helps you out!

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  • Virus that makes all files and folders read-only filesystem on a usb drive

    - by ren florento
    Is there any way on how to remove a virus from Windows that makes the files and folders and the usb drive itself a read-only filesystem as this is an annoying one because the virus keeps copying itself as long as it sees a folder and keeps running which prevents you from creating and deleting files and folders from the usb drive and makes " mount -o remount,rw '/path' " ineffective ? btw i'm not really sure if it is a virus but what makes me think that it is a virus is for the reason the it creates a .exe file within every folder which was named after folder and it also immediately reverts to read-only filesystem which locks the files and folders even after executing the command " mount -o remount,rw '/path' ". i also think the virus is just running only within the usb drive as it is not affecting the folders on ubuntu. I could choose to reformat the usb drive as it only contains few important files but what concerns me is if such virus or whatever you may call it gets into my backup drives that contains many important files.Thanks for any help and advice you could give.

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  • I cannot rename files in bulk using ubuntu's rename feature

    - by user254174
    I cannot rename files in bulk using ubuntu's rename feature. The files are on a NTFS partition. I want to rename files that look like this: whatever pic george.jpg tacoma narrows bridge.jpg green bottle.jpg to: filename (1) filename (2) filename (3) And I cannot do this at all. I don't want to use the command line either. So I can permanently erase files after I have encrypted them without exposing their contents to people who use a file recovery tool. I also don't want a method that takes days or months to rename the file. That is, rename one file at a time. So if I have hundreds of files to rename, this won't be a option. I want to give a each file the same name and numbered in order like shown above. Pyrenamer is not an option for me, unless you can find how to do that in PyRenamer.

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  • How to Share Files Between User Accounts on Windows, Linux, or OS X

    - by Chris Hoffman
    Your operating system provides each user account with its own folders when you set up several different user accounts on the same computer. Shared folders allow you to share files between user accounts. This process works similarly on Windows, Linux, and Mac OS X. These are all powerful multi-user operating systems with similar folder and file permission systems. Windows On Windows, the “Public” user’s folders are accessible to all users. You’ll find this folder under C:\Users\Public by default. Files you place in any of these folders will be accessible to other users, so it’s a good way to share music, videos, and other types of files between users on the same computer. Windows even adds these folders to each user’s libraries by default. For example, a user’s Music library contains the user’s music folder under C:\Users\NAME\as well as the public music folder under C:\Users\Public\. This makes it easy for each user to find the shared, public files. It also makes it easy to make a file public — just drag and drop a file from the user-specific folder to the public folder in the library. Libraries are hidden by default on Windows 8.1, so you’ll have to unhide them to do this. These Public folders can also be used to share folders publically on the local network. You’ll find the Public folder sharing option under Advanced sharing settings in the Network and Sharing Control Panel. You could also choose to make any folder shared between users, but this will require messing with folder permissions in Windows. To do this, right-click a folder anywhere in the file system and select Properties. Use the options on the Security tab to change the folder’s permissions and make it accessible to different user accounts. You’ll need administrator access to do this. Linux This is a bit more complicated on Linux, as typical Linux distributions don’t come with a special user folder all users have read-write access to. The Public folder on Ubuntu is for sharing files between computers on a network. You can use Linux’s permissions system to give other user accounts read or read-write access to specific folders. The process below is for Ubuntu 14.04, but it should be identical on any other Linux distribution using GNOME with the Nautilus file manager. It should be similar for other desktop environments, too. Locate the folder you want to make accessible to other users, right-click it, and select Properties. On the Permissions tab, give “Others” the “Create and delete files” permission. Click the Change Permissions for Enclosed Files button and give “Others” the “Read and write” and “Create and Delete Files” permissions. Other users on the same computer will then have read and write access to your folder. They’ll find it under /home/YOURNAME/folder under Computer. To speed things up, they can create a link or bookmark to the folder so they always have easy access to it. Mac OS X Mac OS X creates a special Shared folder that all user accounts have access to. This folder is intended for sharing files between different user accounts. It’s located at /Users/Shared. To access it, open the Finder and click Go > Computer. Navigate to Macintosh HD > Users > Shared. Files you place in this folder can be accessed by any user account on your Mac. These tricks are useful if you’re sharing a computer with other people and you all have your own user accounts — maybe your kids have their own limited accounts. You can share a music library, downloads folder, picture archive, videos, documents, or anything else you like without keeping duplicate copies.

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