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  • 2 way synchronization of in-house gitosis repositories server with Github repositories

    - by Robert J Berger
    We use gitosis as our local in-house shared repository and also have a private Github account that mirrors our local repository. If one does a git push to the in-house repository the post-update hook updates the Github repository. How can I make this two way without causing "loops"? I.e. if someone pushes to the Github repository, I would like it to also update the in-house gitosis repository. A pointer to an example of how to do it would be greatly appreciated. Or if there are recommendations of alternatives to gitosis that would make this kind of thing easy, I would consider migrating to that.

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  • Problems in Linux Mint 13 MATE with NTFS partitions

    - by usr
    I'm using Linux Mint 13 MATE (32 bit). With Caja file manager (based on Nautilus 2.x) I can create, modify and remove folders and files on NTFS partitions whitout problems. But, for example, when I have a compressed file (.zip, .7z, ...) in a NTFS partition I can't extract it (with File-Roller, Peazip, ...). But I can extract the same file in a ext4 partition perfectly. What might the problem be? Are there (maybe) file permissions for each application?

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  • How can I transfer a file via XMPP using Python?

    - by Enchantner
    I'm using xmpppy library for my jabber remote administration bot, but I can't find how to send/receive a file and save it inside the directory specified. The documentation is poor and there isn't any examples, but I really want to make it. Can anyone show some examples or some links about it? Or maybe I should use an alternative xmpp bindings?

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  • Create PHP DOM xml file and create a save file link/prompt without writing the file to the server wh

    - by Reed Richards
    I've created a PHP DOM xml piece and saved it to a string like this: <?php // create a new XML document $doc = new DomDocument('1.0'); ... ... ... $xmldata = $doc->saveXML(); ?> Now I can't use the headers to send a file download prompt and I can't write the file to the server, or rather I don't want the file laying around on it. Something like a save this file link or a download prompt would be good. How do I do it?

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  • C++: ifstream::getline problem

    - by Jay
    I am reading a file like this: char string[256]; std::ifstream file( "file.txt" ); // open the level file. if ( ! file ) // check if the file loaded fine. { // error } while ( file.getline( string, 256, ' ' ) ) { // handle input } Just for testing purposes, my file is just one line, with a space at the end: 12345 My code first reads the 12345 successfully. But then instead of the loop ending, it reads another string, which seems to be a return/newline. I have saved my file both in gedit and in nano. And I have also outputted it with the Linux cat command, and there is no return on the end. So the file should be fine. Why is my code reading a return/newline? Thanks.

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  • Data synchronization using XMPP

    - by Jason
    Hi: I'm looking for some insight/advice on synchronizing data over XMPP. I've never developed anything for XMPP before so excuse me if some of my questions seem ridiculous. Basically, what I have is a decentralized social network. Each person has it's own Web site (or server) with a unique URI (one domain could host many servers). Each of these servers can have many clients. E.g., a desktop application, mobile application, etc. What I would like to accomplish is near real-time synchronization/communication between client and server, e.g., I update something on my desktop application, I see it change on my Web site. My server and client code is Python. So, I would like to make use of SleekXMPP if possible (it's license seems to have changed to MIT). I was thinking, and here is where I need advice, that each server would register an account at a dedicated XMPP server, e.g., [email protected]. and then I could use different resources for clients [email protected]/client1, [email protected]/client2, etc. If anyone can register any username, then maybe I also need some intermediate service (since it's decentralized, i'm not sure how to control registrations). Another option, I guess, is that each server runs it's own xmpp server. Assuming, that was all worked out, if I want to broadcast messages to all my resources (except the sending one), how do I do that? Do I have to subscribe to myself? This also seems like a good candidate for publish-subscribe, let me know if you think that could work and what the design/flow of that process would be. thanks :)

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  • PHP MINISERVER DOWNLOAD RESUME-ERROR! Resource id # 4

    - by snikolov
    $httpsock = @socket_create_listen("9090"); if (!$httpsock) { print "Socket creation failed!\n"; exit; } while (1) { $client = socket_accept($httpsock); $input = trim(socket_read ($client, 4096)); $input = explode(" ", $input); $range = $input[12]; $input = $input[1]; $fileinfo = pathinfo($input); switch ($fileinfo['extension']) { default: $mime = "text/html"; } if ($input == "/") { $input = "index.html"; } $input = ".$input"; if (file_exists($input) && is_readable($input)) { echo "Serving $input\n"; $contents = file_get_contents($input); $output = "HTTP/1.0 200 OK\r\nServer: APatchyServer\r\nConnection: close\r\nContent-Type: $mime\r\n\r\n$contents"; } else { //$contents = "The file you requested doesn't exist. Sorry!"; //$output = "HTTP/1.0 404 OBJECT NOT FOUND\r\nServer: BabyHTTP\r\nConnection: close\r\nContent-Type: text/html\r\n\r\n$contents"; if(isset($range)) { list($a, $range) = explode("=",$range); str_replace($range, "-", $range); $size2 = $size-1; $new_length = $size-$range; $output = "HTTP/1.1 206 Partial Content\r\n"; $output .= "Content-Length: $new_length\r\n"; $output .= "Content-Range: bytes $range$size2/$size\r\n"; } else { $size2=$size-1; $output .= "Content-Length: $new_length\r\n"; } $chunksize = 1*(1024*1024); $bytes_send = 0; $file = "a.mp3"; $filesize = filesize($file); if ($file = fopen($file, 'r')) { if(isset($range)) $output = 'HTTP/1.0 200 OK\r\n'; $output .= "Content-type: application/octet-stream\r\n"; $output .= "Content-Length: $filesize\r\n"; $output .= 'Content-Disposition: attachment; filename="'.$file.'"\r\n'; $output .= "Accept-Ranges: bytes\r\n"; $output .= "Cache-Control: private\n\n"; fseek($file, $range); $download_rate = 1000; while(!feof($file) and (connection_status()==0)) { $var_stat = fread($file, round($download_rate *1024)); $output .= $var_stat;//echo($buffer); // is also possible flush(); sleep(1);//// decrease download speed } fclose($file); } /** $filename = "dada"; $file = fopen($filename, 'r'); $filesize = filesize($filename); $buffer = fread($file, $filesize); $send = array("Output"=$buffer,"filesize"=$filesize,"filename"=$filename); $file = $send['filename']; */ //@ob_end_clean(); // $output .= "Content-Transfer-Encoding: binary"; //$output .= "Connection: Keep-Alive\r\n"; } socket_write($client, $output); socket_close ($client); } socket_close ($httpsock); hey guys i have create a miniwebserver downloader it can download files from your server, however i am unable to resume my download when i download the file i get Resource id # 4 and also i cant resume the download,i would like to know how i can monitor record the client output how much bandwidth he has downloaded perl has something like this put its hardcore if possible kindly provide me with some pointers thank you :)

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  • Auto populate input based on file name with AngularJS

    - by LouieV
    I am playing around with AngularJS and have not been able to solve this problem. I have a view that has a form to upload a file to a node server. So far I have manage to do this using some directives and a service. I allow the user to send a custom name to the POST data if they desire. What I wan to accomplish is that when the user selects a file the filename models auto populates. My view looks like: <div> <input file-model="phpFile" type="file"> <input name="filename" type="text" ng-model="filename"> <button ng-click="send()">send</button> </div> file-model is my directive that allows the file to be assigned to a scope. myApp.directive('fileModel', ['$parse', function($parse) { return { restrict: 'A', link: function(scope, element, attrs) { var model = $parse.(attrs.fileModel); var modelSetter = model.assign; element.bind('change', function() { scope.$apply(function() { modelSetter(scope, element[0].files[0]); }); }); } }]); The service: myApp.service('fileUpload', ['$http', function($http){ this.uploadFileToUrl = function(file, uploadUrl, optionals) { var fd = new FormData(); fd.append('file', file); for (var key in file) { fd.append(key, file[key]); } for(var i = 0; i < optionals.length; i++){ fd.append(optionals[i].name, optionals[i].data); } }); }]); Here as you can see I pass the file, append its properties, and append any optional properties. In the controller is where I am having the troubles. I have tried $watch and using the file-model but I get the same error either way. myApp.controller('AddCtrl', function($scope, $location, PEberry, fileUpload){ //$scope.$watch(function() { // return $scope.phpFile; //},function(newValue, oldValue) { // $scope.filename = $scope.phpFile.name; //}, true); // if ($scope.phpFiles) { // $scope.filename = $scope.phpFiles.name; // } $scope.send = function() { var uploadUrl = "/files"; var file = $scope.phpFile; //var opts = [{ name: "uname", data: file.name }] fileUpload.uploadFileToUrl(file, uploadUrl); }; }); Thank you for your help!

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  • PHP mini-server download resulme-error! Resource id # 4

    - by snikolov
    <?php $httpsock = @socket_create_listen("9090"); if (!$httpsock) { print "Socket creation failed!\n"; exit; } while (1) { $client = socket_accept($httpsock); $input = trim(socket_read ($client, 4096)); $input = explode(" ", $input); $range = $input[12]; $input = $input[1]; $fileinfo = pathinfo($input); switch ($fileinfo['extension']) { default: $mime = "text/html"; } if ($input == "/") { $input = "index.html"; } $input = ".$input"; if (file_exists($input) && is_readable($input)) { echo "Serving $input\n"; $contents = file_get_contents($input); $output = "HTTP/1.0 200 OK\r\nServer: APatchyServer\r\nConnection: close\r\nContent-Type: $mime\r\n\r\n$contents"; } else { //$contents = "The file you requested doesn't exist. Sorry!"; //$output = "HTTP/1.0 404 OBJECT NOT FOUND\r\nServer: BabyHTTP\r\nConnection: close\r\nContent-Type: text/html\r\n\r\n$contents"; if(isset($range)) { list($a, $range) = explode("=",$range); str_replace($range, "-", $range); $size2 = $size-1; $new_length = $size-$range; $output = "HTTP/1.1 206 Partial Content\r\n"; $output .= "Content-Length: $new_length\r\n"; $output .= "Content-Range: bytes $range$size2/$size\r\n"; } else { $size2=$size-1; $output .= "Content-Length: $new_length\r\n"; } $chunksize = 1*(1024*1024); $bytes_send = 0; $file = "a.mp3"; $filesize = filesize($file); if ($file = fopen($file, 'r')) { if(isset($range)) $output = 'HTTP/1.0 200 OK\r\n'; $output .= "Content-type: application/octet-stream\r\n"; $output .= "Content-Length: $filesize\r\n"; $output .= 'Content-Disposition: attachment; filename="'.$file.'"\r\n'; $output .= "Accept-Ranges: bytes\r\n"; $output .= "Cache-Control: private\n\n"; fseek($file, $range); $download_rate = 1000; while(!feof($file) and (connection_status()==0)) { $var_stat = fread($file, round($download_rate *1024)); $output .= $var_stat;//echo($buffer); // is also possible flush(); sleep(1);//// decrease download speed } fclose($file); } /** $filename = "dada"; $file = fopen($filename, 'r'); $filesize = filesize($filename); $buffer = fread($file, $filesize); $send = array("Output"=>$buffer,"filesize"=>$filesize,"filename"=>$filename); $file = $send['filename']; */ //@ob_end_clean(); // $output .= "Content-Transfer-Encoding: binary"; //$output .= "Connection: Keep-Alive\r\n"; } socket_write($client, $output); socket_close ($client); } socket_close ($httpsock); Hey guys, I haved create a miniwebserver downloader. It can download files from your server. However, I am unable to resume my download when I download the file – I get Resource id # 4 – and I also can't resume the download. I would like to know how I can monitor and record the client output and how much bandwidth he has downloaded. Perl has something like this, but it's hardcore; if possible, kindly provide me with some pointers thank you :)

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  • How do I turn an array of bytes back into a file and open it automatically with C#?

    - by Ace Grace
    Hi, I am writing some code to add file attachments into an application I am building. I have add & Remove working but I don't know where to start to implement open. I have an array of bytes (from a table field) and I don't know how to make it automatically open e.g. If I have an array of bytes which is a PDF, how do I get my app to automatically open Acrobat or whatever the currently assigned application for the extension is using C#?

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  • Ruby: output not saved to file

    - by Sophie
    I'm trying to give a file as input, have it changed within the program, and save the result to a file that is output. But the output file is the same as the input file. :/ Total n00b question, but what am I doing wrong?: puts "Reading Celsius temperature value from data file..." num = File.read("temperature.dat") celsius = num.to_i farenheit = (celsius * 9/5) + 32 puts "Saving result to output file 'faren_temp.out'" fh = File.new("faren_temp.out", "w") fh.puts farenheit fh.close

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  • Symfony and uploadify

    - by Thomas
    Hi! I want to use uploadify with Symfony 1.4, but so far I couldn't. Uploadify loads correctly, I choose my files, it says that the files were successfully uploaded, but the are nowhere. (I'm doing this on localhost) Is there anybody who met this problem before? Thanks, Tom $file = $request->getParameter('file'); $filename = sha1($file->getOriginalName()).$file->getExtension($file->getOriginalExtension()); $file->save(sfConfig::get('sf_upload_dir').'/'.$filename);

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  • Uploading multiple files using Spring MVC 3.0.2 after HiddenHttpMethodFilter has been enabled

    - by Tiny
    I'm using Spring version 3.0.2. I need to upload multiple files using the multiple="multiple" attribute of a file browser such as, <input type="file" id="myFile" name="myFile" multiple="multiple"/> (and not using multiple file browsers something like the one stated by this answer, it indeed works I tried). Although no versions of Internet Explorer supports this approach unless an appropriate jQuery plugin/widget is used, I don't care about it right now (since most other browsers support this). This works fine with commons fileupload but in addition to using RequestMethod.POST and RequestMethod.GET methods, I also want to use other request methods supported and suggested by Spring like RequestMethod.PUT and RequestMethod.DELETE in their own appropriate places. For this to be so, I have configured Spring with HiddenHttpMethodFilter which goes fine as this question indicates. but it can upload only one file at a time even though multiple files in the file browser are chosen. In the Spring controller class, a method is mapped as follows. @RequestMapping(method={RequestMethod.POST}, value={"admin_side/Temp"}) public String onSubmit(@RequestParam("myFile") List<MultipartFile> files, @ModelAttribute("tempBean") TempBean tempBean, BindingResult error, Map model, HttpServletRequest request, HttpServletResponse response) throws IOException, FileUploadException { for(MultipartFile file:files) { System.out.println(file.getOriginalFilename()); } } Even with the request parameter @RequestParam("myFile") List<MultipartFile> files which is a List of type MultipartFile (it can always have only one file at a time). I could find a strategy which is likely to work with multiple files on this blog. I have gone through it carefully. The solution below the section SOLUTION 2 – USE THE RAW REQUEST says, If however the client insists on using the same form input name such as ‘files[]‘ or ‘files’ and then populating that name with multiple files then a small hack is necessary as follows. As noted above Spring 2.5 throws an exception if it detects the same form input name of type file more than once. CommonsFileUploadSupport – the class which throws that exception is not final and the method which throws that exception is protected so using the wonders of inheritance and subclassing one can simply fix/modify the logic a little bit as follows. The change I’ve made is literally one word representing one method invocation which enables us to have multiple files incoming under the same form input name. It attempts to override the method protected MultipartParsingResult parseFileItems(List fileItems, String encoding) {} of the abstract class CommonsFileUploadSupport by extending the class CommonsMultipartResolver such as, package multipartResolver; import java.io.UnsupportedEncodingException; import java.util.HashMap; import java.util.Iterator; import java.util.List; import java.util.Map; import javax.servlet.ServletContext; import org.apache.commons.fileupload.FileItem; import org.springframework.util.StringUtils; import org.springframework.web.multipart.MultipartException; import org.springframework.web.multipart.MultipartFile; import org.springframework.web.multipart.commons.CommonsMultipartFile; import org.springframework.web.multipart.commons.CommonsMultipartResolver; final public class MultiCommonsMultipartResolver extends CommonsMultipartResolver { public MultiCommonsMultipartResolver() { } public MultiCommonsMultipartResolver(ServletContext servletContext) { super(servletContext); } @Override @SuppressWarnings("unchecked") protected MultipartParsingResult parseFileItems(List fileItems, String encoding) { Map<String, MultipartFile> multipartFiles = new HashMap<String, MultipartFile>(); Map multipartParameters = new HashMap(); // Extract multipart files and multipart parameters. for (Iterator it = fileItems.iterator(); it.hasNext();) { FileItem fileItem = (FileItem) it.next(); if (fileItem.isFormField()) { String value = null; if (encoding != null) { try { value = fileItem.getString(encoding); } catch (UnsupportedEncodingException ex) { if (logger.isWarnEnabled()) { logger.warn("Could not decode multipart item '" + fileItem.getFieldName() + "' with encoding '" + encoding + "': using platform default"); } value = fileItem.getString(); } } else { value = fileItem.getString(); } String[] curParam = (String[]) multipartParameters.get(fileItem.getFieldName()); if (curParam == null) { // simple form field multipartParameters.put(fileItem.getFieldName(), new String[] { value }); } else { // array of simple form fields String[] newParam = StringUtils.addStringToArray(curParam, value); multipartParameters.put(fileItem.getFieldName(), newParam); } } else { // multipart file field CommonsMultipartFile file = new CommonsMultipartFile(fileItem); if (multipartFiles.put(fileItem.getName(), file) != null) { throw new MultipartException("Multiple files for field name [" + file.getName() + "] found - not supported by MultipartResolver"); } if (logger.isDebugEnabled()) { logger.debug("Found multipart file [" + file.getName() + "] of size " + file.getSize() + " bytes with original filename [" + file.getOriginalFilename() + "], stored " + file.getStorageDescription()); } } } return new MultipartParsingResult(multipartFiles, multipartParameters); } } What happens is that the last line in the method parseFileItems() (the return statement) i.e. return new MultipartParsingResult(multipartFiles, multipartParameters); causes a compile-time error because the first parameter multipartFiles is a type of Map implemented by HashMap but in reality, it requires a parameter of type MultiValueMap<String, MultipartFile> It is a constructor of a static class inside the abstract class CommonsFileUploadSupport, public abstract class CommonsFileUploadSupport { protected static class MultipartParsingResult { public MultipartParsingResult(MultiValueMap<String, MultipartFile> mpFiles, Map<String, String[]> mpParams) { } } } The reason might be - this solution is about the Spring version 2.5 and I'm using the Spring version 3.0.2 which might be inappropriate for this version. I however tried to replace the Map with MultiValueMap in various ways such as the one shown in the following segment of code, MultiValueMap<String, MultipartFile>mul=new LinkedMultiValueMap<String, MultipartFile>(); for(Entry<String, MultipartFile>entry:multipartFiles.entrySet()) { mul.add(entry.getKey(), entry.getValue()); } return new MultipartParsingResult(mul, multipartParameters); but no success. I'm not sure how to replace Map with MultiValueMap and even doing so could work either. After doing this, the browser shows the Http response, HTTP Status 400 - type Status report message description The request sent by the client was syntactically incorrect (). Apache Tomcat/6.0.26 I have tried to shorten the question as possible as I could and I haven't included unnecessary code. How could be made it possible to upload multiple files after Spring has been configured with HiddenHttpMethodFilter? That blog indicates that It is a long standing, high priority bug. If there is no solution regarding the version 3.0.2 (3 or higher) then I have to disable Spring support forever and continue to use commons-fileupolad as suggested by the third solution on that blog omitting the PUT, DELETE and other request methods forever. Just curiously waiting for a solution and/or suggestion. Very little changes to the code in the parseFileItems() method inside the class MultiCommonsMultipartResolver might make it to upload multiple files but I couldn't succeed in my attempts (again with the Spring version 3.0.2 (3 or higher)).

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  • iPhone Network Share Access

    - by user361988
    I am trying to download a data file from a local network share to an iPhone device. I have placed the file on a computer on the network and can view through browsers such as Chrome or Mozilla, from any computer on the local network. However, Safari on a Mac and the iPhone do not find the file! An example of the URL I use is 'file://computer/SharedDocs/file.csv'. Why do Safari and the iPhone fail to find the file?

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  • Working with multiple GIT severs

    - by th3flyboy
    Hello, I have a question. Is it possible to set up a system so that you have a private GIT server that you host, which automatically syncs with a remote one, hosted by a site like Sourceforge, and then you can commit your local to the private GIT server, and then when you have to merge the changes from your private wip branches that are on your private GIT over to the master/branch/tag from the public GIT, and then push the change to the public GIT? I ask this because I have a lot of personal work I would like to get working before putting it up for the public to see, and I'm shifting between several computers/operating systems in the process. If this is not possible in standard GIT, are there any other options that would allow me to do this? Thanks, Peter

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  • Microsoft Sync Framework - How to reprovision a table (or entire scope) after schema changes?

    - by Rabbi
    I have already setup syncing with Microsoft Sync Framework, and now I need to add fields to a table. How do I re-provision the databases? The setup is exceedingly simple: Two SQL Express 2008 servers The scope includes the entire database Using Microsoft Sync Framework 2.0 Synchronizing by direct access. Using the standard new SqlSyncProvider Do I make the structural changes at both ends? Or do I only change one server and let Sync Framework somehow propagate the change? Do I need to delete the _tracking tables and/or the stored procedures? How about the triggers? Has anyone been using the Sync Framework? Please help.

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  • please give me a solution

    - by user327832
    here is the code i have written so far but ended up giving me error import java.io.File; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStream; public class Main { public static void main(String[] args) throws Exception { File file = new File("c:\\filea.txt"); InputStream is = new FileInputStream(file); long length = file.length(); System.out.println (length); bytes[] bytes = new bytes[(int) length]; try { int offset = 0; int numRead = 0; while (numRead >= 0) { numRead = is.read(bytes); } } catch (IOException e) { System.out.println ("Could not completely read file " + file.getName()); } is.close(); Object[] see = new Object[(int) length]; see[1] = bytes; System.out.println ((String[])see[1]); } }

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