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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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  • 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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  • Does anyone has the experience of using the new p4 replicate command in their Perforce back-up /rest

    - by Thomas Corriol
    Hi all, we recently performed an upgrade of our whole perforce system to 2009.02 During this exercise, we noticed that the back-up /restore process that was installed here by the Perforce consultant a year ago was not completely working. Basically, the verify command has never worked (scary !). As we are obliged to revisit our Back-Up/Restore scripts, I was toying with the idea of using the new p4 replicate command. The idea is to use it alongside an rsync of the data files, so that in case of crash we will lose at worst an hour of work (if we execute them every hour). Does anyone has the experience or an example of back-up/restore scripts using the p4 replicate command of the 2009.02 version ? Thanks, Thomas

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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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  • My database has been deleted suddenly on the server how to recover it?

    - by user2728312
    I'm running an application on windows server that connect to a SQL Server database. Today, when I opened SQL Server Management Studio, I was surprised the database is not in the list of the databases! I don't know what's the reason. I searched in the server files but I can't find the database and also in the recycle bin. I put my database in C:\db\myWeb.mdf and suddenly it's been removed! Can anyone tell me how to recover the database?

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  • How to create a named temporary file in memory?

    - by conradlee
    I would like to use Python's tempfile module to create a temporary file that I will use for communication between processes (use of pipes is awkward). The documentation I've linked to above shows two functions that almost do what I want: tempfile.NamedTemporaryFile # For creating named tempfiles tempfile.SpooledTemporaryFile # For creating tempfiles in memory but actually I want a tempfile that is both named AND in memory. Any ideas?

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  • [NSData dataWithContentsOfFile:path] doesn't work

    - by Felics
    Hello, when I have the fallowing code to read a binary file: NSString* file = [NSString stringWithUTF8String:fileName]; NSString* filePath = resource ? [[NSBundle mainBundle] pathForResource:file ofType:nil] : [[NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES) objectAtIndex:0] stringByAppendingPathComponent: file]; NSData* fileData = [NSData dataWithContentsOfFile:filePath]; Where "fileName" and resource are load function parameters. "resource" is used to know if the file is located in application bundle or in Documents. Sometimes this code works well and sometimes it doesn't. As far I saw this problem is random. I can run the code 10 times in a row and it works fine and after that it gives me nil data without any modification. Does anybody knows what could be the problem? Could it be related with file extension or file name? Thank you. PS: I use this code on iPhone Simulator and the file exists in application bundle.

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  • Find files in a folder using Java

    - by Sii
    Hello, I'm new here so be kind to my stupidity :) What I need to do if Search a folder say C:\example I then need to go through each file and check to see if it matches a few start characters so if files start temp****.txt tempONE.txt tempTWO.txt So if the file starts with temp and has an extension .txt I would like to then put that file name into a File file = new File("C:/example/temp***.txt); so I can then read in the file, the loop then needs to move onto the next file to check to see if it meets as above. I hope this makes sense, thanks for taking the time to view this I do apperciate it :)

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  • time on files differ by 1 sec. FAIL Robocopy sync

    - by csmba
    I am trying to use Robocopy to sync (/IMG) a folder on my PC and a shared network drive. The problem is that the file attributes differ by 1 sec on both locations (creation,modified and access). So every time I run robocopy, it syncs the file again... BTW, problem is the same if I delete the target file and robocopy it from new... still, new file has 1 sec different properties. Env Details: Source: Win 7 64 bit Target: WD My Book World Edition NAS 1TB which takes its time from online NTP pool.ntp.org (I don't know if file system is FAT or not)

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  • Anyone know any good backend user online file manager?

    - by skyhigh
    Hi I'm looking for a backend system where your clients can login and upload files to your server, download files from the server and you can delete the users, create users, etc. I do not know the proper name for this kind of software. Maybe its called online file manager? Any recommendations? My server supports PHP, apache and mysq. Thanks

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  • PHP: Fastest way possible to read contents of a file.

    - by SoLoGHoST
    Ok, I'm looking for the fastest possible way to read all of the contents of a file via php with a filepath on the server, also these files can be huge. So it's very important that it does a READ ONLY to it as fast as possible. Is reading it line by line faster than reading the entire contents? Though, I remember reading up on this some, that reading the entire contents can produce errors for huge files. Is this true?

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  • Is there a way to efficiently yield every file in a directory containing millions of files?

    - by Josh Smeaton
    I'm aware of os.listdir, but as far as I can gather, that gets all the filenames in a directory into memory, and then returns the list. What I want, is a way to yield a filename, work on it, and then yield the next one, without reading them all into memory. Is there any way to do this? I worry about the case where filenames change, new files are added, and files are deleted using such a method. Some iterators prevent you from modifying the collection during iteration, essentially by taking a snapshot of the state of the collection at the beginning, and comparing that state on each move operation. If there is an iterator capable of yielding filenames from a path, does it raise an error if there are filesystem changes (add, remove, rename files within the iterated directory) which modify the collection? There could potentially be a few cases that could cause the iterator to fail, and it all depends on how the iterator maintains state. Using S.Lotts example: filea.txt fileb.txt filec.txt Iterator yields filea.txt. During processing, filea.txt is renamed to filey.txt and fileb.txt is renamed to filez.txt. When the iterator attempts to get the next file, if it were to use the filename filea.txt to find it's current position in order to find the next file and filea.txt is not there, what would happen? It may not be able to recover it's position in the collection. Similarly, if the iterator were to fetch fileb.txt when yielding filea.txt, it could look up the position of fileb.txt, fail, and produce an error. If the iterator instead was able to somehow maintain an index dir.get_file(0), then maintaining positional state would not be affected, but some files could be missed, as their indexes could be moved to an index 'behind' the iterator. This is all theoretical of course, since there appears to be no built-in (python) way of iterating over the files in a directory. There are some great answers below, however, that solve the problem by using queues and notifications. Edit: The OS of concern is Redhat. My use case is this: Process A is continuously writing files to a storage location. Process B (the one I'm writing), will be iterating over these files, doing some processing based on the filename, and moving the files to another location. Edit: Definition of valid: Adjective 1. Well grounded or justifiable, pertinent. (Sorry S.Lott, I couldn't resist). I've edited the paragraph in question above.

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