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  • File Open/Save Dialog always 'Not Responding'

    - by Amanda
    I am aware that this question has been asked once, however the solution for them didn't work with me. Whenever I go to open/save a file in any program, the dialog does not come up, and the application goes to 'Not Responding'. This goes on for about a few minutes, and then stops, but still does not open the dialog. There have been a few occasions where the dialog has suddenly worked for a while, but then the problem comes back. I have tried many solutions given around the internet, I have cleaned it with CCleaner, disk defragged it, sorted the index. Nothing works. Is there anybody who has any idea what the problem is? This is Windows Vista. I'm not quite sure what kind of information you guys would need about my laptop, but I'll give you it if you need it. :) Solutions I have Tried: I have tried deleting the 'Shellicon' folder in the registry, which wasn't even in there. I have looked for mapped network drives lingering around, and I haven't come across any. I have tried rebuilding teh Widnows Search Indexes...no difference.

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  • Windows 7 SSH file server

    - by Siriss
    Hello all- I have looked at the other posts, but have not quite found an answer I have a question about windows file sharing over SSH. I have copssh installed and it is working for Remote desktop connections. I have port 22 forwarded on my router etc. I connect from a Mac or Putty with this address: ssh -l copsshusername 3391:localhost:3389 [external ip] That works fine. I would like to configure Windows 7 to allow my ssh account that I use to login, access to certain shared folders. I have documents and videos and things that I would like to be able to download externally. I have done this before on Linux and a long time ago on XP, but I cannot figure out what I am missing on Windows 7. There is a designated SSH user that copssh uses to run the service and that I use to to login as. I have googled and googled and have not found a solution that does everything I need that is why I am turning here for ideas. I hope I am explaining this correctly. Thank you very much for your help!

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  • How can I tell why I have access to a file share on Windows Server

    - by Joel
    I have a file share on a Windows 2008 R2 server in a AD domain (call it \SECURESERVER\STUFF) and I am not sure if I have the share and folder permissions set up right. I noticed the problem when I set up new server (WORKGROUP\FOREIGNSERVER) that was not joined to the domain and tried to copy some files off of \SECURESERVER\STUFF. I was surprised to find that when I tried to access the files, it did not prompt me for a username and password and proceeded to give me full access to the files. That worried me so I tried the same thing on some workstations that were not in the domain and they did NOT have the same behavior (they did prompt for a username/password as desired/expected). So, I think there is something peculiar about FOREIGNSERVER. I am logging into it with a local admin account, but my domain and SECURESERVER should know nothing of this server. I've carefully gone through the share and folder permissions on the share but I can't find the reason that FOREIGNSERVER has access. How can I find out why FOREIGNSERVER has access to SECURESERVER?

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  • Website and file/directory permissions

    - by mathiass
    I've been given a task to fix this one website. One of its issues is that on one page, the images have broken links - the images are not showing, and clicking on the image (i.e. direct link to the image file) results in a 403 (Forbidden) error. I am looking for some feedback on what could be the possible cause. The directory where the images are stored has the following permissions: drwxrws--- www "group" 10240 Aug 2008 "image directory name" I had to hide the names. I checked the page source code, and everything seems to be in place. The rest of the site, and other images outside that image directory are showing fine. I was told that recently there have been some changes to the server. I'm trying to assume that there is no fault in the source code, and the permissions are - or used to be - correct (since the site has been working before, and no recent changes to the site itself have been made). My only thoughts at the moment is that either: a) the directory permission should be: drwxrws--x (executable) for the other users, or b) there is a change in the server settings that I don't know of. Is there anything else I should check?

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  • .bat file - Nagios v3.2 service check and start if stopped

    - by LbakerIT
    I'm just barely getting into programming so I do apologize for my ignorance. I'm trying to create a .bat file that will check if a service is running on XP Pro. If service is running it will exit 0. If the service is stopped start service wait 10 seconds (via ping i'm guessing) check if service is running if service is running exit 0 if service is stopped start service wait 10 seconds Do this check a total of 3 times. if service does not come up within that time: exit 2 Exit 0 = ok exit 1 = warning exit 3 = critical (and this will alert) I need to do this for 3 different services but i'm expecting that it would be better to create one per service. That way you get notified on the specific service that is not coming back up. The goal is that if the service stops it will start it. If after 30 seconds it is unable to start the service then it will send an alert. The reason I'm trying to do it with a .bat is this is consistent with all other scripts and I did not want to complicate it further by adding different kinds of code. Yay for consistency! Again I do apologize for my ignorance I've been thrown into this project last minute. Thank you for the help and reading my question!

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  • using java.util.Scanner to read a file byte by byte

    - by openidsucks
    I'm trying to read a one line file character by character using java.util.Scanner. However I'm getting this exception": Exception in thread "main" java.util.InputMismatchException: For input string: "contents of my file" at java.util.Scanner.nextByte(Scanner.java:1861) at java.util.Scanner.nextByte(Scanner.java:1814) at p008.main(p008.java:18) <-- line where I do scanner.nextByte() Here's my code: public static void main(String[] args) throws FileNotFoundException { File source = new File("file.txt"); Scanner scanner = new Scanner(source); while(scanner.hasNext()) { System.out.println((char)scanner.nextByte()); } scanner.close() } Does anyone have any ideas as to what I might be doing wrong? Edit: I realized I wrote hasNext() instead of hasNextByte(). However if I do that it doesn't print out anything.

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  • Batch file script to remove special characters from filenames (Windows)

    - by njreed.myopenid.com
    I have a large set of files, some of which contain special characters in the filename (e.g. ä,ö,%, and others). I'd like a script file to iterate over these files and rename them removing the special characters. I don't really mind what it does, but it could replace them with underscores for example e.g. Störung%20.doc would be renamed to St_rung_20.doc In order of preference: A DOS batch file A Windows script file to run with cscript (vbs) A third party piece of software that can be run from the command-line (i.e. no user interaction required) Another language script file, for which I'd have to install an additional script engine Background: I'm trying to encrypt these file with GnuPG on Windows but it doesn't seem to handle special characters in filenames with the --encrypt-files option.

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  • .NET: IOException for permissions when creating new file?

    - by Rosarch
    I am trying to create a new file and write XML to it: FileStream output = File.Create(Path.Combine(PATH_TO_DATA_DIR, fileName)); The argument evaluates to: C:\path\to\Data\test.xml The exception is: The process cannot access the file 'C:\path\to\Data\test.xml' because it is being used by another process. What am I doing wrong here? UPDATE: This code throws the same exception: StreamWriter writer = new StreamWriter(Path.Combine(PATH_TO_DATA_DIR, fileName)); UPDATE 2: The file I am trying to create does not exist in the file system. So how can be it in use?

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  • Effective file permissions tool's api in windows

    - by apoorv020
    Starting from Windows Server 2003, Windows included a new tool which calculates the effective permissions for a user (basically it resolves all groups access and takes care of all "deny" permissions as well). An example in point is that a user A belongs to groups B and C. B has been denied read permissions on a file F, while C has been allowed read and write permissions on the file and I want to calculate the effective permissions user A has on file F. This tool is available on Windows Server 2003,Vista,7 and Server 2008 by right clicking on a file and going to properties - security - advanced - effective permissions. What I need is an API in C# which does the same job. The most common FILE API returns access rules (class FileAccessRules), but there seems to be no direct way to calculate effective permissions from these set of access rules. Note: I do not want to process effective permissions in the code if at all possible, but am ready to do so as a last resort.

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  • How to know a file is finished copying

    - by Yigang Wu
    I’m using ReadDirectoryChangesW to spy a folder, if I’m copying a large file to the folder, I can receive multiple FILE_ACTION_MODIFIED messages, it seems each time windows writes a large chunk of the file, you get the file modified notification for each time. I tried to use CreateFile API to check if the file can open by AP or not, but sometime, some of files are always locked by other AP, for example, if you are opening the Outlook, the PST will update, but my AP can’t access it, we have to start Shadow Copy to open it. So my question is, how to know a file is finished copying?

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  • File I/O OS handling

    - by Albinoswordfish
    This isn't a direct coding question but more of a OS handling mechanism. I was reading somebody's previous question regarding C# and file handling. Apparently C# was throwing an exception regarding a file being locked when trying to access this. So my question is, does C# use an internal lock to handle file I/O between processes, or does the OS use some type of mutual exclusion for file I/O? From what I learned about operating systems, well at least unix, is that the OS doesn't implement any type of mutual exclusion for processes trying to access the same file.

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  • Crossdomain file edit

    - by Misiur
    Hi there. I need to know, where from is my script used (it's for sale, and i don't want any thiefs). I want to write on my server in file, IP of user, domain where from script has been runned, date, etc. I've tried fopen, fwrite, but is_file_writable returned that it isn't. File CHmods are 777, it parent catalog has too 777 chmods. Now i'm trying something like that: <?php $file = 'http://www.misiur.com/security/seal.txt'; $data = date("Y-m-d H:i:s"); $ip = $_SERVER['REMOTE_ADDR']; $svr = $_SERVER['SERVER_NAME']; $str = "[$data] Loaded by $ip at $svr\r\n"; $current = file_get_contents($file); $current .= $str; file_put_contents($file, $current); ?> However - nothing happens. What i've got to do?

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  • Increase file upload size limit in iis6

    - by JustFoo
    Is there any other place besides the metabase.xml file where the file upload size can be modified? I am currently running a staging server with IIS6 and it is setup to allow uploading of files up to 20mb. This works perfectly fine. I have a new production server where I am trying to setup this same available size limit. So I edited the metabase.xml file and set it to 20971520. Then I restarted IIS and that didn't work. So I then restarted the entire server, that also didn't work. I can upload files around 2mb so it is definitely allowing file sizes larger then the standard 200kb default size. I try uploading a 5mb file and my upload.aspx page completely crashes. Is it possible there is something else I need to configure? The production server is located on a server farm, could there be some limits set on there end? Thanks

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  • Rename image file on upload php

    - by blasteralfred
    Hi, I have a form which uploads and re sizes image. The html file file submits data to a php file. The script is as follows; Index.html <form action="resizer.php" method="post" enctype="multipart/form-data"> Image: <input type="file" name="file" /> <input type="submit" name="submit" value="upload" /> </form> Resizer.php <?php require_once('imageresizer.class.php'); $imagename = "myimagename"; //Path To Upload Directory $dirpath = "uploaded/"; //MAX WIDTH AND HEIGHT OF IMAGE $max_height = 100; $max_width = 100; //Create Image Control Object - Parameters(file name, file tmp name, file type, directory path) $resizer = new ImageResizer($_FILES['file']['name'],$_FILES['file']['tmp_name'],$dirpath); //RESIZE IMAGE - Parameteres(max height, max width) $resizer->resizeImage($max_height,$max_width); //Display Image $resizer->showResizedImage(); ?> imageresizer.class.php <?php class ImageResizer{ public $file_name; public $tmp_name; public $dir_path; //Set variables public function __construct($file_name,$tmp_name,$dir_path){ $this->file_name = $file_name; $this->tmp_name = $tmp_name; $this->dir_path = $dir_path; $this->getImageInfo(); $this->moveImage(); } //Move the uploaded image to the new directory and rename public function moveImage(){ if(!is_dir($this->dir_path)){ mkdir($this->dir_path,0777,true); } if(move_uploaded_file($this->tmp_name,$this->dir_path.'_'.$this->file_name)){ $this->setFileName($this->dir_path.'_'.$this->file_name); } } //Define the new filename public function setFileName($file_name){ $this->file_name = $file_name; return $this->file_name; } //Resize the image function with new max height and width public function resizeImage($max_height,$max_width){ $this->max_height = $max_height; $this->max_width = $max_width; if($this->height > $this->width){ $ratio = $this->height / $this->max_height; $new_height = $this->max_height; $new_width = ($this->width / $ratio); } elseif($this->height < $this->width){ $ratio = ($this->width / $this->max_width); $new_width = $this->max_width; $new_height = ($this->height / $ratio); } else{ $new_width = $this->max_width; $new_height = $this->max_height; } $thumb = imagecreatetruecolor($new_width, $new_height); switch($this->file_type){ case 1: $image = imagecreatefromgif($this->file_name); break; case 2: $image = imagecreatefromjpeg($this->file_name); break; case 3: $image = imagecreatefrompng($this->file_name); break; case 4: $image = imagecreatefromwbmp($this->file_name); } imagecopyresampled($thumb, $image, 0, 0, 0, 0, $new_width, $new_height, $this->width, $this->height); switch($this->file_type){ case 1: imagegif($thumb,$this->file_name); break; case 2: imagejpeg($thumb,$this->file_name,100); break; case 3: imagepng($thumb,$this->file_name,0); break; case 4: imagewbmp($thumb,$this->file_name); } imagedestroy($image); imagedestroy($thumb); } public function getImageInfo(){ list($width, $height, $type) = getimagesize($this->tmp_name); $this->width = $width; $this->height = $height; $this->file_type = $type; } public function showResizedImage(){ echo "<img src='".$this->file_name." />"; } public function onSuccess(){ header("location: index.php"); } } ?> Everything is working well. The image will be uploaded in it's original filename and extension with a "_" prefix. But i want to rename the image to "myimagename" on upload, which is a variable in "Resizer.php". How can i make this possible?? Thanks in advance :) blasteralfred

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  • Paint java GUI component to image file

    - by Simon
    Let's say I have JButton test = new JButton("Test Button"); and I want to draw the button into an image object and save it to a file. I tried this: BufferedImage b = new BufferedImage(500, 500, BufferedImage.TYPE_INT_ARGB); test.paint(b.createGraphics()); File output = new File("C:\\screenie.png"); try { ImageIO.write(b, "png", output); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } This code produced an empty 500x500 PNG-file. Does anyone know how I can draw the GUI component to an image file?

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  • Opening a file from a pack URI in WPF

    - by cptmorgan
    Hi All, I am looking to open a .csv file from the application pack to do some unit testing. So what I would really love is some analog to File.ReadAllText(string path) which is instead X.ReadAllText(Uri uri). I haven't as yet been able to find this. Does anyone know if it is possible to read text / bytes (don't mind which) from a file in the pack without compiling this file to disk first? Oh and btw, File.ReadAllText(@"pack://application:,,,/SpreadSheetEngine/Tests/Example.csv") didn't work for me.. Thanks in advance.. Gav

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  • mmap() for large file I/O?

    - by Boatzart
    I'm creating a utility in C++ to be run on Linux which can convert videos to a proprietary format. The video frames are very large (up to 16 megapixels), and we need to be able to seek directly to exact frame numbers, so our file format uses libz to compress each frame individually, and append the compressed data onto a file. Once all frames are finished being written, a journal which includes meta data for each frame (including their file offsets and sizes) is written to the end of the file. I'm currently using ifstream and ofstream to do the file i/o, but I am looking to optimize as much as possible. I've heard that mmap() can increase performance in a lot of cases, and I'm wondering if mine is one of them. Our files will be in the tens to hundreds of gigabytes, and although writing will always be done sequentially, random access reads should be done in constant time. Any thoughts as to whether I should investigate this further, and if so does anyone have any tips for things to look out for? Thanks!

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  • Attaching HTML file as email in VB 6.0

    - by Shax
    Hi, I am trying to attach an html file file to email using Visual Basic 6.0. when the cursor is comes on Open strFile For Binary Access Read As #hFile line it gives error "Error encoding file - Bad file name or number". Please all your help and support would be highly appreciated. Dim handleFile As Integer Dim strValue As String Dim lEventCtr As Long handleFile = FreeFile Open strFile For Binary Access Read As #handleFile Do While Not EOF(hFile) ' read & Base 64 encode a line of characters strValue = Input(57, #handleFile) SendCommand EncodeBase64String(strValue) & vbCrLf ' DoEvents (occasionally) lEventCtr = lEventCtr + 1 If lEventCtr Mod 50 = 0 Then DoEvents Loop Close #handleFile Exit Sub File_Error: Close #handleFile m_ErrorDesc = "Error encoding file - " & Err.Description Err.Raise Err.Number, Err.Source, m_ErrorDesc End Sub

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  • Specific 'boot file' definition

    - by Jazz
    Hey, I have been given a general explanation of how a computer boots up. However a very loose definition to the term 'boot file' was given. Could someone explain 'boot file' to me in a very simple but concise manner? I have read about the POST, the clearing of registers, BIOS in the CMOS, etc. What I understand is that the boot file is different to the boot program. the boot program gets the system ready to accept an OS while the boot file contains some of the parameters by which the system will operate. The boot program is stored on ROM and the boot file isnt? cheers, jazz

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  • Recognizing file - Python

    - by Francisco Aleixo
    Ok, so the title may trick you a bit, and I'm sorry for that but didn't find a better title. This question might be a bit hard to understand so I'll try my best. I have no idea how this works or if it is even possible but what I want to do is for example create a file type (lets imagine .test (in which a random file name would be random.test)). Now before I continue, its obviously easy to do this using for example: filename = "random.test" file = open(filename, 'w') file.write("some text here") But now what I would like to know is if it is possible to write the file .test so if I set it to open with a wxPython program, it recognizes it and for example opens up a Message Dialog automatically. I'm sorry if I'm being vague and in case you don't understand, let me know so I can try my best to explain you.

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  • How to code a batch file to copy and rename the most recently dated file?

    - by david.murtagh.keltie.com
    I'm trying to code a batch file to copy only the most recently dated file in a given folder to another directory on the local machine, and simultaneously rename it as it does. I've found a very similar question here http://stackoverflow.com/questions/97371/batch-script-to-copy-newest-file and have managed to cobble together the below code from other forums too, but have hit a brick wall as it only results in the batch file itself being copied to the destination folder. It doesn't matter to me where the batch file itself sits in order for this to run. The source folder is C:! BATCH and the destination folder is C:\DROP The code is below, apologies if this is a glaringly obvious answer but it's literally the first foray into coding batch files for me... Thanks! @echo off setLocal EnableDelayedExpansion pushd C:\! BATCH for /f "tokens=* delims= " %%G in ('dir/b/od') do (set newest=%%G) copy "!newest!" C:\DROP\ PAUSE

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  • Best way to choose a random file from a directory in a shell script

    - by jhs
    What is the best way to choose a random file from a directory in a shell script? Here is my solution in Bash but I would be very interested for a more portable (non-GNU) version for use on Unix proper. dir='some/directory' file=`/bin/ls -1 "$dir" | sort --random-sort | head -1` path=`readlink --canonicalize "$dir/$file"` # Converts to full path echo "The randomly-selected file is: $path" Anybody have any other ideas? Edit: lhunath makes a good point about parsing ls. I guess it comes down to whether you want to be portable or not. If you have the GNU findutils and coreutils then you can do: find "$dir" -maxdepth 1 -mindepth 1 -type f -print0 \ | sort --zero-terminated --random-sort \ | sed 's/\d000.*//g/' Whew, that was fun! Also it matches my question better since I said "random file". Honsetly though, these days it's hard to imagine a Unix system deployed out there having GNU installed but not Perl 5.

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  • Export GridView to TXT, then upload file to server

    Basically what I want to do is export an array (or GridView) to a file called "getpathin.dat". That is easy, but the method I am using downloads the file to my computer, which is what I don't want. I want to write an array to either a PRE-EXISTING file that is on the server OR create a new file on the server in a folder, and this new file will contain either the array or the gridview, which I have stored in the array. i'll be doing this in... Visual Studio 2008/SQL Server using C#

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  • How to use application config file in C#?

    - by badpanda
    I am trying to use a config file in my C# console application. I created the file within the project by going New -- Application Configuration File, and naming it myProjectName.config. My config file looks like this: <?xml version="1.0" encoding="utf-8" ?> <configuration> <appSettings> <add key="SSDirectory" value="D:\Documents and Settings\****\MyDocuments\****" /> </appSettings> </configuration> The code to access it looks like this: private FileValidateUtil() { sSDirFilePath = ConfigurationSettings.AppSettings["SSDirectory"]; if (sSDirFilePath == null) Console.WriteLine("config file not reading in."); } Can anyone lend a hint as to why this is not working? (I am getting the error message.) Thanks!! badPanda

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  • Need to get the uploaded file to my local PC

    - by Suhail
    Hi, I have created a test form which will ask users to enter a name and upload the image file: <html lang="en"> <head> <title>Testing image upload</title> </head> <body> <form action="/services/upload" method="POST" enctype="multipart/form-data"> File Description: <input name='fdesc' type='text'><br> File name: <input type="file" name="fname"><br> <div><input type="submit"></div> </form> </body> </html> i need to get the file uploaded by the user and store it on my local PC. can this be done in python ? please let me know.

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