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  • How to configure sudoers with path wildcards?

    - by C. Lee
    I need sudo for a command for any path under a particular area. Example: sudo mycommand /opt/apps/myapp/... What is the sudoers syntax to allow this command to run in any path that falls under /opt/apps/myapp? This is Solaris 10 sudo. Thank you for your reply, but I don't need wildcards for the path to the commands, but wildcards for the arguments for the commands. For example, we want to do something like... sudo mycmd /opt/userarea/area1 sudo mycmd /opt/userarea/area1/area2 sudo mycmd /opt/userarea/area1/area2/area3 So far, using wildcards for the arguments in sudoers look like this: /opt/userarea/* /opt/userarea/*/* And it seems like if we want to have N levels of directories, then we need N lines in sudoers! Is there a better way to include all N levels in one line in sudoers? Thanks.

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  • How do I use a URL path instead of a file path in an Open File dialog in Mac OSX or ChromiumOS?

    - by Chris
    In Windows 7 (and perhaps earlier), the default "Open File" dialog box allows you to type a full URL into the "File name" section as if it were a file path, e.g. "http://www.example.com/pic.gif" instead of "C:/windows/pictures/pic.gif". When uploading a file to a website on the client side - say, an image - this allows the client to upload a picture located on a server accessible via the URL instead of downloading the image, saving it locally, then referencing the local image in the "Open File" dialog. It's a great option for Windows users. I have three separate questions: What is this procedure formally called? How do I describe this succinctly so that my searches for more information are fruitful? Can something similar be done in Mac OSX, Chromium OS, or a Linux environment? If so, how? Thanks!

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  • Is there such a thing as a file hosted container which deduplicates data held within?

    - by Mallow
    Background I have backups of a website which stores all of it's data into a single file. This file is several gigs large and I have many different backups of this file. Most of the data within is mostly the same plus whatever was added or changed to it. I want to keep all the concurrent backups I've made through the years in case I find a horrible surprise of data corruption along the line. However storing a 10gig file every month gets expensive. Seeking Solution I've often thought about different ways of alleviating this problem. One thought that comes up very often combines the idea of a duplicating file system which doesn't require it's own partitioned volume on a hard drive. Something like what truecrypt does, what it calls, "file hosted containers" which when using the truecrypt program allows you to mount and dismount that volume as a regular hard drive. Question Is there a virtual hard drive mounter which uses file-based container which uses data deduplicaiton file system? (This question is a little awkward to put into words, if you have a better idea on how to ask this question please feel free to help out.)

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  • Change windows default path or last used path for saving files? Is that possible?

    - by Mark
    I'm trying to change the default saving path to my specfied path for example if the file is picture (.jpg .gif etc..) and using google chrome I save an image the path can be anything from default or last used so I want it to be mine specfied for example I choose "C:\SaveHerePics" I don't know how you can change it maybe Registry? Dll? I would like to create vb.net program with textbox you choose the new location to be located every time for example "C:\SaveHerePics" and it should not have anything to do with this program even if you close the program the default location for pictures saved via any program or source saved to "C:\SaveHerePics" Thanks a lot for your help!

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  • How to Upload a file from client to server using OFBIZ?

    - by SIVAKUMAR.J
    Hi all, Im new to ofbiz.So is my question is have any mistake forgive me for my mistakes.Im new to ofbiz so i did not know some terminologies in ofbiz.Sometimes my question is not clear because of lack of knowledge in ofbiz.So try to understand my question and give me a good solution with respect to my level.Because some solutions are in very high level cannot able to understand for me.So please give the solution with good examples. My problem is i created a project inside the ofbiz/hot-deploy folder namely "productionmgntSystem".Inside the folder "ofbiz\hot-deploy\productionmgntSystem\webapp\productionmgntSystem" i created a .ftl file namely "app_details_1.ftl" .The following are the coding of this file <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1"> <title>Insert title here</title> <script TYPE="TEXT/JAVASCRIPT" language=""JAVASCRIPT"> function uploadFile() { //alert("Before calling upload.jsp"); window.location='<@ofbizUrl>testing_service1</@ofbizUrl>' } </script> </head> <!-- <form action="<@ofbizUrl>testing_service1</@ofbizUrl>" enctype="multipart/form-data" name="app_details_frm"> --> <form action="<@ofbizUrl>logout1</@ofbizUrl>" enctype="multipart/form-data" name="app_details_frm"> <center style="height: 299px; "> <table border="0" style="height: 177px; width: 788px"> <tr style="height: 115px; "> <td style="width: 103px; "> <td style="width: 413px; "><h1>APPLICATION DETAILS</h1> <td style="width: 55px; "> </tr> <tr> <td style="width: 125px; ">Application name : </td> <td> <input name="app_name_txt" id="txt_1" value=" " /> </td> </tr> <tr> <td style="width: 125px; ">Excell sheet &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;: </td> <td> <input type="file" name="filename"/> </td> </tr> <tr> <td> <!-- <input type="button" name="logout1_cmd" value="Logout" onclick="logout1()"/> --> <input type="submit" name="logout_cmd" value="logout"/> </td> <td> <!-- <input type="submit" name="upload_cmd" value="Submit" /> --> <input type="button" name="upload1_cmd" value="Upload" onclick="uploadFile()"/> </td> </tr> </table> </center> </form> </html> the following coding is present in the file "ofbiz\hot-deploy\productionmgntSystem\webapp\productionmgntSystem\WEB-INF\controller.xml" ...... ....... ........ <request-map uri="testing_service1"> <security https="true" auth="true"/> <event type="java" path="org.ofbiz.productionmgntSystem.web_app_req.WebServices1" invoke="testingService"/> <response name="ok" type="view" value="ok_view"/> <response name="exception" type="view" value="exception_view"/> </request-map> .......... ............ .......... <view-map name="ok_view" type="ftl" page="ok_view.ftl"/> <view-map name="exception_view" type="ftl" page="exception_view.ftl"/> ................ ............. ............. The following are the coding present in the file "ofbiz\hot-deploy\productionmgntSystem\src\org\ofbiz\productionmgntSystem\web_app_req\WebServices1.java" package org.ofbiz.productionmgntSystem.web_app_req; import javax.servlet.http.HttpServletRequest; import javax.servlet.http.HttpServletResponse; import java.io.DataInputStream; import java.io.FileOutputStream; import java.io.IOException; public class WebServices1 { public static String testingService(HttpServletRequest request, HttpServletResponse response) { //int i=0; String result="ok"; System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- Start"); String contentType=request.getContentType(); System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- contentType : "+contentType); String str=new String(); // response.setContentType("text/html"); //PrintWriter writer; if ((contentType != null) && (contentType.indexOf("multipart/form-data") >= 0)) { System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) after if (contentType != null)"); try { // writer=response.getWriter(); System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - try Start"); DataInputStream in = new DataInputStream(request.getInputStream()); int formDataLength = request.getContentLength(); byte dataBytes[] = new byte[formDataLength]; int byteRead = 0; int totalBytesRead = 0; //this loop converting the uploaded file into byte code while (totalBytesRead < formDataLength) { byteRead = in.read(dataBytes, totalBytesRead,formDataLength); totalBytesRead += byteRead; } String file = new String(dataBytes); //for saving the file name String saveFile = file.substring(file.indexOf("filename=\"") + 10); saveFile = saveFile.substring(0, saveFile.indexOf("\n")); saveFile = saveFile.substring(saveFile.lastIndexOf("\\")+ 1,saveFile.indexOf("\"")); int lastIndex = contentType.lastIndexOf("="); String boundary = contentType.substring(lastIndex + 1,contentType.length()); int pos; //extracting the index of file pos = file.indexOf("filename=\""); pos = file.indexOf("\n", pos) + 1; pos = file.indexOf("\n", pos) + 1; pos = file.indexOf("\n", pos) + 1; int boundaryLocation = file.indexOf(boundary, pos) - 4; int startPos = ((file.substring(0, pos)).getBytes()).length; int endPos = ((file.substring(0, boundaryLocation)).getBytes()).length; //creating a new file with the same name and writing the content in new file FileOutputStream fileOut = new FileOutputStream("/"+saveFile); fileOut.write(dataBytes, startPos, (endPos - startPos)); fileOut.flush(); fileOut.close(); System.out.println("\n\n\t**********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - try End"); } catch(IOException ioe) { System.out.println("\n\n\t*********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - Catch IOException"); //ioe.printStackTrace(); return("exception"); } catch(Exception ex) { System.out.println("\n\n\t*********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) - Catch Exception"); return("exception"); } } else { System.out.println("\n\n\t********************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response) else part"); result="exception"; } System.out.println("\n\n\t*************************************\n\tInside WebServices1.testingService(HttpServletRequest request, HttpServletResponse response)- End"); return(result); } } I want to upload a file to the server.The file is get from user "<input type="file"..> tag in the "app_details_1.ftl" file & it is updated into the server by using the method "testingService(HttpServletRequest request, HttpServletResponse response)" in the class "WebServices1".But the file is not uploaded. Give me a good solution for uploading a file to the server. Thanks & Regards, Sivakumar.J

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  • Java compiler error: "cannot find symbol" when trying to access local variable

    - by HH
    $ javac GetAllDirs.java GetAllDirs.java:16: cannot find symbol symbol : variable checkFile location: class GetAllDirs System.out.println(checkFile.getName()); ^ 1 error $ cat GetAllDirs.java import java.util.*; import java.io.*; public class GetAllDirs { public void getAllDirs(File file) { if(file.isDirectory()){ System.out.println(file.getName()); File checkFile = new File(file.getCanonicalPath()); }else if(file.isFile()){ System.out.println(file.getName()); File checkFile = new File(file.getParent()); }else{ // checkFile should get Initialized at least HERE! File checkFile = file; } System.out.println(file.getName()); // WHY ERROR HERE: checkfile not found System.out.println(checkFile.getName()); } public static void main(String[] args) { GetAllDirs dirs = new GetAllDirs(); File current = new File("."); dirs.getAllDirs(current); } }

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  • Visual Studio 2008 project file does not load because of an unexpected encoding change.

    - by Xenan
    In our team we have a database project in visual Studio 2008 which is under source control by Team Foundation Server. Every two weeks or so, after one co-worker checks in, the project file won't load on the other developers machines. The error message is: The project file could not be loaded. Data at the root level is invalid. Line 1, position 1. When I look at the project file in Notepad++, the file looks like this: ??<NUL?NULxNULmNULlNUL NULvNULeNULrNULsNULiNULoNULnNUL ... and so on (you can see <?xml version in this) whereas an normal project file looks like: <?xml version="1.0" encoding="utf-16"?> ... So probably something is wrong with the encoding of the file. This is a problem for us because it turns out to be impossible to get the file encoding correct again. The 'solution' is to throw away the project file an get the last know working version from source control. According to the file, the encoding should be UTF-16. According to Notepad++, the corrupted file is actually UTF-8. My questions are: Why is Visual Studio messing up the encoding of the project file, apparently at random times and at random machines? What should we do to prevent this? When it has happened, is there a possibility to restore the current file in the correct encoding instead of pulling an older version from source control? As a last note: the problem is with one single project file, all other project files don't expose this problem.

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  • File.Move does not inherit permissions from target directory?

    - by Joseph Kingry
    In case something goes wrong in creating a file, I've been writing to a temporary file and then moving to the destination. Something like: var destination = @"C:\foo\bar.txt"; var tempFile = Path.GetTempFileName(); using (var stream = File.OpenWrite(tempFile)) { // write to file here here } string backupFile = null; try { var dir = Path.GetDirectoryName(destination); if (!Directory.Exists(dir)) { Directory.CreateDirectory(dir); Util.SetPermissions(dir); } if (File.Exists(destination)) { backupFile = Path.Combine(Path.GetTempPath(), new Guid().ToString()); File.Move(destination, backupFile); } File.Move(tempFile, destination); if (backupFile != null) { File.Delete(backupFile); } } catch(IOException) { if(backupFile != null && !File.Exists(destination) && File.Exists(backupFile)) { File.Move(backupFile, destination); } } The problem is that the new "bar.txt" in this case does not inherit permissions from the "C:\foo" directory. Yet if I create a file via explorer/notepad etc directly in the "C:\foo" there's no issues, so I believe the permissions are correctly set on "C:\foo". Update Found Inherited permissions are not automatically updated when you move folders, maybe it applies to folders as well. Now looking for a way to force an update of file permissions. Is there a better way overall of doing this?

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  • Can't access my files in ASP.NET web site

    - by jumbojs
    I'm having a very difficult time. I am running windows 2008 server, I have an Able Commerce site using ASP.NET with C#. I'm writing an automated task that will ftp some xml files down into a local directory on our web server and then the program parses the xml file and saves information to our database. The problem, once I save the files to our local directory, my program has no access to the files. The NETWORK SERVICE user permissions isn't being inherited by the xml files so my program can't do anything with them. I can manually change the permissions, but this wouldn't be automated and won't work. How can I get this to work? help please, it's very frustrating.

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  • How to get file path using FileUpload to be read by FileStream?

    - by john ryan
    I have a Method that open excel file and read it through exceldatareaderClass that i have downloaded in codeplex by using filestream. Currently I just declared the exact directory where the filestream open an excel file.And it works fine. Stream stream = new FileStream("C:\\" + FileUpload.PostedFile.FileName, FileMode.Open, FileAccess.Read, FileShare.Read); Now i need to read the excel file wherever location the user place it like on windows forms fileupload.FileStream needs the exact location where the file is located. How to do this.? Example: Sample.xls is located on My Documents the file path should be like : C:\Documents and Settings\user\My Documents\ string openpath ="" ;//filepath Stream stream = new FileStream(openpath+ FileUpload.PostedFile.FileName, FileMode.Open, FileAccess.Read, FileShare.Read); Thanks in Regards

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  • XML File as Excel file.

    - by FrustratedWithFormsDesigner
    I have a number of reports that I run against my database that need to eventually go to the end-users as Excel spreadsheets. Initially, I was creating text reports, but the steps to convert the text to a spreadsheet were a bit cumbersome. There were too many steps to import text to the spreadsheet, and multi-line text rows were imported as individual rows in Excel (which was incorrect). Currently, I am generating simple XML saving the file with an ".xls" extension. This works better, but there is still the problem of Excel prompting the user with an XML import dialogue every time they open the file, and then having to save a new file if they add notes or change the layout to the file (which they almost certainly will be doing). Sample "xls" file: <?xml version="1.0" standalone="yes"?> <report_rows> <row> <NAME>Test Data</NAME> <COUNT>345</COUNT> </row> <!-- many more row elements... --> </report_rows> Is there any way to add markup to the file to hint to Excel how it should import and handle the file? Ideally, the end user should be able to open and save the file like any othe spreadsheet they create directly from Excel. Is this even possible? UPDATE: We are running Office 2003 here. UPDATE: The XML is generated from a sqlplus script, no option to use C#/.NET here.

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  • General question: Filesystem or database?

    - by poeschlorn
    Hey guys, i want to create a small document management system. there are several users who store their files. each file which is uploaded contains an info which user uploaded it and the document content itself. In a view there are displayed all files of ONE specific user, ordered by date. What would be better: 1) giving the documents a name or metadata(XML) which contain the date and user (and iterate through them to get the metadata) or 2) giving the files a random/unique name and store metadata in a DB? something like this: date | user | filename What would you say and why? The used programming language is java and the DB is MySQL.

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  • What would happen if a same file being read and appended at the same time(python programming)?

    - by Shane
    I'm writing a script using two separate thread one doing file reading operation and the other doing appending, both threads run fairly frequently. My question is, if one thread happens to read the file while the other is just in the middle of appending strings such as "This is a test" into this file, what would happen? I know if you are appending a smaller-than-buffer string, no matter how frequently you read the file in other threads, there would never be incomplete line such as "This i" appearing in your read file, I mean the os would either do: append "This is a test" - read info from the file; or: read info from the file - append "This is a test" to the file; and such would never happen: append "This i" - read info from the file - append "s a test". But if "This is a test" is big enough(assuming it's a bigger-than-buffer string), the os can't do appending job in one operation, so the appending job would be divided into two: first append "This i" to the file, then append "s a test", so in this kind of situation if I happen to read the file in the middle of the whole appending operation, would I get such result: append "This i" - read info from the file - append "s a test", which means I might read a file that includes an incomplete string?

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  • CouchDB crashes at startup when path to config file has space(s)

    - by Barry Wark
    I'm hoping to run CouchDB as a per-user Launch Agent on OS X. I'm using the coucdbx-core folder from the CouchDB Server.app as the base of my CouchDB deployment. I'd like each user to have their own couch instance (on a different port), necessitating separate config files for each instance. The logical place to put these files is in ~/Library/Application Support/ for each user. I can put the entire distribution in ~/Library/Application Support/my-app/coucdbx, and put the .ini at ~/Library/Application Support/my-app/local.ini. Starting couchdb as bin/couchdb -a ../local.ini (from ~/Library/Application Support/my-app/coucdbx) works great. But I'd like to save every user the ~50MB couchdbx and install the couchdbx-core in a shared location (e.g. within my app's .app bundle). When I do this, the path to the per-user config file contains a space, and I get the following error when starting CouchDB: $ bin/couchdb -n -a ~/Library/Application\ Support/us.physion.ovation/default.ini {"init terminating in do_boot",{{badmatch,{error,{bad_return,{{couch_app,start,[normal,["/Users/hs/prj/build-couchdb/build/etc/couchdb/default.ini","/Users/hs/prj/build-couchdb/build/etc/couchdb/local.ini"]]},{'EXIT',{{badmatch,{error,{error,enoent}}},[{couch_server_sup,start_server,1,[{file,"/Users/hs/prj/build-couchdb/dependencies/couchdb/src/couchdb/couch_server_sup.erl"},{line,56}]},{application_master,start_it_old,4,[{file,"application_master.erl"},{line,274}]}]}}}}}},[{couch,start,0,[{file,"/Users/hs/prj/build-couchdb/dependencies/couchdb/src/couchdb/couch.erl"},{line,18}]},{init,start_it,1,[]},{init,start_em,1,[]}]}} Is there any way to provide a config file at the command line, if that config file's path includes space(s)? Despite my best efforts in the mailing list archives, wiki and google, I haven't been able to find a solution or a definitive "it can't work". Any help greatly appreciated.

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  • htaccess rewrite rules in Nginx: setting the rewrite path

    - by ct2k7
    I have a htaccess file I'm trying to convert into an nignx config file. Here's my htaccess file. RewriteEngine on RewriteCond %{SCRIPT_FILENAME} !-f RewriteCond %{SCRIPT_FILENAME} !-d RewriteRule !\.(jpg|css|js|gif|png)$ public/ [L] RewriteRule !\.(jpg|css|js|gif|png)$ public/index.php?url=$1 And the rules I have in my nginx config file: location / { if ($request_uri !~ "-f"){ rewrite !\.(jpg|css|js|gif|png)$ public/ break; } rewrite !\.(jpg|css|js|gif|png)$ public/index.php?url=$1; } # pass the PHP scripts to FastCGI server listening on 127.0.0.1:9000 location ~ \.php$ { # Move to the @missing part when the file doesn't exist try_files $uri @missing; # Fix for server variables that behave differently under nginx/$ fastcgi_split_path_info ^(.+\.php)(/.+)$; # Include the standard fastcgi_params file included with ngingx include fastcgi_params; fastcgi_param PATH_INFO $fastcgi_path_info; fastcgi_index index.php; # Pass to upstream PHP-FPM; This must match whater you name you$ #fastcgi_pass phpfpm; fastcgi_pass 127.0.0.1:9000; } location @missing { rewrite ^(.*)$ public/index.php?url=$1 break; } However, when I hit /, I get a 403 Forbidden, but I can get to /public/index.php, thus the rewrite isn't working. Any ideas on what I'm doing wrong?

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  • Efficient way to check for changes to the contents of folders

    - by MrVimes
    I am creating an application that maintains a database of files of a certain type in a given folder (and all subfolders) Initially the program will recurse the folders and add any file it finds of that type to the database. I want the application to have the ability to re-scan the folder and add any files that were not there the last time the folders were scanned. It can't use the date created property of the file because there is a high chance of a file being added to the folders that isn't a new file. I am wondering what the most efficient way of doing this is, and if there is a way that doesn't involve checking each file is in the database already (which, if there are 5000 files would mean 5000 queries of a list 5000 items in size, or 25 million 'checks' for the sql engine to perform) I suppose a more specific question to acheive the same goal would be - is there a property of a file (in Microsoft Windows) that will reliably tell you when that file arrived in that folder.

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  • Best way to rename existing unique field names in database?

    - by Rajdeep Siddhapura
    I have a database table that contains id, filename, userId id is unique identifier filename should also be unique table may contain 10000 records When a user uploads a file it should be entered in database with given rules: If there is no record with same filename, it should be added as it is (Ex. foobar.pdf) If there is record with same filename, it should be added as uploadedName(2).ext (foobar(2).pdf) If there are n records with same base filename (foobar), it should be added as uploadedName(n+1).ext (foobar(20).pdf) Now if foobar(2).pdf is uploaded, it should be added as foobar(2)(2).pdf & so on This pattern needs to be followed because the file is already being uploaded at client side using ajax before sending the details to server and the file hosting service follows the above rules to name the files. My solution: maintain a file that contains all the names and the number of times it has occurred. if a filename that exists in file is entered, increase occurrence count and new name is generated, else add to it to file if the new name generated is in database, add it to file and generate new name

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  • Relative path reference in WebConfig.ConnectionString

    - by Ngu Soon Hui
    Is it possible to specify a relative path reference in connectionstring, attachDbFileName property in a web.config? For example, In my database is located in the App_data folder, I can easily specify the AttachDBFilename as|DataDirectory|\mydb.mdf and the |Datadirectory| will automatically resolve to the correct path. Now, suppose that web.config file is located in A folder, but the database is located in B\App_data folder, where A and B folder is located in the same folder. Is there anyway to use relative path reference to resolve to the correct path?

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  • Do we need seperate file path for window and linux in java

    - by Kishor Sharma
    I have a file on linux ubuntu server hosted with path name /home/kishor/project/detail/. When I made a web app in window to upload and download file from specified location i used path "c:\kishor\projects\detail\" for saving in window. For my surprise when i used window file path name in my server i am still able to get files and upload them, i.e, "c:\kishor\projects\detail\". Can anyone explain why it is working (as window and linux both use different file path pattern).

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  • use absolute or relative path?

    - by ajsie
    in my config.php where i have all constants i set the PATH to a absolute path. but this means that when i move my application folder i have to change this path. i wondered if its better to set a relative path, in that way whenever i move my application between production and development folder, i dont have to change it. how do you guys do when you move between folders?

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  • sys.path() and PYTHONPATH issues

    - by Justin
    I've been learning Python, I'm working in 2.7.3, and I'm trying to understand import statements. The documentation says that when you attempt to import a module, the interpreter will first search for one of the built-in modules. What is meant by a built-in module? Then, the documentation says that the interpreter searches in the directories listed by sys.path, and that sys.path is initialized from these sources: the directory containing the input script (or the current directory). PYTHONPATH (a list of directory names, with the same syntax as the shell variable PATH). the installation-dependent default. Here is a sample output of a sys.path command from my computer using python in command-line mode: (I deleted a few so that it wouldn't be huge) ['', '/usr/lib/python2.7', '/usr/lib/python2.7/lib-old', '/usr/lib/python2.7/lib-dynload', '/usr/local/lib/python2.7/dist-packages', '/usr/lib/python2.7/dist-packages', '/usr/lib/python2.7/dist-packages/PIL', '/usr/lib/python2.7/dist-packages/gst-0.10', '/usr/lib/python2.7/dist-packages/gtk-2.0', '/usr/lib/pymodules/python2.7', '/usr/lib/python2.7/dist-packages/ubuntuone-couch', '/usr/lib/python2.7/dist-packages/ubuntuone-storage-protocol'] Now, I'm assuming that the '' path refers to the directory containing the 'script', and so I figured the rest of them would be coming from my PYTHONPATH environmental variable. However, when I go to the terminal and type env, PYTHONPATH doesn't exist as an environmental variable. I also tried import os then os.environ, but I get the same output. Do I really not have a PYTHONPATH environmental variable? I don't believe I ever specifically defined a PYTHONPATH environmental variable, but I assumed that when I installed new packages they automatically altered that environment variable. If I don't have a PYTHONPATH, how is my sys.path getting populated? If I download new packages, how does Python know where to look for them if I don't have this PYTHONPATH variable? How do environment variables work? From what I understand, environment variables are specific to the process for which they are set, however, if I open multiple terminal windows and run env, they all display a number of identical variables, for example, PATH. I know there file locations for persistent environment variables, for example /etc/environment, which contains my PATH variable. Is it possible to tell where a persistent environment variable is stored? What is the recommended location for storing new persistent environment variables? How do environment variables actually work with say, the Python interpreter? The Python interpreter looks for PYTHONPATH, but how does it work at the nitty-gritty level?

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  • mysql not in my PATH for some reason

    - by Dr.Dredel
    I've installed mysql on several macs and on one of them mysql is not in the path. If I export it it shows up in the path correctly, but upon reboot, disappears. What should I do to get the machine to keep it in the path and what are the machines that DO have it in their path doing differently? Any thoughts appreciated.

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  • how to find file's path according to filename?

    - by Phsika
    how to find file's path according to filename? i need to find file's path according to file name sample code: string path= System.IO.Directory.GetDirectories(@"c:\", "kategori",System.IO.SearchOption.AllDirectories).First(); But i need : string path= System.IO.Directory.GetDirectories(@"anyFolder", "kategori",System.IO.SearchOption.AllDirectories).First();

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  • Visual C# 2008 Control to set a path.

    - by Eyla
    Greetings, What control in Visual C# 2008 would allow me to set a path and get the value of that path. For example: I want the user to click a button then select a path where he/she would do the operation such as save a file in selected path.

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  • vb6 Open File For Append issue Path Not Found

    - by Schwayday
    Open App.Path & "\Folder\" & str(0) For Output Seems to get a path not found however if directly before that I do MsgBox App.Path & "\Folder\" & str(0) It Provides the correct directory/filename that I want and if I replace that string with the direct path in quotes it works fine however that won't be very good for other users of my app :( Anyone know why this doesn't work?

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