Search Results

Search found 92226 results on 3690 pages for 'file access'.

Page 377/3690 | < Previous Page | 373 374 375 376 377 378 379 380 381 382 383 384  | Next Page >

  • I can't delete a file - even when using unlocker

    - by chipperyman573
    So I have a 64 bit windows 7 ultimate computer. On my desktop, I have a .iso file that should NOT be used by anything. However, when I use lockhunter (unlocker doesn't work for me), it says it can't unlock or delete it. When I check what it's being used by, it says by the system. Here's a picture when I click delete: Here's a picture of when I click unlock it: When I click Other Close locking process: I've had this file on my computer for a LONG time now and haven't been able to delete it. I've also tried eraser erase on startup, but that doesn't work, either. How can I remove this file?

    Read the article

  • Detect whether the loading image is taken from camera directly, when using smartphones

    - by Eitan
    I am using html, tag: <input type = "file" /> On android and on many cellulars I have the ability to get the file directly by taking a picture and save it. How can I know (by javascript code) how did I get the picture (direcly by the camera, or by some files that on my cellular)? I did some workarround, and found exif (http://www.nihilogic.dk/labs/exif/exif.js), but I didn't succeed using it, either I don't know whether exif is the right solution. Thanks :)

    Read the article

  • Is there a java api to access bugzilla?

    - by Mauli
    Is there a (standalone!) java api which wraps the XML-RPC interface to bugzilla? I don't want to program my own api for it, and I can't actually find a library which does this (and only this). Update: I'm looking for something like this http://oss.dbc.dk/bugzproxy/ only written in Java

    Read the article

  • Exception when access inner class reflectively

    - by MikeJiang
    Hi Folk, Here is a sample java program. I wonder why the two approaches reslut different stories. Is it a bug or kind of bitter java feature? And I run the sample upon java 1.5 package test; public class TestOut{ public static void main(String[] args){ new TestIn();//it works Class.forName("test.TestOut$TestIn").newInstance();// throw IllegalAccessException } private static class TestIn{} }

    Read the article

  • Accessing Password Protected Network Drives in Windows in C#?

    - by tkeE2036
    Hi Everyone, So in C# I am trying to access a file on a network, for example at "//applications/myapp/test.txt", as follows: const string fileLocation = @"//applications/myapp/test.txt"; using (StreamReader fin = new StreamReader(FileLocation)) { while(!fin.EndOfStream()){ //Do some cool stuff with file } } However I get the following error: System.IO.IOException : Logon failure: unknown user name or bad password. I figure its because I need to supply some network credentials but I'm not sure how to get those to work in this situation. Does anyone know the best way (or any way) to gain access to these files that are on a a password protected location? Thanks in advance!!

    Read the article

  • how to access database through javascript?

    - by nectar
    I am creating one admin page where I have multiple textboxes.when I enter the userid in one textbox I want to display user name in next textbox when admin moves to next text box.for this I can use ajax or javascript? which one will be better?how can I do it through javascript.

    Read the article

  • Save output of a php file in a html file..

    - by piemesons
    I am having a php file which executes some code to generate a html file.Its like I m having a form from which some data will be posted to x.php file, which gives a output(a webpage). I want to save that output in a html file.What the most efficient way of doing this.?

    Read the article

  • deny direct access to a php file by typing the link in the url

    - by aeonsleo
    hi, I am using php session for a basic login without encryption for my site. I want to prevent a user from directly accessing a php page by typing the url when he/she is not signed in. But this is not happening. I am using session_start(), initializing session variables and aslo unsetting and destroying sesssion during logout. Also if I type the link in a different browser the page is getting displayed. I am not very well versed with php , only a beginner. I googled for such problem and found few alternatives as keeping all files in a seperate folder from the web root, using .htaccess etc. Can someone explain in simple terms what could be a good solution.thanks in advance.

    Read the article

  • Only allow the POST method for a specific file in a directory

    - by Dave Chen
    I have one file that should only be accessible via the POST method. /var/www/folder/index.php The document root is /var/www/ and index.php is nested inside a folder. My configurations are as follows: <Directory "/var/www/folder"> <Files "index.php"> order deny,allow Allow from all <LimitExcept POST> Deny from all </LimitExcept> </Files> </Directory> I visit my server at 127.0.0.1/folder but I can GET and POST the file just like normal. I've also tried reversing the order, order allow,deny, require, limitexcept and limit. How can I only allow POST requests to be processed by one file in a folder?

    Read the article

  • How to check whether given file is in PROPER word file format?

    - by shekhar
    Hi, I am developing one application using C# for processing MSWord files. My application gets hang when I pass invalid .doc file as an input. For example, if I have one foo.pdf file and I pass it to my application after changing its extension (foo.doc). Is it possible to check whether file is valid doc file before trying to open it? Please enlighten !!!! Thanks in advance

    Read the article

  • Downloading file from FTP using cURL

    - by Josiah
    I'm trying to use a cURL command to download a file from an FTP server to a local drive on my computer. I've tried curl "ftp://myftpsite" --user name:password -Q "CWD /users/myfolder/" -O "myfile.raw" But it returns an error that says: curl: Remote file name has no length! curl: try 'curl --help' or 'curl --manual' for more information curl: (6) Could not resolve host: myfile.raw; No data record of requested type I've tried some other methods, but nothing seems to work. Also, I'm not quite sure how to specify which folder I want the file to be downloaded to. How would I do that?

    Read the article

  • Access varialbe from code behind via jQery

    - by Morron
    Hi, I have the following code that I want to return to a variable "t" in jQery: Code behind: Public Shared Function GetSomeText() As String Dim result = "This is from code behind" Return result End Function Caller variable in jQuery: //This is not working like that, I think var t = GetSomeText(); So, how can I make variable "t" get the "result" from Function GetSomeText from code-behind? Thank you.

    Read the article

  • WSDL file generated on the fly?

    - by Valer
    Hi everybody, for a web service, depending on a XML file, there are a couple af classes in C# generated. Depending on these classes, there is at compile time then the WSDL file generated. Is there a possibility at runtime to simply replace the XML file and to have the WSDL file generated on the fly? Best regards, Valer

    Read the article

  • Efficient way to maintain a sorted list of access counts in Python

    - by David
    Let's say I have a list of objects. (All together now: "I have a list of objects.") In the web application I'm writing, each time a request comes in, I pick out up to one of these objects according to unspecified criteria and use it to handle the request. Basically like this: def handle_request(req): for h in handlers: if h.handles(req): return h return None Assuming the order of the objects in the list is unimportant, I can cut down on unnecessary iterations by keeping the list sorted such that the most frequently used (or perhaps most recently used) objects are at the front. I know this isn't something to be concerned about - it'll make only a miniscule, undetectable difference in the app's execution time - but debugging the rest of the code is driving me crazy and I need a distraction :) so I'm asking out of curiosity: what is the most efficient way to maintain the list in sorted order, descending, by the number of times each handler is chosen? The obvious solution is to make handlers a list of (count, handler) pairs, and each time a handler is chosen, increment the count and resort the list. def handle_request(req): for h in handlers[:]: if h[1].handles(req): h[0] += 1 handlers.sort(reverse=True) return h[1] return None But since there's only ever going to be at most one element out of order, and I know which one it is, it seems like some sort of optimization should be possible. Is there something in the standard library, perhaps, that is especially well-suited to this task? Or some other data structure? (Even if it's not implemented in Python) Or should/could I be doing something completely different?

    Read the article

  • Android Access inbox

    - by rahul
    I want to show few sms available in inbox in my application. And how to identify or to distinguish to sms from others? Is there any message ID or something to identify sms?

    Read the article

  • Is it better to check if file exists before deleting it

    - by Kevin Fegan
    Sometimes when I want to delete a file (from within a script), I will just delete it rather than checking if it exists first. So I do this: $ rm "temp.txt" 2>/dev/null Instead of this: [ -f "temp.txt" ] && rm "temp.txt" I just feel it's a waste of time to go and check if the file exists and return an exit code. So, perhaps it's quicker to do it the first way, especially if most of the time, the file is likely to be present. Are there any other advantages (or downsides) to do it one way or the other? Am I wrong to think it will ever be quicker?

    Read the article

  • iPhone Core Data - Access deep attributes with to many relationships

    - by ncohen
    Hi everyone, Let say I have an entity user which has a one to many relationship with the entity menu which has a one to many relationship with the entity meal which has a many to one relationship with the entity recipe which has a one to many relationship with the entity element. What I would like to do is to select the elements which belong to a particular user (username = myUsername) and particular menu*s* (minDate < menu.date < maxDate). Does anyone have an idea how to get them? Thanks

    Read the article

  • friend declaration block an external function access to the private section of a class

    - by MiP
    I'm trying to force function caller from a specific class. For example this code bellow demonstrate my problem. I want to make 'use' function would be called only from class A. I'm using a global namespace all over the project. a.h #include "b.h" namespace GLOBAL{ class A{ public: void doSomething(B); } } a.cpp #include "a.h" using namespace GLOBAL; void A::doSomething(B b){ b.use(); } b.h namespace GLOBAL{ class B{ public: friend void GLOBAL::A::doSomething(B); private: void use(); } Compiler says: ‘GLOBAL::A’ has not been declared ‘void GLOBAL::B::use()’ is private Can anyone help here ? Thanks a lot, Mike.

    Read the article

  • How to count the length (number of lines) of a csv file in Rails?

    - by Mathias
    Hello, I have a form (Rails) which allows me to load a .csv file using the file_field. In the view: <% form_for(:upcsv, :html => {:multipart => true}) do |f| %> <table> <tr> <td><%= f.label("File:") %></td> <td><%= f.file_field(:filename) %></td> </tr> </table> <%= f.submit("Submit") %> <% end %> Clicking Submit redirects me to another page (create.html.erb). The file was loaded fine, and I was able to read the contents just fine in this second page. I am trying to show the number of lines in the .csv file in this second page. My controller (semi-pseudocode): class UpcsvController < ApplicationController def index end def create file = params[:upcsv][:filename] ... #params[:upcsv][:file_length] = file.length # Show number of lines in the file #params[:upcsv][:file_length] = file.size ... end end Both file.length and file.size returns '91' when my file only contains 7 lines. From the Rails documentation that I read, once the Submit button is clicked, Rails creates a temp file of the uploaded file, and the params[:upcsv][:filename] contains the contents of the temp/uploaded file and not the path to the file. And I don't know how to extract the number of lines in my original file. What is the correct way to get the number of lines in the file? My create.html.erb: <table> <tr> <td>File length:</td> <td><%= params[:upcsv][:file_length] %></td> </tr> </table> I'm really new at Rails (just started last week), so please bear with my stupid questions. Thank you!

    Read the article

  • Random access view in boost::multi_array

    - by linai
    Here is a boost example: typedef boost::multi_array<double, 1> array_type; typedef array_type::index index; array_type A(boost::extents[100]); for(index i = 0; i != A.size(); ++i) { A[i] = (double)i; } // creating view array_type::index_gen indices; typedef boost::multi_array_types::index_range range; array_type::array_view<1>::type myview = A[ indices[range(0,50)] ]; What this code does is creating a subarray or view mapping onto the original array. This view is continuous and covers from 0th to 50th elements of an original array. What if I need to explicitly define elements I'd like to see in the view? How can I create a view with indices like [1, 5, 35, 23] ? Any ideas?

    Read the article

< Previous Page | 373 374 375 376 377 378 379 380 381 382 383 384  | Next Page >