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PHP how to list all files in directory in a combobox - only file names and not with full paths
$files = glob('images/items/Done/*.jpg'); echo "<select>"; foreach ($files as $file) { echo "<option>".pathinfo($file, PATHINFO_BASENAME)."</option>"; } echo "</select>";

Categories : PHP

finding most recent file version from list of file path names with jumbled file names
you can try: find $WORK.../.history -type f -printf '%T@ %p ' | sort -nr | cut -f2- | xargs grep 'your_pattern' Decomposed: the find finds all plain files and prints their modification time and path the sort sort sort them numerically - and reverse, so highest number comes first (the latest modified) the cut removes the time from each line the xargs run its argument for each file what get to it input, in this case will run the grep command, so the 1st file what the grep find - was the lastest modified The above not works when the filenames containing spaces, but hopefully this is not your case... The -printf works only with GNU find. For the repetative work, you can split the command to two parts: find $WORK.../.history -type f -printf '%T@ %p ' | sort -nr | cut -f2- > /somewhe

Categories : Eclipse

How to remove specific directory names from path list
You can do the following (treating the path as a string): 'http://my-cdn-path.com/' + '/'.join(i.split('/')[-4:])) Or, using os.path: 'http://my-cdn-path.com/' + '/'.join(os.path.abspath(i).split(os.sep)[-4:])

Categories : Python

How to get full directory path of the file on browse in Firefox and IE using javascript?
You do not need full file path from the client browser since it will have nothing to do with the uploaded destination on the server. Many browsers restrict full local path availability due to security concerns. IE will show full path if you add site to "Trusted Sites", but again there is no need for that.

Categories : Javascript

Function for getting full directory structure
You don't have to check. os.walk always returns a tuple in a form: (current_directory, directory_list, files_list) So you always know that the second element of the tuple is a list of directories found in the current_directory, and that the third element is a list of files found in the current_directory. You should use os.walk in that manner: for current, directories, files in os.walk(path): # do your stuff...

Categories : Python

get filenames from a directory without full path
Use Path.GetFileName which returns file name and extension from full path var files = Directory.GetFiles(directory).Select(f => Path.GetFileName(f)); Or even var files = Directory.GetFiles(directory).Select(Path.GetFileName);

Categories : C#

Find the root directory without full path in C#
If the subfolder is under the same root as you application you can use Directory.GetDirectoryRoot See: http://msdn.microsoft.com/en-us/library/system.io.directory.getdirectoryroot.aspx

Categories : C#

Getting just the current directory without the full path in python
Like this: os.path.split(os.getcwd())[1] Although os.path.split returns a tuple, you don't need to unpack it. You can simply select the item that you need and ignore the one that you don't need.

Categories : Python

Directory.GetFiles() returns full path
You can use Path.GetFileName: string[] ImageNames = Directory.GetFiles(path) .Select(p => Path.GetFileName(p)).ToArray(); This will produce a list with only the names of the files.

Categories : C#

Xtend how to get full path of current working directory
Yes, Java has a File class. You can create one by calling this constructor which takes a String. Then you can call getAbsolutePath() on it. You can call it like this: package com.sandbox; import java.io.File; public class Sandbox { public static void main(String[] args) { File file = new File("relative path"); String absolutePathString = file.getAbsolutePath(); } }

Categories : Java

How do I update a link with a full directory path with jQuery URI?
OK, I figured this out with a whole bunch more googling. The function that works for what I needed is: var handler = function (data, href, target) { // Insert data into div $( '#' + target ).html(data); console.log("href = ", href); // Convert relative URLs in #right to include directory var base_path = URI(href).directory(); //var base_path = uri.directory() + '/'; console.log(base_path); $('#right a:uri(is: relative)').each(function() { var cur_href = $(this).attr("href"); $(this).attr("href", base_path+"/"+cur_href); }); }; The change is to use .each to allow an anonymous function to do the work on each relative link.

Categories : Javascript

how to get the full path of sdcard directory in android phonegap with javascript?
Please try this code.. There is window.appRootDir.fullPath through you get full path but before that you must use dirReady(); <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1"> <title> Insert title here </title> <script type="text/javascript" src="lib/android/cordova-1.7.0.js"> </script> <script type="text/javascript"> window.appRootDirName = "download_test"; document.addEventListener("deviceready", onDeviceReady, false); function onDeviceReady() { console.log("device is ready");

Categories : Javascript

Probing Path - How to Force Calls to Remain Within a Directory Structure
Further investigation has revealed that this was not a reflection or assembly loading problem. The problems I was experiencing was due to build issues inside the assemblies I was reflecting on. The errors were obscure and it made the problem look like something it wasn't. Thanks to SWeko for noting to look into the inner exceptions. That comment helped a lot!!

Categories : C#

List of files from SVN with full path
Maybe the following command line already helps you. It removes all lines from the output, that end with the character '/', i.e. it removes directories. svn ls -R | grep -v '/$' If you need more control, I recommend using the option --xml and processing the output with a script. For example, the following python script prints the full pathname of all files, when invoked with svn ls -R --xml | /tmp/filter.pz: #! /usr/bin/env python3 import sys, lxml.etree document = lxml.etree.parse(sys.stdin.buffer) for entry in document.xpath('//entry[@kind="file"]'): print(entry.xpath('string(name)'))

Categories : Svn

C docs: help Doxygen recognize class structure
Take a look at the Exclude series of tags in your configuration file. Since your source is in the ./src/ directory, one option might be to use one of the following: Exclude a directory: EXCLUDE = /src Exclude a pattern of file names: EXCLUDE_PATTERNS = *.c These are off of the top of my head. You might have to add some path information. I expect that using the first or both will produce output that does not have any of the comments in your C files. As a side note, do you think using a structure for the interface for your 'class' would meet your requirements?

Categories : C

Nested
    - display full path of structured list?
Not quite what your after, but close. Try this: <style> ol { list-style-type:upper-latin; } </style> That will print your index item labels as letters. If you wanted something like "1a...2a...2a.a" you'd have to use either javascript / server side code to build up your list in a loop programatically.

Categories : Jquery

Bash script to grep through one file for a list names, then grep through a second file to match those names to get a lookup value
awk -v search="$search_string" '$0 ~ search { gsub(/"/, "", $5); print $1" "$5; }' "$filename" | while read line do result=$(awk -v search="$line" '$0 ~ search { print $3; } ' "$lookup_file"); # Do "something" with $result done

Categories : Bash

php file search issue - not returning full path to file
I was unfamiliar with the RecursiveDirectoryIterator function, so I decided to check it out. Neat stuff! I wrote the following, it seems to do what you want: Method #1 ... $path = $_SERVER['DOCUMENT_ROOT']; $search = rawurldecode($_GET['f']); $it = new RecursiveDirectoryIterator( $path ); foreach (new RecursiveIteratorIterator($it) as $file){ $pathfile = str_replace($path,'',$file); if (strpos($pathfile, $search) !== false) { echo 'file = '. basename($file) .'<br>'; echo 'link = http://www.mydomain.com'. $pathfile .'<br>'; echo 'size = '. round(filesize($file)/1024,2) .' KB<br><br>'; } } You run it via http://www.mydomain.com/search.php?f=whatever ... assuming you name the script search.php, this will find any file containing

Categories : PHP

open file by its full path
Actually that code works as it is – at least in the current version, C++11. Before that, you need to convert the string to a C-style string: ifstream file(address.c_str()); Although you should beware of spaces in the file’s path as CapelliC mentioned in his (now-deleted) answer; in order to ensure that the user can enter paths with spaces (such as “~/some file.txt”), use std::getline instead of the stream operator: getline(cin, address);

Categories : C++

Git whatchanged - how to see full path of a file with changes
whatchanged can take the same commands as diff-tree and rev-list which it uses internally. you can try adding --stat=200 --stat-name-width=150 to your command from diff-tree which will adjust the width of the stats column to 200, and the file name to 150 which should give you all of it, or you can use --pretty to adjust the log output to your needs in other ways

Categories : GIT

Get full path from an opened file - C#
The only way for you to find open file handles is to use the NtQuerySystemInformation call. Here is a project that has this done as an explorer context menu. In this guy's case, he looks for files open in a specific folder. You would then have to match the file name to the file you have in your print job. By the way, this is not C# but you can wrap and call the same calls he is using. The rest is really up to you to figure out. ;)

Categories : C#

Get full path and long file name from short file name
@echo off setlocal rem this need to be a short name to avoid collisions with dir command bellow cd C:BALBAL~1BLBALB~1 set "curr_dir=%cd%" set "full_path=" :repeat for /f "delims=" %%f in ('for %%d in ^(.^) do @dir /a:d /n /b "..*%%~snd"') do ( set "full_path=%%f\%full_path%" ) cd .. if ":" NEQ "%cd:~-2%" ( goto :repeat ) else ( set "full_path=%cd%%full_path%" ) echo --%full_path%-- cd %curr_dir% endlocal The path is hardcoded at the beginning but you can change it or parameterizied it.As you can easy get the full name of a file here is only a solution for directories. EDIT now works for file and directory and a parameter can be passed: @echo off rem --------------------------------------------- rem ---------------------- TESTS ---------------- rem --------------

Categories : Batch File

how to open a file by its full path given by user?
Your application is failing because you aren't properly handling spaces in the filename. Try this instead of cin >> address;: getline(cin,address); See this question for the difference between cin and getline.

Categories : C++

Is there any way to find out full file path from another process?
If notepad received document's file name from command line, use GetCommandLine function.

Categories : C++

Any way to not be required to use full path to file in Xcode? [C/C++]
The problem isn't that "XCode" doesn't accept a relative path. The problem is that your code has a different current working directory than you think. You can, as the comment says, use chdir to get to a place where your file is, or use a relative path that takes into account the currend working directory (you can use getcwd() to get the current working directory). When you are using gcc or g++ on the command line, you are in control over what directory the final application runs in.

Categories : C++

Full name of uploaded path with file uploader
The method System.Drawing.Image.FromFile expect the file to be presented on the server, but when you use _uploadImageFile.PostedFile.FileName this is the file which was picked up on the client. So obviously System.Drawing.Image.FromFile(); will not find it because you pass it a client path. I would recomment you to use other methods like Image.FromStream(Stream);, you can pass input file stream and you can get the image.

Categories : C#

Get full path image into input file
MOZILLA DEVELOPER NETWORK show us an example to do that: <input type="file" id="fileElem" multiple accept="image/*" style="display:none" onchange="handleFiles(this.files)"> <a href="#" id="fileSelect">Select some files</a> <div id="fileList"> <p>No files selected!</p> </div> <script> window.URL = window.URL || window.webkitURL; var fileSelect = document.getElementById("fileSelect"), fileElem = document.getElementById("fileElem"), fileList = document.getElementById("fileList"); fileSelect.addEventListener("click", function (e) { if (fileElem) { fileElem.click(); } e.preventDefault(); // prevent navigation to "#" }, false); function handleFiles(files) { if (!files.length) { fileList.innerHTML = "<p>No files

Categories : Javascript

Can't take uploaded image file full path in JSP
In Internet Expolorer it works more-or-less by accident, because you're probably running the application on the same machine as your accessing it from. Internet Explorer sends the real filename along when you upload a file, for example C:UsersAdministratorDesktopImage.jpg. Your web application runs on the same machine so it can read that file from disk. However, Chrome and Firefox don't want to expose the full path names from your client to the application, and they use the fakepathImage.jpg. There is no such file on your disk, and that's why the web appliaction can't read the image. What you should do is extract the different parts from the request, find the part that contains the uploaded file, and read the data from the request (instead of from file). The good news is that you don't h

Categories : Java

How to parse xml-file with directory structure
Do you create this file-schema on your own? If you can change it, i would definitly. Try to make something like this: <?xml version="1.0" encoding="UTF-8"?> <Directory id="regular"> <Directory id="case_10_some_description"> <Directory id="english"> <Directory id="images"> <file id="screenshot_1.png"/> <file id="screenshot_2.png"/> <file id="screenshot_3.png"/> <file id="screenshot_4.png"/> <file id="screenshot_5.png"/> <file id="screenshot_6.png"/> </Directory> </Directory> </Directory> <Directory id="case_12_some_description"> <Directory id="e

Categories : Python

How to change file and directory names with find?
First create a shell script like this: #!/bin/bash dir=$(dirname "$1") mv "$1" "$dir/$2" Then this find should work for you: find . -name "old_name" -exec ./ren '{}' 'new_name' ; This will find file with the name old_name from the current dir in all subdir and rename them to new_name

Categories : Linux

How can I quickly reference a file from git status without the full path
There isn't a way to do this on the CLI (unless you wrote it yourself). Those sorts of use cases are what GUIs excel at. Try a bunch until you find one you like.

Categories : GIT

filesystem directory_iterator not returning full file path
For full path you need to use start->path().directory_string() directory_iterator end; for (directory_iterator start(p); start != end; ++start) { if (is_directory(start->status())) { //A directory } if (is_regular_file(start->status())) { std::cout << start->path().directory_string()<< " "; } } Play around with following based on your need, refer boost docs for more details, however names their suggest all. start->path().branch_path(); start->path().directory_string(); start->path().file_string(); start->path().root_path(); start->path().parent_path(); start->path().string(); start->path().basename(); start->path().filename(); start->path().root_name(); start->path().relative_path(

Categories : C++

maven assembly, avoiding full path in zip file?
First i would suggest to create dist-packaging module which contains the resulting package. Furthermore it sounds like your assembly-descriptor for creating the final archive which contains the zip files of others is wrong. <id>proj1-assembly</id> <formats> <format>zip</format> </formats> <includeBaseDirectory>false</includeBaseDirectory> <dependencySets> <dependencySet> <outputDirectory>/</outputDirectory> <useProjectArtifact>false</useProjectArtifact> <unpack>false</unpack> <scope>runtime</scope> </dependencySet> </dependencySets> But you must be aware that you have to maintain the dependencies in

Categories : Maven

move up and down in directory structure to access file
Since currentDir is a DirectoryInfo currentDir.Parent gives you the DirectoryInfo of the parent directory including the full-path and all other informations of DirectoryInfo. You have to use it's FullName property: string fullPath = parent.FullName; If you want to find the file: FileInfo[] files = currentDir.Parent.Parent.GetFiles("FileName.txt", System.IO.SearchOption.TopDirectoryOnly);

Categories : C#

Editing file names and saving to new directory in python
If I'am not wrong, you are only opening the file and then you are immediately closing it again? With out any writing to the file it is surely empty. Have a look here: http://docs.python.org/2/library/shutil.html shutil.copyfile(src, dst) ;)

Categories : Python

find and replace a word with another in all file names of a directory
Use find with the -exec option to call rename on every file and subdirectory containing "owner" in its name: find path/to/my/directory/ -depth -name "*owner*" -exec /usr/bin/rename owner user {} + If you don't have rename, you can use a mv command with bash parameter expansion: find path/to/my/directory/ -depth -name "*owner*" -exec bash -c 'mv "{}" $(old="{}"; new="${old##*/}"; echo "${old%/*}/${new/owner/user}")' ; bash -c '...' invokes the bash shell on the one-liner surrounded by single-quotes. The one-liner acts as a mv command that renames all occurrences of "owner" in the basename of the matching filename/subdirectory to "user". Specifically, find substitutes every {} after the -exec with the matching file/subdirectory's pathname. Here mv accepts two arguments; the first i

Categories : Linux

How to get the full Path of the file upload control in a string variable in asp .net?
It sounds like you're actually asking for the original path to the file on the client machine. This is (a) useless (it's on a different computer) and (b) impossible to get (the browser doesn't tell you it). What are you trying to do?

Categories : C#

Extracting zip file on host by PHP destroys directory structure
PHP does not actually provide a function that extracts a ZIP including its directory structure. I found the following code in a user comment in the manual: function unzip($zipfile) { $zip = zip_open($zipfile); while ($zip_entry = zip_read($zip)) { zip_entry_open($zip, $zip_entry); if (substr(zip_entry_name($zip_entry), -1) == '/') { $zdir = substr(zip_entry_name($zip_entry), 0, -1); if (file_exists($zdir)) { trigger_error('Directory "<b>' . $zdir . '</b>" exists', E_USER_ERROR); return false; } mkdir($zdir); } else { $name = zip_entry_name($zip_entry); if (file_exists($name)) { trigger_error('File "<b>' . $name

Categories : PHP

How to set hierarchical structure values through the Web API using friendly names and without exposing the structure itself to the user?
If you collection elements had aliases then you would be able to use a natural URI path structure, which is intended to map hierarchies. <ROOT> <COLLECTION_ELEMENT> //alias = foo <SIMPLE_ELEMENT /> // alias = nail <SIMPLE_ELEMENT /> // alias = hammer <COLLECTION_ELEMENT> /alias = bar <SIMPLE_ELEMENT /> // red <SIMPLE_ELEMENT /> // green </COLLECTION_ELEMENT> </COLLECTION_ELEMENT> .... <COLLECTION_ELEMENT> //baz <SIMPLE_ELEMENT /> // alias = nail <SIMPLE_ELEMENT /> // alias = hammer <COLLECTION_ELEMENT> //blunk <SIMPLE_ELEMENT /> // red <SIMPLE_ELEMENT /> // green

Categories : C#

Allowing punctuation characters in directory and file names in bash
Always quote your variable substitutions. I.e. not cp $source $target, but cp "$source" "$target". This way they won't be subject to word splitting and pathname expansion. Specify "--" before positional arguments to file operation commands. I.e. not cp "$source" "$target", but cp -- "$source" "$target". This prevents interpreting file names starting with dash as options. And yes, "/" is not a valid character for file/directory names.

Categories : Bash



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