How can one with minimal effort (using some already existing facility, if possible) convert paths like `c:\aaa\bbb\..\ccc` to `c:\aaa\ccc`?

How can one get an absolute or normalized file path in .NET?

How to get full path of file while selecting file using `&lt;input type=‘file’&gt;`

    &lt;input type=&quot;file&quot; id=&quot;fileUpload&quot;&gt;
    &lt;script type=&quot;text/javascript&quot;&gt;
    function getFilePath(){
         $(&#39;input[type=file]&#39;).change(function () {
             var filePath=$(&#39;#fileUpload&#39;).val(); 
         });
    }
    &lt;/script&gt;

but the filePath var contains `only name` of selected file, not the `full path`. &lt;br/&gt;
I searched it on net, but it seems that for security reasons browsers (FF,chrome) just give name of file.&lt;br/&gt;
Is there any other way to get full path of selected file?

How to get full path of selected file on change of &lt;input type=‘file’&gt; using javascript, jquery-ajax?

I have a file path String of the form &quot;e:\\...\\xxx.jpg&quot;
How do I create a drawable from it?

android drawable from file path

`String` variable contains a file name, `C:\Hello\AnotherFolder\The File Name.PDF`. How do I only get the file name `The File Name.PDF` as a String?

I planned to split the string, but that is not the optimal solution.

How do I get the file name from a String containing the Absolute file path?

I need to extract the name of the parent directory of a certain path. This is what it looks like: 

    C:\stuff\directory_i_need\subdir\file.jpg

I would like to extract `directory_i_need`.

Extract a part of the filepath (a directory) in Python

Until now I have not been specifying a MongoDB data directory and have had only one 30 GB primary partition.

I just ran out of space and added a new hard disk. How can I transfer my data (that is apparently in `/var/lib/mongodb/`) and configure MongoDB so that everything runs off of the new disk without affecting my existing installation?

Changing MongoDB data store directory

I need to get the last part of current directory, for example  from `/Users/smcho/filegen_from_directory/AIRPassthrough`, I need to get `AIRPassthrough`. 

With python, I can get it with this code.

    import os.path
    
    path = &quot;/Users/smcho/filegen_from_directory/AIRPassthrough&quot;
    print os.path.split(path)[-1]

Or

    print os.path.basename(path)

How can I do the same thing with C#?

## ADDED
With the help from the answerers, I found what I needed.

    using System.Linq;
    string fullPath = Path.GetFullPath(fullPath).TrimEnd(Path.DirectorySeparatorChar);
    string projectName  = fullPath.Split(Path.DirectorySeparatorChar).Last();

or 

    string fullPath = Path.GetFullPath(fullPath).TrimEnd(Path.DirectorySeparatorChar);
    string projectName = Path.GetFileName(fullPath);

Get the (last part of) current directory name in C#

I was wondering about the difference between `\` and `/` in file paths. I have noticed that sometimes a path contains `/ `and sometimes it is with `\`.

It would be great if anyone can explain when to use `\` and `/`.


Difference between forward slash (/) and backslash (\) in file path

Every now and then when I build my solution here (with 7 projects in it) I get the dreaded &#39;Command copy exited with code 4&#39; error, in Visual Studio 2010 Premium ed.

This is because of the post-build event not being able to go through.

Here&#39;s what solves the problem, temporarily

- Sometimes: A restart of Visual Studio and I&#39;m able to build the solution
- Sometimes:  Both a restart of Visual Studio and my file manager of choice (Q-Dir 4.37) solves it.

Here&#39;s what the post-build event looks like:

    xcopy &quot;$(SolutionDir)Solution Items\References\*.dll&quot; &quot;$(TargetDir)&quot; /Y

When you get the command copy exited with code [insert value] error, it&#39;s normally because of the following:

- read / write permissions
- missing files
- wrong directories

However - obviously at times when I build the solution, there&#39;s no problem.

FYI, I uninstalled ReSharper 5.1.1 two weeks ago and Visual Studio&#39;s been giving me some errors since then (among them not being able to debug). I re-installed Visual Studio and it&#39;s working better since then, but still get this problem. Could it have to do with some ReSharper stuff being somewhere?

Have you had the same problem and solved it? Or do you have any possible solution to it?

Command copy exited with code 4 when building - Visual Studio restart solves it

I have been trying to get the URI path for an asset file. 

    uri = Uri.fromFile(new File(&quot;//assets/mydemo.txt&quot;));

When I check if the file exists I see that file doesn&#39;t exist
                         
   

    File f = new File(filepath);
    if (f.exists() == true) {
        Log.e(TAG, &quot;Valid :&quot; + filepath);
    } else {
        Log.e(TAG, &quot;InValid :&quot; + filepath);
    }

Can some one tell me how I can mention the absolute path for a file existing in the asset folder


How to get URI from an asset File?

I have a path (including directory and file name).  
I need to test if the file-name is a valid, e.g. if the file-system will allow me to create a file with such a name.  
The file-name *has some unicode characters* in it.

It&#39;s safe to assume the directory segment of the path is valid and accessible (&lt;sup&gt;I was trying to make the question more gnerally applicable, and apparently I wen too far&lt;/sup&gt;).

I very much do not want to have to escape anything unless I *have* to. 

I&#39;d post some of the example characters I am dealing with, but apparently they get automatically removed by the stack-exchange system. Anyways, I want to keep standard unicode entities like `&#246;`, and only escape things which are invalid in a filename.

---

Here is the catch. **There may (or may not) already be a file at the target of the path.** I need to keep that file if it does exist, and not create a file if it does not.

Basically I want to check if I *could* write to a path **without actually opening the path for writing** (and the automatic file creation/file clobbering that typically entails).

As such:

    try:
        open(filename, &#39;w&#39;)
    except OSError:
        # handle error here
[from here](https://stackoverflow.com/questions/4102896/check-whether-a-path-is-valid-in-python)

Is not acceptable, because it will overwrite the existent file, which I do not want to touch (if it&#39;s there), or create said file if it&#39;s not.

I know I can do:

    if not os.access(filePath, os.W_OK):
        try:
            open(filePath, &#39;w&#39;).close()
            os.unlink(filePath)
        except OSError:
            # handle error here

But that will *create* the file at the `filePath`, which I would then have to `os.unlink`.

In the end, it seems like it&#39;s spending 6 or 7 lines to do something that should be as simple as `os.isvalidpath(filePath)` or similar.

---

As an aside, I need this to run on (at least) Windows and MacOS, so I&#39;d like to avoid platform-specific stuff.


``

Check whether a path is valid in Python without creating a file at the path&#39;s target

So I have a web project, and I&#39;m trying to get the root directory of the website using the c# method `Directory.GetCurrentDirectory()`.  I don&#39;t want to be using a static path as the file locations will be changing in the future.  This method is running in my imageProcess.aspx.cs file, but where I thought it would return:

    C:\Users\tcbl\documents\visual studio 2010\Projects\ModelMonitoring\ModelMonitoring\imageProcess.aspx.cs

I&#39;m instead getting:
    
    C:\Program Files\Common Files\Microsoft Shared\DevServer\10.0\

Can anyone explain why this is happening and what a possible solution might be? Thanks a lot.

Getting current directory in .NET web application

I am learning golang(beginner) and I have been searching on both google and stackoverflow but I could not find an answer so excuse me if already asked, but how can I mkdir if not exists in golang.

For example in node I would use fs-extra with the function [ensureDirSync] [1] (if blocking is of no concern of course)

    fs.ensureDir(&quot;./public&quot;);

  [1]: https://github.com/jprichardson/node-fs-extra#ensuredirdir-callback


mkdir if not exists using golang

My program will take arbitrary strings from the internet and use them for file names. Is there a simple way to remove the bad characters from these strings or do I need to write a custom function for this?

Is there a way of making strings file-path safe in c#?

Given a folder path (like `C:\Random Folder`), how can I find a file in it that holds a certain extension, like `txt`? I assume I&#39;ll have to do a search for `*.txt` in the directory, but I&#39;m not sure how I&#39;m supposed to start this search in the first place.

Find a file with a certain extension in folder

So, an *absolute path* is a way to get to a certain file or location describing the full route to it, the full path, and it&#39;s OS dependent (the absolute paths for Windows and Linux, for example, are different). A *relative path*, on the other hand, is a route to a file or location which is described from the current location `..` (two dots) indicating a superior level in the directories tree. That has been clear to me for several years now.

When searching I&#39;ve even seen that there are canonicalized files too!
All I know is that CANONICAL means something like &quot;according to the rules&quot; or something.

Can somebody enlighten me in therms of theory about canonical stuff?

What&#39;s a &quot;canonical path&quot;?

I&#39;m not able to see the bigger picture here I think; but basically I have no idea why you would use `os.path.join` instead of just normal string concatenation?

I have mainly used VBScript so I don&#39;t understand the point of this function.

Why use os.path.join over string concatenation?

Does C++ have any equivalent to python&#39;s function `os.path.join`? Basically, I&#39;m looking for something that combines two (or more) parts of a file path so that you don&#39;t have to worry about making sure the two parts fit together perfectly. If it&#39;s in Qt, that would be cool too.

Basically I spent an hour debugging some code and at least part of it was because `root + filename` had to be `root/ + filename`, and I&#39;m looking to avoid that in the future.

How to build a full path string (safely) from separate strings?

**Situation**

I want to use gulp and related front-end tool chains in Windows-hosted development environments. I&#39;m hitting a wall trying to use gulp plug-ins like Browser-Sync, because the node_modules folder graph fans out making the windows file paths too long to copy the files.  I&#39;d like a pragmatic approach for handling this problem right now on Windows, irrespective of what the Node community may or may not provide to improve npm usability on the Windows in the future.

**2 Questions**

1. Is there an npm workflow for Windows that just works the way it was intended? &quot;run the command and the files install&quot; (e.g. comparable to npm on OSX, npm on Linux, ruby gems or even nuget)  I don&#39;t want to fiddle with a bunch of manual file edits, symlinks, etc. every time I use npm on Windows.

2. Is there a well-documented, stable Cygwin workflow for npm and node execution to workaround the Windows API file path limits?

Gory details listed below...

**General Problem** 

 - Running npm install from a standard Windows command prompt fails on deeply nested node_modules hierarchies.
 - Per Joyent&#39;s github repo thread, [this is an acknowledged issue][1] with no palatable workarounds for developers in Windows-centric environments. (**Really?**)
 - NT Kernel supports file path lengths up to 32,767 characters. 
 - Windows API&#39;s MAXPATH is limited to 260 characters. 
 - Windows API handles file operations for all major Windows shells and whatnot including: Explorer, CMD, Powershell,MYSgit bash, etc. (**MS really?  How long has NTFS been around?**)
 - Cygwin supports long file paths, but npm.cmd doesn&#39;t work out-of-box due to crlf formatting.  I tried the DOS2Unix transform on npm to get it working with Cygwin, but there seem to be other issues with this.

**My Current Hack**

 - Create an &quot;n&quot; folder as a staging area on the root of C:\, because
   this shortens my folder path. 
 - Run npm inside the &quot;n&quot; folder to install modules for whatever I need. 
 - Fire up Cygwin and use cp to copy the node_modules folder into a destination project. 
 - Rinse and repeat when dependencies change or when I need to spin up a new project.

**Other Unpalatable Workarounds**

**Symbolic Links** can be used to shorten file paths, but these are kludgy hacks.  As the npm ecosystem grows, nested dependency chains will become too long and this workaround become unusable.

**Adding ALL dependences to the root folder&#39;s package.json** file was mentioned in one thread I came across. Although this approach will flatten the folder structure and prevent loading of duplicate modules, this workaround feels unnatural.  It also kills the usability, durability, and productivity of npm, because you have to fiddle with files and folders post-install either manually or with some hacky scripts. The approach is also vulnerable to the same fate that Symbolic Links approach may eventually suffer. 



  [1]: https://github.com/joyent/node/issues/6960#issuecomment-46704998

Node npm windows file paths are too long to install packages

I run R on Windows, and have a csv file on the Desktop. I load it as follows,

    x&lt;-read.csv(&quot;C:\Users\surfcat\Desktop\2006_dissimilarity.csv&quot;,header=TRUE)


but the R gives the following error message

&gt; Error: &#39;\U&#39; used without hex digits in character string starting &quot;C:\U&quot;

So what&#39;s the correct way to load this file. I am using Vista

File path issues in R using Windows (&quot;Hex digits in character string&quot; error)

What does the following file path mean? 

    $(Services_Jobs_Drop_Path)\**\*.config

The variable just holds some path, nothing interesting. I&#39;m a lot more concerned, what the hell the `**` mean.
Any ideas?

P.S. The following path is used in msbuild scripts, if it helps.

What does two asterisks together in file path mean?

&gt; **Possible Duplicate:**  
&gt; [How to get only filenames within a directory using c#?](https://stackoverflow.com/questions/7140081/how-to-get-only-filenames-within-a-directory-using-c)  

&lt;!-- End of automatically inserted text --&gt;

Using C#, I want to get the list of files in a folder.  
My goal: `[&quot;file1.txt&quot;, &quot;file2.txt&quot;]`

So I wrote this:

    string[] files = Directory.GetFiles(dir);

Unfortunately, I get this output: `[&quot;C:\\dir\\file1.txt&quot;, &quot;C:\\dir\\file2.txt&quot;]`

I could strip the unwanted &quot;C:\\dir\\&quot; part afterward, but is there a more elegant solution?

Directory.GetFiles: how to get only filename, not full path?

I&#39;m using `io/ioutil` to read a small text file:

    fileBytes, err := ioutil.ReadFile(&quot;/absolute/path/to/file.txt&quot;)

And that works fine, but this isn&#39;t exactly portable. In my case, the files I want to open are in my GOPATH, for example:

    /Users/matt/Dev/go/src/github.com/mholt/mypackage/data/file.txt

Since the `data` folder rides right alongside the source code, I&#39;d love to just specify the relative path:

    data/file.txt

But then I get this error:

&gt; panic: open data/file.txt: no such file or directory

How can I open files using their relative path, especially if they live alongside my Go code?

_(**Note that my question is specifically about opening files relative to the GOPATH.** Opening files using any relative path in Go is as easy as giving the relative path instead of an absolute path; files are opened relative to the compiled binary&#39;s working directory. In my case, I want to open files relative to where the binary was compiled. In hindsight, this is a bad design decision.)_

How can I open files relative to my GOPATH?

I am looking for an efficient way to convert back slash to forward slash in R. Sometime I copy the link of the directory in Windows and I get something like this:

`C:\Users\jd\Documents\folder\file.txt`

How can I quickly change this to `C:/Users/jd/Documents/folder/file.txt` ? I cannot even read the above expression as character. It throws an error 

&gt;&quot;\u used without hex digits in character string starting &quot;&quot;C:\u&quot;.

I know TAB function in R helps to find the location fast, but was just wondering if there was any other work around. I could change the working directory to the location of folder also. I was just playing around and tried to convert backslash to forward slash and was not straight forward so asked this just because of curiosity.

Efficiently convert backslash to forward slash in R

I recently installed phpunit on my server via the pear installer.

When I go to run a test I get the following error:

  &gt;PHP Warning:  require_once(PHPUnit/Util/Filter.php): failed to open stream: No such file or directory in /usr/bin/phpunit on line 44
  &gt;
 &gt; PHP Fatal error:  require_once(): Failed opening required &#39;PHPUnit/Util/Filter.php&#39; (include_path=&#39;.:/usr/bin/php&#39;) in /usr/bin/phpunit on line 44

After doing some searching, I tried making some modifications to the include_path in my php.ini file on the server. But that hasn&#39;t done a thing.

Any idea what might be causing this?

phpunit require_once() error

I found a reference to a file in a log that had the following format:

    \\?\C:\Path\path\file.log

I cannot find a reference to what the sequence of \\?\ means. I believe the part between the backslashes refers to a hostname.

For instance, on my Windows computer, the following works just fine:

    dir \\?\C:\

and also, just fine with same result:

    dir \\.\C:\

Questions:

1. Is there a reference to what the question mark means in this particular path format?
2. What might generate a file path in such a format?


What does \\?\ mean when prepended to a file path

I tried to get the directory of the console application using the below code,

    Assembly.GetExecutingAssembly().Location

but this one gives me where the assemble resides. This may be different from where I executed the application.

My console application parses logs with no parameters. It must go to the `logs/` folder inside of the executable&#39;s folder or if I give it a path to `logs/` it parses it.

How to get execution directory of console application

I&#39;m building a path string in PHP. I need it to work across platforms (i.e., Linux, Windows, OS X). 
I&#39;m doing this:

    $path = $someDirectory.&#39;/&#39;.$someFile;

Assume `$someDirectory` and `$someFile` are formatted correctly at run-time on the various platforms. This works beautifully on Linux and OS X, but not on Windows. The issue is the `/` character, which I thought would work for Windows.

Is there a PHP function or some other trick to switch this to `\` at runtime on Windows?

&lt;b&gt;EDIT:&lt;/b&gt; Just to be clear, the resultant string is 

    c:\Program Files (x86)\Sitefusion\Sitefusion.org\Defaults\pref/user.preferences

on Windows. Obviously the mix of slashes confuses Windows.

How to get a platform independent directory separator in PHP?

Is there any way to determine current filesystem location of executed class from the code of this class, in runtime, given that it&#39;s executed using `java` command? 

For example for following class:

    public class MyClass{
     public static void main(String[] args){\
       new MyClass().doStuff();
     }
    
     private void doStufF(){
       String path = ... // what to do here?!
       System.out.println(&quot;Path of class being invoked: &quot; + path);
     }
    
    }

Is there any way to get and print the correct path in filesystem of MyClass.class file just being invoked?

Similarly, if I invoke jar file from `java -jar` command, is there any way to get a path of jar file from within classes inside that jar?


P.S.:Please, **do not make any assumptions about working directory** from which `java` command is invoked!

Getting filesystem path of class being executed

I have two Bash scripts in the same folder (saved somewhere by the user who downloads the entire repository):

 * `script.sh` is run by the user
 * `helper.sh` is required and run by `script.sh`

The two scripts should be in the same directory. I need the first script to call the second one, but there are two problems:

1. Knowing the current working directory is useless to me, because I don&#39;t know how the user is executing the first script (could be with `/usr/bin/script.sh`, with `./script.sh`, or it could be with `../Downloads/repo/scr/script.sh`)
2. The script `script.sh` will be changing to a different directory before calling `helper.sh`.

I can definitely hack together Bash that does this by storing [the current directory](https://stackoverflow.com/a/246128/617937) in a variable, but that code seems needlessly complicated for what I imagine is a very common and simple task.

Is there a standard way to _reliably_ call `helper.sh` from within `script.sh`? And will work in any Bash-supported operating system?

Run a script in the same directory as the current script

I tried a solution (see below) that works fine, except in Android 4.4 the call to `startActivityForResult()` brings up an activity titled &quot;Open from&quot;, which has &quot;Recent&quot;, &quot;Images&quot;, &quot;Downloads&quot; as well as several apps to pick from. When I choose &quot;Images&quot; and try to resolve the returned content URI (using the code below), the call to `cursor.getString()` returns null. If I choose the exact same file using the Gallery app, `cursor.getString()` returns a file path. I&#39;ve only tested this in API levels 16 and 19. Everything works as expected in 16. As far as 19 goes, I have to choose the Gallery or other app or it doesn&#39;t work.

    private String getRealPathFromURI(Context context, Uri contentUri) {
        Cursor cursor = null;
        try { 
            String[] proj = { MediaStore.Images.Media.DATA };
            cursor = context.getContentResolver().query(contentUri,  proj, null, null, null);
            int column_index = cursor.getColumnIndexOrThrow(MediaStore.Images.Media.DATA);
            cursor.moveToFirst();
            String path = cursor.getString(column_index);
            
            return path;
        } finally {
            if (cursor != null) {
                cursor.close();
            }
        }
    }

All Filepath Solutions on Gang of Coders

Total of 31 Filepath Solutions