How to get the file name from a full path using JavaScript?

Question

Is there a way that I can get the last value (based on the '\' symbol) from a full path?

Example:

C:\Documents and Settings\img\recycled log.jpg

With this case, I just want to get recycled log.jpg from the full path in JavaScript.

Answer 1

var filename = fullPath.replace(/^.*[\\\/]/, '')

This will handle both \ OR / in paths

Answer 2

Just for the sake of performance, I tested all the answers given here:

var substringTest = function (str) {
    return str.substring(str.lastIndexOf('/')+1);
}

var replaceTest = function (str) {
    return str.replace(/^.*(\\|\/|\:)/, '');
}

var execTest = function (str) {
    return /([^\\]+)$/.exec(str)[1];
}

var splitTest = function (str) {
    return str.split('\\').pop().split('/').pop();
}

substringTest took   0.09508600000000023ms
replaceTest   took   0.049203000000000004ms
execTest      took   0.04859899999999939ms
splitTest     took   0.02505500000000005ms

And the winner is the Split and Pop style answer, Thanks to bobince !

Answer 3

In Node.js, you can use Path's parse module...

var path = require('path');
var file = '/home/user/dir/file.txt';

var filename = path.parse(file).base;
//=> 'file.txt'

Answer 4

What platform does the path come from? Windows paths are different from POSIX paths are different from Mac OS 9 paths are different from RISC OS paths are different...

If it's a web app where the filename can come from different platforms there is no one solution. However a reasonable stab is to use both '\' (Windows) and '/' (Linux/Unix/Mac and also an alternative on Windows) as path separators. Here's a non-RegExp version for extra fun:

var leafname= pathname.split('\\').pop().split('/').pop();

Answer 5

Ates, your solution doesn't protect against an empty string as input. In that case, it fails with TypeError: /([^(\\|\/|\:)]+)$/.exec(fullPath) has no properties.

bobince, here's a version of nickf's that handles DOS, POSIX, and HFS path delimiters (and empty strings):

return fullPath.replace(/^.*(\\|\/|\:)/, '');

Answer 6

The following line of JavaScript code will give you the file name.

var z = location.pathname.substring(location.pathname.lastIndexOf('/')+1);
alert(z);

Answer 7

Not more concise than nickf's answer, but this one directly "extracts" the answer instead of replacing unwanted parts with an empty string:

var filename = /([^\\]+)$/.exec(fullPath)[1];

Answer 8

Another one

var filename = fullPath.split(/[\\\/]/).pop();

Here split have a regular expression with a character class
The two characters have to be escaped with '\'

Or use array to split

var filename = fullPath.split(['/','\\']).pop();

It would be the way to dynamically push more separators into an array, if needed.
If fullPath is explicitly set by a string in your code it need to escape the backslash!
Like "C:\\Documents and Settings\\img\\recycled log.jpg"

Answer 9

A question asking "get file name without extension" refer to here but no solution for that. Here is the solution modified from Bobbie's solution.

var name_without_ext = (file_name.split('\\').pop().split('/').pop().split('.'))[0];

Answer 10

I use:

var lastPart = path.replace(/\\$/,'').split('\\').pop();

It replaces the last \ so it also works with folders.

Answer 11

<script type="text/javascript">
    function test()
    {
        var path = "C:/es/h221.txt";
        var pos =path.lastIndexOf( path.charAt( path.indexOf(":")+1) );
        alert("pos=" + pos );
        var filename = path.substring( pos+1);
        alert( filename );
    }
</script>
<form name="InputForm"
      action="page2.asp"
      method="post">
    <P><input type="button" name="b1" value="test file button"
    onClick="test()">
</form>

Answer 12

Little function to include in your project to determine the filename from a full path for Windows as well as GNU/Linux & UNIX absolute paths.

/**
 * @param {String} path Absolute path
 * @return {String} File name
 * @todo argument type checking during runtime
 * @see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/includes
 * @see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/slice
 * @see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/lastIndexOf
 * @example basename('/home/johndoe/github/my-package/webpack.config.js') // "webpack.config.js"
 * @example basename('C:\\Users\\johndoe\\github\\my-package\\webpack.config.js') // "webpack.config.js"
 */
function basename(path) {
  let separator = '/'

  const windowsSeparator = '\\'

  if (path.includes(windowsSeparator)) {
    separator = windowsSeparator
  }

  return path.slice(path.lastIndexOf(separator) + 1)
}

Answer 13

There’s no need to handle backslashes specially; most answers don’t handle search parameters.

The modern approach is to simply use the URL API and get the pathname property. The API normalizes backslashes to slashes.

In order to parse the resulting %20 to a space, simply pass it to decodeURIComponent.

const getFileName = (fileName) => new URL(fileName).pathname.split("/").pop();

// URLs need to have the scheme portion, e.g. `file://` or `https://`.
console.log(getFileName("file://C:\\Documents and Settings\\img\\recycled log.jpg")); // "recycled%20log.jpg"
console.log(decodeURIComponent(getFileName("file://C:\\Documents and Settings\\img\\recycled log.jpg"))); // "recycled log.jpg"
console.log(getFileName("https://example.com:443/path/to/file.png?size=480")); // "file.png"

.as-console-wrapper { max-height: 100% !important; top: 0; }

Add a .filter(Boolean) before the .pop() if you always want the last non-empty part of the path (e.g. file.png from https://example.com/file.png/).

If you only have a relative URL but still simply want to get the file name, use the second argument of the URL constructor to pass a base origin. "https://example.com" suffices: new URL(fileName, "https://example.com"). It’s also possible to prepend "https://" to your fileName — the URL constructor accepts https://path/to/file.ext as a valid URL.

Answer 14

The complete answer is:

<html>
    <head>
        <title>Testing File Upload Inputs</title>
        <script type="text/javascript">

        function replaceAll(txt, replace, with_this) {
            return txt.replace(new RegExp(replace, 'g'),with_this);
        }

        function showSrc() {
            document.getElementById("myframe").href = document.getElementById("myfile").value;
            var theexa = document.getElementById("myframe").href.replace("file:///","");
            var path = document.getElementById("myframe").href.replace("file:///","");
            var correctPath = replaceAll(path,"%20"," ");
            alert(correctPath);
        }
        </script>
    </head>
    <body>
        <form method="get" action="#"  >
            <input type="file"
                   id="myfile"
                   onChange="javascript:showSrc();"
                   size="30">
            <br>
            <a href="#" id="myframe"></a>
        </form>
    </body>
</html>

Answer 15

<html>
    <head>
        <title>Testing File Upload Inputs</title>
        <script type="text/javascript">
            <!--
            function showSrc() {
                document.getElementById("myframe").href = document.getElementById("myfile").value;
                var theexa = document.getElementById("myframe").href.replace("file:///","");
                alert(document.getElementById("myframe").href.replace("file:///",""));
            }
            // -->
        </script>
    </head>
    <body>
        <form method="get" action="#"  >
            <input type="file" 
                   id="myfile" 
                   onChange="javascript:showSrc();" 
                   size="30">
            <br>
            <a href="#" id="myframe"></a>
        </form>
    </body>
</html>

Answer 16

Successfully Script for your question ,Full Test

<script src="~/Scripts/jquery-1.10.2.min.js"></script>

<p  title="text" id="FileNameShow" ></p>
<input type="file"
   id="myfile"
   onchange="javascript:showSrc();"
   size="30">

<script type="text/javascript">

function replaceAll(txt, replace, with_this) {
    return txt.replace(new RegExp(replace, 'g'), with_this);
}

function showSrc() {
    document.getElementById("myframe").href = document.getElementById("myfile").value;
    var theexa = document.getElementById("myframe").href.replace("file:///", "");
    var path = document.getElementById("myframe").href.replace("file:///", "");
    var correctPath = replaceAll(path, "%20", " ");
   alert(correctPath);
    var filename = correctPath.replace(/^.*[\\\/]/, '')
    $("#FileNameShow").text(filename)
}

Answer 17

This solution is much simpler and generic, for both 'file name' and 'path'.

const str = 'C:\\Documents and Settings\\img\\recycled log.jpg';

// regex to split path to two groups '(.*[\\\/])' for path and '(.*)' for file name
const regexPath = /^(.*[\\\/])(.*)$/;

// execute the match on the string str
const match = regexPath.exec(str);
if (match !== null) {
    // we ignore the match[0] because it's the match for the hole path string
    const filePath = match[1];
    const fileName = match[2];
}

Answer 18

function getFileName(path, isExtension){

  var fullFileName, fileNameWithoutExtension;

  // replace \ to /
  while( path.indexOf("\\") !== -1 ){
    path = path.replace("\\", "/");
  }

  fullFileName = path.split("/").pop();
  return (isExtension) ? fullFileName : fullFileName.slice( 0, fullFileName.lastIndexOf(".") );
}

Answer 19

var file_name = file_path.substring(file_path.lastIndexOf('/'));