JS: Most optimized way to remove a filename from a path in a string?
JavascriptOptimizationJavascript Problem Overview
I have strings formatted as follows:
path/to/a/filename.txt
Now I'd like to do some string manipulation which allows me to very efficiently remove the "filename.txt" part from this code. In other words, I want my string to become this:
path/to/a/
What's the most efficient way to do this? Currently I'm splitting the string and reconnecting the seperate elements except for the last one, but I get the feeling this is a really, REALLY inefficient way to do it. Here's my current, inefficient code:
res.getPath = function(file)
{
var elem = file.split("/");
var str = "";
for (var i = 0; i < elem.length-1; i++)
str += elem[i] + "/";
return str;
}
Javascript Solutions
Solution 1 - Javascript
Use lastIndexOf() to find the position of the last slash and get the part before the slash with substring().
str.substring(0, str.lastIndexOf("/"));
Solution 2 - Javascript
If you're using Node.js:
const path = require("path")
const removeFilePart = dirname => path.parse(dirname).dir
removeFilePart("/a/b/c/d.txt")
// Returns "/a/b/c"
Solution 3 - Javascript
How about this:
"path/to/a/filename.txt".split("/").slice(0, -1).join("/")+"/"
Solution 4 - Javascript
In node, you can use path.dirname
const path = require('path')
fullFilePath = '/some/given/path/to-a-file.txt'
directoryPath = path.dirname(fullFilePath)
console.log(directoryPath) // ===> '/some/given/path'
Solution 5 - Javascript
str = str.split('/')
str.pop()
str.join('/') + '/'
Solution 6 - Javascript
function splitPath(path) {
var dirPart, filePart;
path.replace(/^(.*\/)?([^/]*)$/, function(_, dir, file) {
dirPart = dir; filePart = file;
});
return { dirPart: dirPart, filePart: filePart };
}
there that's better
Solution 7 - Javascript
If this is to process a filename from a file upload form, the HTML5 spec recommends the following code:
function extractFilename(path) {
if (path.substr(0, 12) == "C:\\fakepath\\")
return path.substr(12); // modern browser
var x;
x = path.lastIndexOf('/');
if (x >= 0) // Unix-based path
return path.substr(x+1);
x = path.lastIndexOf('\\');
if (x >= 0) // Windows-based path
return path.substr(x+1);
return path; // just the filename
}
Reference: http://www.w3.org/TR/html5/number-state.html#file-upload-state</strike>
http://www.w3.org/TR/html5/forms.html#file-upload-state-(type=file)
Solution 8 - Javascript
function getDirname(pathname, separator) {
var parts = pathname.split(separator);
if (parts[parts.length - 1].indexOf('.') > -1) {
return parts.slice(0, -1).join(separator)
}
return pathname;
}
Usage:
var dir = getDirname(url.parse(request.url).pathname, '/');
.
var dir = getDirname(path.join('foo', 'bar', 'text.txt'), path.sep);
Solution 9 - Javascript
test/dir/lib/file- _09.ege.jpg
- Will be to - test/dir/lib/
file- _09.ege.jpg
- Will be to - file- _09.ege.jpg
console.log("test - "+getPath('test/dir/lib/file- _09.ege.jpg'));
function getPath(path){
path = path.match(/(^.*[\\\/]|^[^\\\/].*)/i);
if(path != null){
return path[0];
}else{
return false;
}
}
console.log("test - "+getPath('test/dir/lib/file- _09.ege.jpg'));
function getPath(path){
path = path.match(/(^.*[\\\/]|^[^\\\/].*)/i);
if(path != null){
return path[0];
}else{
return false;
}
}