How to get the directory of a file?
For example, I pass in a string
C:\Program Files\nant\bin\nant.exe
I want a function that returns me
C:\Program Files\nant\bin
I would prefer a built in function that does the job, instead of having manually split the string and exclude the last one.
Edit: I am running on Windows
If you use Node.js, path module is quite handy.
path.dirname("/home/workspace/filename.txt") // '/home/workspace/'
I don't know if there is any built in functionality for this, but it's pretty straight forward to get the path.
path = path.substring(0,path.lastIndexOf("\\")+1);
Use:
var dirname = filename.match(/(.*)[\/\\]/)[1]||'';
*The answers that are based on lastIndexOf('/') or lastIndexOf('\') are error prone, because path can be "c:\aa/bb\cc/dd".
(Matthew Flaschen did took this into account, so my answer is a regex alternative)
There's no perfect solution, because this functionality isn't built-in, and there's no way to get the system file-separator. You can try:
path = path.substring(0, Math.max(path.lastIndexOf("/"), path.lastIndexOf("\\")));
alert(path);
Yes, the inbuilt module path has dirname() function, which would do the job for you.
const path = require("path");
file_path = "C:\\Program Files\\nant\\bin\\nant.exe" \\windows path
file_path = "C:/Program Files/nant/bin/nant.exe" \\linux path
path.dirname(file_path); \\gets you the folder path based on your OS
I see that your path is neither windows nor Linux compatible. Do not hardcode path; instead, take a reference from a path based on your OS.
I generally tackle such situations by creating relative paths using path.join(__dirname, "..", "assets", "banner.json");.
This gives me a relative path that works regardless of the OS you are using.
function getFileDirectory(filePath) {
if (filePath.indexOf("/") == -1) { // windows
return filePath.substring(0, filePath.lastIndexOf('\\'));
}
else { // unix
return filePath.substring(0, filePath.lastIndexOf('/'));
}
}
console.assert(getFileDirectory('C:\\Program Files\\nant\\bin\\nant.exe') === 'C:\\Program Files\\nant\\bin');
console.assert(getFileDirectory('/usr/bin/nant') === '/usr/bin');
Sorry to bring this back up but was also looking for a solution without referencing the variable twice. I came up with the following:
var filepath = 'C:\\Program Files\\nant\\bin\\nant.exe';
// C:\Program Files\nant\bin\nant.exe
var dirpath = filepath.split('\\').reverse().splice(1).reverse().join('\\');
// C:\Program Files\nant\bin
This is a bit of a walk through manipulating a string to array and back but it's clean enough I think.
filepath.split("/").slice(0,-1).join("/"); // get dir of filepath
such that
"/path/to/test.js".split("/").slice(0,-1).join("/") == "/path/to"
And this?
If isn't a program in addressFile, return addressFile
function(addressFile) {
var pos = addressFile.lastIndexOf("/");
pos = pos != -1 ? pos : addressFile.lastIndexOf("\\");
if (pos > addressFile.lastIndexOf(".")) {
return addressFile;
}
return addressFile.substring(
0,
pos+1
);
}
console.assert(getFileDirectory('C:\\Program Files\\nant\\bin\\nant.exe') === 'C:\\Program Files\\nant\\bin\\');
console.assert(getFileDirectory('/usr/bin/nant') === '/usr/bin/nant/');
console.assert(getFileDirectory('/usr/thisfolderhaveadot.inhere') === '/usr/');
The core Javascript language doesn't provide file/io functions. However if you're working in a Windows OS you can use the FileSystemObject (ActiveX/COM).
Note: Don't use this in the client script-side script of a web application though, it's best in other areas such as in Windows script host, or the server side of a web app where you have more control over the platform.
This page provides a good tutorial on how to do this.
Here's a rough example to do what you want:
var fso, targetFilePath,fileObj,folderObj;
fso = new ActiveXObject("Scripting.FileSystemObject");
fileObj = fso.GetFile(targetFilePath);
folderObj=fileObj.ParentFolder;
alert(folderObj.Path);
path.dirname("path/to/file")? It will be adjusted to linux or windows path based on your environment