For example, assuming that x = filename.jpg, I want to get filename, where filename could be any file name (Let\'s assume the file nam
If you have to process a variable that contains the complete path (ex.: thePath = "http://stackoverflow.com/directory/subdirectory/filename.jpg") and you want to return just "filename" you can use:
theName = thePath.split("/").slice(-1).join().split(".").shift();
the result will be theName == "filename";
To try it write the following command into the console window of your chrome debugger:
window.location.pathname.split("/").slice(-1).join().split(".").shift()
If you have to process just the file name and its extension (ex.: theNameWithExt = "filename.jpg"):
theName = theNameWithExt.split(".").shift();
the result will be theName == "filename", the same as above;
But I can't say nothing about neither performances comparison with other answers nor for browser or OS compatibility.
working snippet 1: the complete path
var thePath = "http://stackoverflow.com/directory/subdirectory/filename.jpg";
theName = thePath.split("/").slice(-1).join().split(".").shift();
alert(theName);
working snippet 2: the file name with extension
var theNameWithExt = "filename.jpg";
theName = theNameWithExt.split("/").slice(-1).join().split(".").shift();
alert(theName);
working snippet 2: the file name with double extension
var theNameWithExt = "filename.tar.gz";
theName = theNameWithExt.split("/").slice(-1).join().split(".").shift();
alert(theName);