For example, assuming that x = filename.jpg
, I want to get filename
, where filename
could be any file name (Let’s assume the file name only contains [a-zA-Z0-9-_] to simplify.).
I saw x.substring(0, x.indexOf('.jpg'))
on DZone Snippets, but wouldn’t x.substring(0, x.length-4)
perform better? Because, length
is a property and doesn’t do character checking whereas indexOf()
is a function and does character checking.
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Answer
If you know the length of the extension, you can use x.slice(0, -4)
(where 4 is the three characters of the extension and the dot).
If you don’t know the length @John Hartsock regex would be the right approach.
If you’d rather not use regular expressions, you can try this (less performant):
filename.split('.').slice(0, -1).join('.')
Note that it will fail on files without extension.