javascript regular expression for -999x999

Question

I need a regular expression for:

-[n digits]x[n digits]

I already tried this:

var s = "path/path/name-799x1024.jpg"; 
s.replace(/\d/g, "");

But this gets only the digits. Here is a small jsfiddle: http://jsfiddle.net/aq6dp49n/

The outcome I try to get is:

pfad/pfade/name.jpg

How do I add the - and the small x between the two digits?

Your current expression doesn't seem to match any digits. Are you sure you've shown us the right one? — Bergi, Jul 29 '15 at 16:20
are you trying to store the number of just remove them from `s`? — depperm, Jul 29 '15 at 16:25

Vivin Paliath · Answer 1 · 2015-07-29T16:24:02.963

5

The regular expression that would match that is /-\d+x\d+/. Hence:

s.replace(/-\d+x\d+/, "")

Should work.

Here's what the regex means: the first - tells it that it should look for a - character. Then you have \d+ which means "one or more of \d", where \d is short-hand for the character class [0-9], i.e., all digits. After that you have x, which means it will look for the character x, and finally you have \d+ again, which is the same as before.

edited Jul 29 '15 at 16:24

answered Jul 29 '15 at 16:20

Vivin Paliath

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One fateful day many years ago I decided to learn perl, which led me to learning regex. :) I've also added an explanation about the meaning of the regex. – Vivin Paliath Jul 29 '15 at 16:22

Adam Katz · Answer 2 · 2015-07-31T22:39:32.807

To match

-[n digits]x[n digits]

You would want

match(/-[0-9]{n}x[0-9]{n}\b/)

Though if you want an arbitrary (one or more) number of digits, use + in place of {n}. In the case of your example, you'd want 3 and 4 for your values of n.

Here's a step-by-step explanation of what this does:

/-[0-9]{3}x[0-9]{4}\b/

    - matches the character - literally
    [0-9]{3} match a single character present in the list below
        Quantifier: {3} Exactly 3 times
        0-9 a single character in the range between 0 and 9
    x matches the character x literally (case sensitive)
    [0-9]{4} match a single character present in the list below
        Quantifier: {4} Exactly 4 times
        0-9 a single character in the range between 0 and 9
    \b assert position at a word boundary (^\w|\w$|\W\w|\w\W)

To remove the last size-like part of a string, this should do:

"path/path/name-799x1024.jpg".replace(/(.*)-[0-9]+x[0-9]+/, "$1");
// "path/path/name.jpg"
"path/path/name-10x12-799x1024.jpg".replace(/(.*)-[0-9]+x[0-9]+/, "$1");
// "path/path/name-10x12.jpg"

This takes advantage of the fact that regexps are greedy, so the (.*) absorbs (and saves) as much preceding text as possible before finding the next match.

(I prefer to use [0-9] in place of \d because it's more specific (\d also matches non-latin numerals) and therefore slightly faster, though in this case it shouldn't matter.)

javascript regular expression for -999x999

2 Answers2