Javascript regular expression: match anything up until something (if there it exists)

Question

Hi I am new to regular expression and this may be a very easy question (hopefully).

I am trying to use one solution for 3 kind of string

• "45%", expected result: "45"
• "45", expected result: "45"
• "", expected result: ""

What I am trying (let the string be str):

str.match(/(.*)(?!%*)/i)[1]

This in my head would sound like "match any instance of anything up until '%' if it is found, or else just match anything"

In firebug's head it seems to sound more like "just match anything and completely disregard the negative lookahead". Also to make it lazy - (.*)? - doesn't seem to help.

Let's forget for a second that in this specific situation I am only matching numbers, so a /\d*/ would do. I am trying to understand a general rule so that I can apply it whenever.

Anybody would be so kind to help me out?

Answer 1

How about the simpler

str.match(/[^%]*/i)[0]

Which means, match zero-or-more character, which is not a %.

---

Edit: If need to parse until </a>, then you could parse a sequence pf characters, followed by </a>, then then discard the </a>, which means you should use positive look-ahead instead of negative.

str.match(/.*?(?=<\/a>|$)/i)[0]

This means: match zero-or-more character lazily, until reaching a </a> or end of string.

Note that *? is a single operator, (.*)? is not the same as .*?.

(And don't parse HTML with a single regex, as usual.)

Answer 2

I think this is what you're looking for:

/(?:(?!%).)*/

The . matches any character, but only after the negative lookahead, (?!%), confirms that the character is not %. Note that when the sentinel is a single character like %, you can use a negated character class instead, for example:

/[^%]*/

But for a multi-character sentinel like </a>, you have to use the lookahead approach:

/(?:(?!</a>).)*/i

This is actually saying "Match zero or more characters one at a time, but if the next character turns out to be the beginning of the sequence </a> or </A>, stop without consuming it".

Answer 3

The easiest way with an exact search string is to skip regular expressions and just use indexOf, e.g.:

// String to be searched
var s = "Here is a <a>link</a>."

// String to find
var searchString = "</a>";

// Final match
var matched = "";

var c = s.indexOf(searchString);
if (c >= 0)
{
    // Returns the portion not including the search string;
    // in this example, "Here is a <a>link". If you want the
    // search string included, add the length of the search
    // string to c.
    matched = s.substring(c);
}

Answer 4

I just wrote it exactly how you said it:

str.match(/(^[^%]*$)|^([^%]*)%.*/i)

This will match any string without a '%' or the first part of a string that contains a %. You have to get the result from the 1st or 2nd group.

EDIT: This is exactly what you want below

str.match(/(?:^[^%]*$)|^(?:[^%]*)(?=%)/)

• The ?: removes all grouping
• The ?= is a lookahead to see if the string contains %
• and [^%] matches any character that is not a %

so the regex reads match any string that doesnt contain %, OR (otherwise match) all of the characters before the first %

Answer 5

to match 45, 45%, and any number of any length use this (182%, 18242, etc)

str.match(/([0-9]+)([%]?)/)[1];

if you need to match the empty string also include it as ^$, note match("...")[1] will be undefined for the empty string, so you will need to test for match and then check [0] or see if [1] is undefined.

str.match(/([0-9]+)([%]?)|^$/)

if you need to match exactly two digits use {2,2} anchor the expression between begin and end line characters: "^(exp)$"

str.match(/^([0-9]{2,2})([%]?)$/)[1];