How to extract text using Regular expression (RegEx)?

Question

I need some help with regular expressions in JavaScript.

I have the following string.

var str = "SOme TEXT #extract1$ Some more text #extract2$ much more text #extract3$ some junk";

I want to extract all the text between # and $ into an array or object.

Final output would be:

var result = [extract1,extract2,extract3]

extract1, extract2,extract3 can contain any characters like _,@,&,*

dheerosaur · Accepted Answer · 2011-04-28 06:17:52Z

2

You can use JavaScript Regular expressions' exec method.

var regex = /\#([^\$]+)\$/g,
    match_arr,
    result = [],
    str = "some #extract1$ more #extract2$ text #extract3$ some junk";
while ((match_arr = regex.exec(str)) != null)
{
   result.push(match_arr[1]);
}
console.log(result);

Reference: https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/RegExp/exec

answered Apr 28, 2011 at 6:17

dheerosaur

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Shekhar · Accepted Answer · 2011-04-28 06:07:23Z

2

Regex for you will be like #(?<extract>[\s\S]*?)\$

Named group 'extract'will contain the values you want.

As Alex mentioned, if javascript doesnt support named group, we can use numbered group. So modified regex will be #([\s\S]*?)\$ and desired values will be in group number 1.

edited Apr 28, 2011 at 6:07

answered Apr 28, 2011 at 5:55

Shekhar

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7 Comments

yoavmatchulsky Over a year ago

nice touch with the [\s\S]. how does it perform against [^\$] ?

Shekhar Over a year ago

@yoavmatchulsky, certainly [^\$] is faster than [\s\S]

alex Over a year ago

JavaScript doesn't have named capturing groups. Why do you escape #? Also, why the ? after the *?

alex Over a year ago

@Shekhar Also, I understand the ? now. Ungreedy. How did I forget that?

James Khoury Over a year ago

@alex the *? means its a non-greedy * which means it will capture as little as possible. EG a regex of .*b on a string "aaaababa" will capture "aaaabab" but a non greedy .*?b will capture "aaaab". EDIT beat me to it alex. ;)

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alex · Accepted Answer · 2011-04-28 06:06:39Z

0

var matches = str.match(/#(.+?)\$/gi);

jsFiddle.

answered Apr 28, 2011 at 6:06

alex

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