Javascript regex to extract all characters between quotation marks following a specific word

Question

I have the text:

s.events="event3"
s.pageName="Forum: Index"
s.channel="forum"
s.prop1="Forum: Index"
s.prop2="Index Page"
s.prop36=""
s.prop37=""
s.prop38=""
s.prop39=""
s.prop40="53"
s.prop41="Anonymous"
s.prop42="username"
s.prop43=""
s.prop47=""
s.eVar1="Forum: Index"
s.eVar2="Index Page"
s.eVar36=""
s.eVar37=""

saved in a var in javascript and I want to extract the text between the quotes of s.prop42 giving me the result:

"username"

what I have right now is

    var regex = /\?prop.42="([^']+)"/;
    var test = data.match(regex);

but it doesnt seem to work, can someone help me out?

Why are you trying to match a literal question mark? Why are you not allowing *single* quotes in a *double*-quote terminated string? — Bergi, Jul 23 '14 at 02:59

score 2 · Accepted Answer · answered Jul 23 '14 at 02:54

2

Use this:

var myregex = /s\.prop42="([^"]*)"/;
var matchArray = myregex.exec(yourString);
if (matchArray != null) {
    thematch = matchArray[1];
}

In the regex demo, look at the capture group in the right pane.

Explanation

s\.prop42=" matches s.prop42=" (but we won't retrieve it)
The parentheses in ([^"]*) capture any chars that are not a " to Group 1: this is what we want
The code gets the Group 1 capture

answered Jul 23 '14 at 02:54

zx81

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You're welcome, and thanks for the green checkmark! :) – zx81 Jul 23 '14 at 03:23

score 0 · Answer 2 · answered Jul 23 '14 at 03:06

0

Can't comment on above answer, but I think the regex is better with .* like so:

var myregex = /s\.prop42="(.*)"/;
var matchArray = myregex.exec(yourString);
if (matchArray != null) {
    thematch = matchArray[1];
}

answered Jul 23 '14 at 03:06

Michael Y.

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12

Javascript regex to extract all characters between quotation marks following a specific word

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