I need to catch word in some standard string (link id actually), but not to catch all string.
This
(?:color_form_submit_)(\w+)
works in ruby and catches yellow in color_form_submit_yellow
But in JS it seems to catch all string, I can't figure out why.
I wanted to achieve it one regex just for elegance
If you want elegance, then you could try something like this
var str = 'color_form_submit_yellow',
reg = /color_form_submit_(\w+)/;
(str.match(reg) || [])[1]; // "yellow"
and you also have
('fail'.match(reg) || [])[1]; // undefined
This works because
[1, 2, 3] || []; // [1, 2, 3]
null || []; // []
array[1] === (array)[1];
Unfortunately, JS doesn't support lookbehinds and failed match returns null, not an empty array. So the only way to do that in one expression, without if conditions, is something like
color = (str.match(/color_form_submit_(\w+)/) || []).pop()
Example:
> ("color_form_submit_black".match(/color_form_submit_(\w+)/) || []).pop()
"black"
> ("foo bar".match(/color_form_submit_(\w+)/) || []).pop()
undefined
The advantage of this construct compared to the if statement is that it can be used in an expression context, for example, in a function call:
myRegex = /color_form_submit_(\w+)/
showColor((myLabel.match(myRegex) || []).pop())
[1] instead of .pop(), which is what I did in my (now deleted) answer.You need to explicitly look for the match to the first capturing group:
var myregexp = /color_form_submit_(\w+)/;
var match = myregexp.exec(subject);
if (match != null) {
result = match[1];
}
(?<color_form_submit_)\w+, but not JavaScript.
var match = string.match(/color_form_submit_(\w+)/);
match && match[1];
would be the shortest (unnecessarily obfuscated) way you can get in javascript.
'color_form_submit_yellow'.match(/color_form_submit_(\w+)/)[1]