Javascript RegEx Question

Question

Hello,

Suppose I have this:

$('ul.child').slideUp('slow');

What would be the regex to find 'ul.child' and 'slow' (including quotes) in the above expression.

Answer 1

Try

/('[^']*?')/

e.g.

var test = "$('ul.child').slideUp('slow');";
matches = test.match(/('[^']*?')/g);

That's the closest I could get in a hurry. Hope this helps. When I come back, I'll tackle this

Answer 2

Don't know why you possibly will need that but that will do the trick.

var str = "$('ul.child').slideUp('slow');";
var matches = str.match(/\$\((.+)\)\.slideUp\((.+)\);/);
console.log(matches[1]); // 'ul.child' (with quotes)
console.log(matches[2]); // 'slow' (with quotes)

And yes, matches indices are correct, matches[0] contains the whole matched part, while indices >= 1 contain matched groups

Answer 3

This should do it:

var a = "$('ul.child').slideUp('slow');";  
var matches = a.match(/'[\w.]*'/g));
// matches[0] = 'ul.child'
// matches[1] = 'slow'

g is a modifier and matches every occurrence of the expression.

If you want to match more expressions, like ul li, ul + li or ul, li, you have to put those additional characters into the character class.

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Update 1 was not helping.

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Update2:

You had a slight mistake in one of your regex. Change this:

// single quote string
the_code = the_code.replace(/('.+')/g,'<span class="code_string">$1</span>');

to

// single quote string
the_code = the_code.replace(/('.+?')/g,'<span class="code_string">$1</span>')

You have to make it non-greedy (with the ?) in order to not match to the last occurrence of a ', but to the next one.

See here: http://jsbin.com/azovo3/4

If you want to match single and double quote, chage it to this:

the_code = the_code.replace(/(('|").+?("|'))/g,'<span class="code_string">$1</span>');

Answer 4

/^[^']*?('[\w\.]*')*$/

I did this on the following regex tester page and it found both of your strings:

RegexPal

You can use capture groups to get the matches easily. The expression goes:

1. Anything not a single quote, lazy match all [^']*?
2. Capture anything starting with a quote, containing any number of word characters or periods, followed by a single quote greedy match all ('[\w\.]*')*

Is that close to what you want?