php用正则表达式获取css

Question

I'm trying to get css selector + styles with a regexp but there is always a problem...

So do you have an idea (exemple ?) of a regexp do get this:

array(
[0] => "div {
    background: red;
    color: blue;
}",
[1] => "section {
    /* comment with bad idea } <- this break regexp [^}]+ */
    background: blue;
    color: red;
}"
);

fom this css:

div {
    background: red;
    color: blue;
}

section {
    /* comment with bad idea } <- this break regexp [^}]+ */
    background: blue;
    color: red;
}

Thx

Answer 1

I finally did it by another way. First, I start by quoting the undesired symbols with these regex

string = string
.replace(/(')([^']*)(')/ig, function(found){

    return found.replace(/([\[\]\s\,\#\.\:\>\~])/ig, "\/\/$1\\\\");

})
.replace(/(\")([^\"]*)(\")/ig, function(found){

    return found.replace(/([\[\]\s\,\#\.\:\>\~])/ig, "\/\/$1\\\\");

});

Then, I split the string with another regexp:

string = string.split(/(?!\/\/)([\s\>\,])(?!\\\\)|(?=(?!\/\/)[\#\.\[\:](?!\\\\))/ig);

And, to finish, if( string[x].match(/^[a-z]+$/ig) ){..}; it is a tagName

If you see something bad, I'm listening to you.

Edit :

Oups ! I forgot my answer was about php, sorry. But I apply the same method for php, except lookBefore simplify and reduce the regexp to only one, then split...

J