I have a string
string = "width: 10px; height: 20px; whatever: 0px;";
I wish to change the height part to 50px and leave the rest in tact so that the end result looks like
string = "width: 10px; height: 50px; whatever: 0px;";
I have tried string.replace("height: .*px;", "height: 50px;"); however this doesn't appear to work.
Can someone help me please
Guilherme's answer is partially correct: you do need to pass in a regular expression. There's another problem with your regex, though: your repetition (.* right before px) is greedy: it will match everything it can. So if you have height: 20px; whatever: 0px, it will greedily match 20px; whatever: 0px (see how it consumed everything up to the last px?). You will have to change it to a lazy match by appending the repetition with a ?:
string.replace( /height: .*?px;/, "height: 50px;" );
To make it more robust still, you might want to account for a variable amount of whitespace, or if the height occurs at the end of the string (which I use a lookahead for):
string.replace( /height:\s*\d+px(?=;|$)/, 'height: 50px' );
Note I also used a digit specifier (\d) instead of the wildcard metacharacter (.). Obviously this will only work if height will always be specified in integer units.
You need to change the regexp to make it non-greedy
"width: 10px; height: 20px; whatever: 0px;".replace(/height: .*?px;/, "height: 50px;")
Note the question mark after star - that tells the regexp engine to find the shortest possible match, i.e. height: 20px; instead of the longest, height: 20px; whatever: 0px;
If you pass a string as the first argument of your replace method, it will try to replace your string literally. To use regular expressions, you must wrap it with the / character:
yourString = yourString.replace(/height: .*?px;/, "height: 50px;");
I agree with @JamesMontagne that maybe there is a better solution for what you are trying to do. It seems a bit hacky.
