JavaScript: How to pull out a string from a URL using a regular expression

Question

In java, I have this URL as a string:

window.location.href = 
"http://localhost:8080/bladdey/shop/c6c8262a-bfd0-4ea3-aa6e-d466a28f875/hired-3";

I want to create a javascript regular expression to pull out the following string:

c6c8262a-bfd0-4ea3-aa6e-d466a28f875

To find left hand marker for the text, I could use the regex:

window\.location\.href \= \"http\:\/\/localhost:8080\/bladdey\/shop\/

However, I don't know how to get to the text between that and /hired3"

What is the best way to pull out that string from a URL using javascript?

• bas: is the base fixed and the end always like /*?

Answer 1

You could split the string in tokens and look for a string that has 4 occurrences of -.

Or, if the base is always the same, you could use the following code:

String myString = window.location.href;
myString = myString.substring("http://localhost:8080/bladdey/shop/".Length());
myString = myString.subString(0, myString.indexOf('/'));

Answer 2

Use a lookahead and a lookbehind,

(?<=http://localhost:8080/bladdey/shop/).+?(?=/hired3)

Check here for more information.

Also, there is no need to escape the : or / characters.

• Breako Breako: does not compile. I get: Invalid escape sequence (valid ones are \b \t \n \f \r \" \' \ \ )
• Song Gao: Sorry, I copied your text for the first part of the regex. Like I said in the last part, there is no need to escape : or /. Edited.

Answer 3

You need a regex, and some way to use it...

 String theLocation = "http://localhost:8080/bladdey/shop/c6c8262a-bfd0-4ea3-aa6e-d466a28f8752/hired-3";
 String pattern = "(?</bladdey/shop/).+?(?=/hired3)";
 // Create a Pattern object
 Pattern r = Pattern.compile(pattern);
 // Now create matcher object.
 Matcher m = r.matcher(line);
 if (m.find( )) {
 System.out.println("Found value: " + m.group(0) );
 } else {
 System.out.println("NO MATCH");
 }

note - this will still work when you change the host (it only looks for bladdey/shop/)

• Breako Breako: does not compile. I get: invalid escape sequence (valid ones are \b \t \n \f \r \" \' \\ )
• Floris: Ah... I hate it when I get those wrong. Too many differences between different platforms. Since it didn't like my \/ I just took them out - apparently forward slashes are OK, only the backslash needs escaping. I think it should be fixed now.

Answer 4

You can use capturing groups to pull out some content of your string. In your case :

 Pattern pattern = Pattern.compile("(http://localhost:8080/bladdey/shop/)(.+)(/hired-3)");
 Matcher matcher = pattern.matcher(string);
 if(matcher.matches()){
 String value = matcher.group(2);
 }

• Breako Breako: how come you don't need to escape anything there?
• Paddle: @BreakoBreako There is no need to add escape characters in this case. You can try it, it works fine.

Answer 5

String param = html.replaceFist("(?s)^.*http://localhost:8080/bladdey/shop/([^/]+)/hired-3.*$", "1ドル");
if (param.equals(html)) {
 throw new IllegalStateException("Not found");
}
UUID uuid = new UUID(param);

In regex:

• (?s) let the . char wildcard also match newline characters.
• ^ begin of text
• $ end of text
• .* zero or more (*) any (.) characters
• [^...]+ one or more (+) of characters not (^) being ...

Between the first parentheses substitutes 1ドル.

• Breako Breako: Thanks but it does not include the hired-3 part.
• Song Gao: Actually the regex ([^/]+)/ means any sequence of characters that is not a forward slash, and is followed by a forward slash, is also valid.
• Joop Eggen: @BreakoBreako added forgotten hired-3.

Answer 6

Well if you want to pull out GUID from anything:

var regex = /[a-f0-9]{8}-[a-f0-9]{4}-[a-f0-9]{4}-[a-f0-9]{4}-[a-f0-9]{11,12}/i

It should really be {12} but in your url it is malformed and has just 15.5 bytes of information.