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?

is the base fixed and the end always like /*?

bas
– bas

2013-07-29 14:55:31 +00:00
Commented Jul 29, 2013 at 14:55 — bas
– bas, Commented Jul 29, 2013 at 14:55

Eric Leschinski · Accepted Answer · 2013-07-29 15:23:41Z

2

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('/'));

edited Jul 29, 2013 at 15:23

Eric Leschinski

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answered Jul 29, 2013 at 14:57

bas

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Comments

Song Gao · Accepted Answer · 2013-07-29 15:33:05Z

1

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.

edited Jul 29, 2013 at 15:33

answered Jul 29, 2013 at 14:56

Song Gao

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2 Comments

Breako Breako Over a year ago

does not compile. I get: Invalid escape sequence (valid ones are \b \t \n \f \r \" \' \ \ )

Song Gao Over a year ago

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.

Floris · Accepted Answer · 2013-07-29 16:04:10Z

1

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/)

edited Jul 29, 2013 at 16:04

answered Jul 29, 2013 at 15:01

Floris

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2 Comments

Breako Breako Over a year ago

does not compile. I get: invalid escape sequence (valid ones are \b \t \n \f \r \" \' \\ )

Floris Over a year ago

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.

Paddle · Accepted Answer · 2013-07-29 16:19:01Z

1

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);
    }

edited Jul 29, 2013 at 16:19

answered Jul 29, 2013 at 15:00

Paddle

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2 Comments

Breako Breako Over a year ago

how come you don't need to escape anything there?

Paddle Over a year ago

@BreakoBreako There is no need to add escape characters in this case. You can try it, it works fine.

Joop Eggen · Accepted Answer · 2013-07-29 16:45:27Z

1

String param = html.replaceFist("(?s)^.*http://localhost:8080/bladdey/shop/([^/]+)/hired-3.*$", "$1");

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.

edited Jul 29, 2013 at 16:45

answered Jul 29, 2013 at 15:05

Joop Eggen

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3 Comments

Breako Breako Over a year ago

Thanks but it does not include the hired-3 part.

Song Gao Over a year ago

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 Over a year ago

@BreakoBreako added forgotten hired-3.

Esailija · Accepted Answer · 2013-07-29 15:53:27Z

0

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.

answered Jul 29, 2013 at 15:53

Esailija

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