regular expression java for URL parameter string

Question

I am trying to verify if the string match a regular expression or not. The URL format is : key=value&key=value&....

Key or value can be empty.

My code is :

Pattern patt = Pattern.compile("\\w*=\\w*&(\\w *=\\w*)* ");
Matcher m = patt.matcher(s);
if(m.matches()) return true;
else return false;

when i enter one=1&two=2, it shows false whereas it should show true. Any idea !

By the way: Replace `if A return true; else return false;` with `return A;` — Florian Albrecht, Jun 23 '17 at 09:50
@GiladGreen Nope, his regex is good. He just needs to escape \, as it is a Java string. — Mistalis, Jun 23 '17 at 09:53
See http://ideone.com/QjYyZG. You need `"\\w+=\\w+(?:&\\w+=\\w+)*"` regex. The `\w+` will match 1 or more word chars, `\w*` would allow an empty string. Well, maybe you really want to allow `=&=&=&=`, no idea. Then use `"\\w*=\\w*(?:&\\w*=\\w*)*"`. *What are the requirements?*. BTW, added another dupe - [Validate URL query string with regex](https://stackoverflow.com/questions/23959352/validate-url-query-string-with-regex). — Wiktor Stribiżew, Jun 23 '17 at 10:22
Regarding the update: did you really want to match the spaces before the second `=`? — Henry, Jun 23 '17 at 10:29
key=value&key=value&.... Empty key or value are accepted but '=' must be present (e.g. "=value", "key="). — Amir Choubani, Jun 23 '17 at 10:31
@WiktorStribiżew I tried your suggestion but it does not work — Amir Choubani, Jun 23 '17 at 10:36
@AmirChoubani: With your input string, [it works](http://ideone.com/QjYyZG) as it outputs *true*. — Wiktor Stribiżew, Jun 23 '17 at 10:40
Look at the code. `String s = "one=1&two=2";`. Just what *you* provided. — Wiktor Stribiżew, Jun 23 '17 at 10:42
Well, you may also try http://ideone.com/Sa7WNv. It will match key-values where either a key or value can be empty. Is that better? `(?:\\w+=\\w*|=\\w+)(?:&(?:\\w+=\\w*|=\\w+))*` — Wiktor Stribiżew, Jun 23 '17 at 10:48

score 4 · Accepted Answer · answered Jun 23 '17 at 10:54

The regex you need is

Pattern.compile("(?:\\w+=\\w*|=\\w+)(?:&(?:\\w+=\\w*|=\\w+))*");

See the regex demo. It will match:

(?:\\w+=\\w*|=\\w+) - either 1+ word chars followed with = and then 0+ word chars (obligatory key, optional value) or = followed with 1+ word chars (optional key)
(?:&(?:\\w+=\\w*|=\\w+))* - zero or more of such sequences as above.

Java demo:

String s = "one=1&two=2&=3&tr=";
Pattern patt = Pattern.compile("(?:\\w+=\\w*|=\\w+)(?:&(?:\\w+=\\w*|=\\w+))*");
Matcher m = patt.matcher(s);
if(m.matches()) {
    System.out.println("true");
} else {
    System.out.println("false");
}
//  => true

To allow whitespaces, add \\s* where needed. If you need to also allow non-word chars, use, say, [\\w.-] instead of \w to match word chars, . and - (keep the - at the end of the character class).

regular expression java for URL parameter string

1 Answers1