Java Regex Pattern to extract data

Question

I have incoming data something like this

http://localhost:1111/search?id=10&time=3200&type=abc
http://localhost:1111/search?time=3200&id=11&type=abc
http://localhost:1111/search?id=12
http://localhost:1111/search?id=13&time=3200&type=abc

The data is varying but not something completely random or unpredictable.

So basically how do we extract what are the IDs that are incoming in each string ignoring rest of the junk?

@DaveNewton, I gave sample data. It's not the exact data. :) More or less same phenomenon — RaceBase
– RaceBase, Commented Jan 7, 2014 at 16:39
Then it's misleading to show only URLs. Without knowing any of the data's constraints it's difficult to provide concrete advice. — Dave Newton
– Dave Newton, Commented Jan 7, 2014 at 16:41
i don't know how to do substring matching in Java but I'm sure it's there and easy to do with a regex such as the following: .+?id=(\d+).*? — tenub
– tenub, Commented Jan 7, 2014 at 16:42
"Some people, when confronted with a problem, think "I know, I'll use regular expressions." Now they have two problems." Jamie Zawinski — Filburt
– Filburt, Commented Jan 7, 2014 at 16:42
@Reddy your sample data seems to suggest you are working with parameters... not sure if you could just work with ServletRequest or HttpServletRequest objects directly? — user1766760
– user1766760, Commented Jan 7, 2014 at 16:46

arshajii · Accepted Answer · 2014-01-07 16:40:47Z

2

You can try using the regex id=(\d+) and extract the value of the first capturing group:

String url = "http://localhost:1111/search?id=10&time=3200&type=abc";

Pattern id = Pattern.compile("id=(\\d+)");

Matcher m = id.matcher(url);
if (m.find())
    System.out.println(m.group(1));

See Pattern and Matcher.

answered Jan 7, 2014 at 16:40

arshajii

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4 Comments

RaceBase Over a year ago

This is awesome. Simply works. I always thought Pattern works in some other ways where we need to include complex regex pattern :)

arshajii Over a year ago

@Reddy The regex you want seems to be pretty straightforward :) Glad I could help.

RaceBase Over a year ago

@arashajii, one more help, if I want extract 'type'? I tried similar, is not working.

arshajii Over a year ago

@Reddy Well \d matches digits, and it looks like type is composed of non-digits as well. You can try "type=(\\w+)".

Алексей · Accepted Answer · 2014-01-07 16:43:45Z

2

What if there are several ID's that are passed (which is valid)?

IMHO I wouild rather do somethis more like this:

URL url = new URL(<your link>);
String queryString = url.getQuery();

parse queryString into map for example of <String,List<String>> and get the value of ID key

answered Jan 7, 2014 at 16:43

Алексей

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Comments

aliteralmind · Accepted Answer · 2014-01-07 16:42:29Z

1

(?<=[?&])id=(\d+)(?=(?:\&|$))

works in Regex Buddy under the Java and Perl flavor, but not in TextPad, which uses the Boost regex engine. Boost has issues with back-references.

(?<=(?:
   [?&]    //PRECEDED BY a question-mark or ampersand
))          
   id=(\d+) //"id=[one-or-more-digits]"
(?=(?:
   \&|$     //FOLLOWED BY an ampersand or the end of the input
))

This captures the digits only, and avoids issues such as capturing incorrect fields like

anotherid=123sometext

answered Jan 7, 2014 at 16:42

aliteralmind

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Comments

gtgaxiola · Accepted Answer · 2014-01-07 16:51:37Z

Expanding on @user1631616's answer:

Here is a sample code:

public static void main(String[] args) throws MalformedURLException {         
    URL aURL = new URL("http://localhost:1111/search?id=10&time=3200&type=abc");

    HashMap<String, String> params = new HashMap<>();
    String[] query = aURL.getQuery().split("&");
    for(String s: query) {
        String[] split = s.split("=");
        params.put(split[0],split[1]);
    }
    System.out.println(params.get("id")); 
    System.out.println(params.get("type")); 
    System.out.println(params.get("time")); 

}

That way if your HashMap param returns null you know that value was not set on the query string.

And also don't have to worry about the ordering of the parameters.

Manas Bajaj · Accepted Answer · 2014-01-07 16:58:49Z

1

Why exactly do you want to use a regular expression to do this?

I would do it like this:

String url = "http://localhost:1111/search?id=13&time=3200&type=abc";
     String[] split = url.split("&");
     String id = "";    
     for (String s : split){
         if (s.contains("id")){
             id = s.substring(s.indexOf("id=")+3, s.length());
         }
     }

     System.out.println(id);

13

edited Jan 7, 2014 at 16:58

answered Jan 7, 2014 at 16:49

Manas Bajaj

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Comments

Joao Raposo · Accepted Answer · 2014-01-07 16:44:23Z

0

Something like this should do what you want:

(?<=id=)\d+

answered Jan 7, 2014 at 16:44

Joao Raposo

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