CODEWITHSUNDEEP
javaregex
iceDice
iceDice

Reputation: 79

Java regex - parts of words

Is it possible to write regex in Java that matches word parts? For example I want to search for strings 'ab', 'cd' and 'ef'. A match should be returned in following examples:

[lab stef ecde], [lecde effe aaaab]

So a match should be returned if all strings ('ab', 'cd', 'ef') are parts of words anywhere in the text - order is not imported. However match shouldn't be returned if any of strings are missing

[lab stef]

Upvotes: 0

Views: 199

Answers (3)

Pshemo
Pshemo

Reputation: 124275

If it doesn't have to be regex then Tichodroma's answer is the one you are looking for.

But if you really need to complicate your life and use regex you can try to use look-around mechanisms like look ahead and create something like

"lab stef ecde".matches("(?=.*ab)(?=.*cd)(?=.*ef).*") //true
"lab stef".matches("(?=.*ab)(?=.*cd)(?=.*ef).*") //false

to explain it more clearly: in

(?=.*ab)(?=.*cd)(?=.*ef).*
  • (?=.*ab) will check if your string contains .*ab where .* will match any characters before ab part.
  • also look-ahead (?=...) is zero-width which means that it will reset cursor to the position where it was before look-ahead started so in our case it will still be at start of string
  • this way we can use again (?=.*cd) and (?=.*ef)
  • but we also need to include in our regex .* at the end, because matches check if entire string matches our regex, so we need to somehow iterate over entire string.

Upvotes: 6

Multithreader
Multithreader

Reputation: 878

This will do:

^.*(ab.*cd.*ef|ab.*ef.*cd|cd.*ab.*ef|cd.*ef.*ab|ef.*ab.*cd|ef.*cd.*ab).*$

You can test it here: http://www.regexplanet.com/advanced/java/index.html

I believe it is an overkill though. Another optimized solution would be better.

Upvotes: 0

user1907906
user1907906

Reputation:

Find every substring in the input and && the resulting boolean values.

String s = "lab stef ecde";
boolean ab = s.indexOf("ab") > -1;
boolean cd = s.indexOf("cd") > -1;
boolean ef = s.indexOf("ef") > -1;
boolean match = ab && cd && ef; // true

Edit

In Germany there is a proverb:

Warum einfach wenn es auch kompliziert geht?

"Why simple when you can do it complicated?"

That's what I think about regular expressions in this case.

Upvotes: 5

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