CODEWITHSUNDEEP
javaregex
John Daly
John Daly

Reputation: 480

How to use regular expressions to match everything before a certain type of word

I am new to regular expressions.

Is it possible to match everything before a word that meets a certain criteria:

E.g.

THIS IS A TEST - - +++ This is a test

I would like it to encounter a word that begins with an uppercase and the next character is lower case. This constitutes a proper word. I would then like to delete everything before that word.

The example above should produce: This is a test

I only want to this processing until it finds the proper word and then stop.

Any help would be appreciated.

Thanks

Upvotes: 20

Views: 55932

Answers (5)

Bill K
Bill K

Reputation: 62759

I really don't get why people go to regular expressions so quickly.

I've done a lot of string parsing (Used to screen-scrape vt100 menu screens) and I've never found a single case where Regular Expressions would have been much easier than just writing code. (Maybe a couple would have been a little easier, but not much).

I kind of understand they are supposed to be easier once you know them--but you see someone ask a question like this and realize they aren't easy for every programmer to just get by glancing at it. If it costs 1 programmer somewhere down the line 10 minutes of thought, it has a huge net loss over just coding it, even if you took 5 minutes to write 5 lines.

So it's going to need documentation--and if someone who is at that same level comes across it, he won't be able to modify it without knowledge outside his domain, even with documentation.

I mean if the poster had to ask on a trivial case--then there just isn't such thing as a trivial case.

public String getRealText(String scanMe) {
    for(int i=0 ; i < scanMe.length ; i++)
        if( isUpper(scanMe[i]) && isLower(scanMe[i+1]) )
            return scanMe.subString(i);
return null; }

I mean it's 5 lines, but it's simple, readable, and faster than most (all?) RE parsers. Once you've wrapped a regular expression in a method and commented it, the difference in size isn't measurable. The difference in time--well for the poster it would have obviously been a LOT less time--as it might be for the next guy that comes across his code.

And this string operation is one of the ones that are even easier in C with pointers--and it would be even quicker since the testing functions are macros in C.

By the way, make sure you look for a space in the second slot, not just a lower case variable, otherwise you'll miss any lines starting with the words A or I.

Upvotes: 1

Jon Skeet
Jon Skeet

Reputation: 1499860

Having woken up a bit, you don't need to delete anything, or even create a sub-group - just find the pattern expressed elsewhere in answers. Here's a complete example:

import java.util.regex.*;

public class Test
{
    public static void main(String args[])
    {
        Pattern pattern = Pattern.compile("[A-Z][a-z].*");

        String original = "THIS IS A TEST - - +++ This is a test";
        Matcher match = pattern.matcher(original);
        if (match.find())
        {
            System.out.println(match.group());
        }
        else
        {
            System.out.println("No match");
        }        
    }
}

EDIT: Original answer

This looks like it's doing the right thing:

import java.util.regex.*;

public class Test
{
    public static void main(String args[])
    {
        Pattern pattern = Pattern.compile("^.*?([A-Z][a-z].*)$");

        String original = "THIS IS A TEST - - +++ This is a test";
        String replaced = pattern.matcher(original).replaceAll("$1");

        System.out.println(replaced);
    }
}

Basically the trick is not to ignore everything before the proper word - it's to group everything from the proper word onwards, and replace the whole text with that group.

The above would fail with "*** FOO *** I am fond of peanuts" because the "I" wouldn't be considered a proper word. If you want to fix that, change the [a-z] to [a-z\s] which will allow for whitespace instead of a letter.

Upvotes: 3

hhafez
hhafez

Reputation: 39750

then you can do something like this

'.*([A-Z][a-z].*)\s*'

.* matches anything
( [A-Z] #followed by an uper case char 
  [a-z] #followed by a lower case 
  .*)   #followed by anything
  \s*   #followed by zeror or more white space

Which is what you are looking for I think

Upvotes: 0

Tomalak
Tomalak

Reputation: 338148

Replace

^.*?(?=[A-Z][a-z])

with the empty string. This works for ASCII input. For non-ASCII input (Unicode, other languages), different strategies apply.

Explanation

.*?    Everything, until
(?=    followed by
[A-Z]  one of A .. Z and
[a-z]  one of a .. z
)

The Java Unicode-enabled variant would be this:

^.*?(?=\p{Lu}\p{Ll})

Upvotes: 49

Maiku Mori
Maiku Mori

Reputation: 7469

([A-Z][a-z].+)

would match:

This is a text

Upvotes: 0

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