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up vote 2 down vote favorite

I have a String that looks like this 

The#red#studio#502#4

I need to split it into 3 different Strings in the array to be

s[0] = "The red studio"
s[1] = "502"
s[2] = "4"

The problem is the first one should have only words and the second and third should have only numbers...

I was trying to play with the s.split() Method, but no luck.

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have you checked out javadocs for string.split() method ... –  WeloSefer Jan 18 at 3:48
 
Have you tried StringTokenizer .docs.oracle.com/javase/1.4.2/docs/api/java/util/… –  SAN3 Jan 18 at 3:50
5  
You need to understand the specific rules governing the input data. Is it always Some#text#with#words#number#number? –  Jim Garrison Jan 18 at 3:51
1  
What exactly went wrong while using s.split? –  Swapnil Jan 18 at 3:54
1  
Lookahead mechanism might be useful in your case, although it will give you only split like The#red#studio 502 4 so you will have to replace # in first part. –  Pshemo Jan 18 at 3:57
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4 Answers

up vote 7 down vote accepted 
String s= "The#red#studio#502#4";
String[] array = s.split("#(?=[0-9])");
for(String str : array)
{
  System.out.println(str.replace('#',' '));
}

Output: 

The red studio  
502  
4

Ideone link.

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Can you explain that regex? Or at least tell us what that term is called so I can google it. Thanks –  tieTYT Jan 18 at 4:16
1  
I knew the keywords 'Lookahead' and 'Lookbehind'. Quick search gave me this. Basically, I split when there is a hash character and look for a decimal but I don't split on it, but rather split only on # –  Srinivas Jan 18 at 4:19
up vote 0 down vote

Simply try: 'String s[]= yourString.split("#")' it will return string array....

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up vote 0 down vote

I've decided to edit out my impl because I think that @Srinivas's is more elegant. I'm leaving the rest of my answer though because the tests are still useful. It passes on @Srinivas's example too. 

package com.sandbox;

import com.google.common.base.Joiner;
import org.apache.commons.lang.StringUtils;
import org.junit.Test;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;

import static org.junit.Assert.assertEquals;

public class SandboxTest {

    @Test
    public void testQuestionInput() {
        String[] s = makeResult("The#red#studio#502#4");
        assertEquals(s[0], "The red studio");
        assertEquals(s[1], "502");
        assertEquals(s[2], "4");
    }

    @Test
    public void testAdditionalRequirement() {
        String[] s = makeResult("The#red#studio#has#more#words#502#4");
        assertEquals(s[0], "The red studio has more words");
        assertEquals(s[1], "502");
        assertEquals(s[2], "4");
    }

    private String[] makeResult(String input) {
        // impl inside
    }
}
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Can it be made more concise by only using split? –  Srinivas Jan 18 at 4:05
 
It is using split –  tieTYT Jan 18 at 4:05
 
only using split –  Srinivas Jan 18 at 4:07
1  
I don't know how, but it seems that you do. Your example also passes my tests. –  tieTYT Jan 18 at 4:16
up vote 0 down vote

I dont undestand. Your pattern is fixed. You final array always has 3 elements, then why dont just do it manually

Btw, my 2-cent code :D: 

String  a = "The#red#studio#502#4";
String d [] = {a.split("#[0-9]+#[0-9]+")[0].replace("#", " "), a.split("#")[(a.split("#").length - 2)], a.split("#")[(a.split("#").length - 1)]};
System.out.println(Arrays.toString(d));
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