Java: method to get position of a match in a String?

JavaStringMatch

Java Problem Overview


String match = "hello";
String text = "0123456789hello0123456789";

int position = getPosition(match, text); // should be 10, is there such a method?

Java Solutions


Solution 1 - Java

The family of methods that does this are:

> Returns the index within this string of the first (or last) occurrence of the specified substring [searching forward (or backward) starting at the specified index].


String text = "0123hello9012hello8901hello7890";
String word = "hello";

System.out.println(text.indexOf(word)); // prints "4"
System.out.println(text.lastIndexOf(word)); // prints "22"

// find all occurrences forward
for (int i = -1; (i = text.indexOf(word, i + 1)) != -1; i++) {
	System.out.println(i);
} // prints "4", "13", "22"

// find all occurrences backward
for (int i = text.length(); (i = text.lastIndexOf(word, i - 1)) != -1; i++) {
	System.out.println(i);
} // prints "22", "13", "4"

Solution 2 - Java

This works using regex.

String text = "I love you so much";
String wordToFind = "love";
Pattern word = Pattern.compile(wordToFind);
Matcher match = word.matcher(text);

while (match.find()) {
	 System.out.println("Found love at index "+ match.start() +" - "+ (match.end()-1));
}

Output :

> Found 'love' at index 2 - 5

General Rule :

  • Regex search left to right, and once the match characters has been used, it cannot be reused.

Solution 3 - Java

text.indexOf(match);

See the String javadoc

Solution 4 - Java

#Finding a single index

As others have said, use text.indexOf(match) to find a single match.

String text = "0123456789hello0123456789";
String match = "hello";
int position = text.indexOf(match); // position = 10

#Finding multiple indexes

Because of @StephenC's comment about code maintainability and my own difficulty in understanding @polygenelubricants' answer, I wanted to find another way to get all the indexes of a match in a text string. The following code (which is modified from this answer) does so:

String text = "0123hello9012hello8901hello7890";
String match = "hello";

int index = text.indexOf(match);
int matchLength = match.length();
while (index >= 0) {  // indexOf returns -1 if no match found
    System.out.println(index);
    index = text.indexOf(match, index + matchLength);
}

Solution 5 - Java

You can get all matches in a file simply by assigning inside while-loop, cool:

$ javac MatchTest.java 
$ java MatchTest 
1
16
31
46
$ cat MatchTest.java 
import java.util.*;
import java.io.*;

public class MatchTest {
	public static void main(String[] args){
		String match = "hello";
		String text = "hello0123456789hello0123456789hello1234567890hello3423243423232";
		int i =0;
		while((i=(text.indexOf(match,i)+1))>0)
			System.out.println(i);
	}
}

Solution 6 - Java

int match_position=text.indexOf(match);

Solution 7 - Java

import java.util.StringTokenizer;

public class Occourence {

  public static void main(String[] args) {
	String key=null,str ="my name noorus my name noorus";		 
	int i=0,tot=0;

	StringTokenizer st=new StringTokenizer(str," ");
	while(st.hasMoreTokens())
	{   
		tot=tot+1;
		key = st.nextToken();
		while((i=(str.indexOf(key,i)+1))>0)
		{
			System.out.println("position of "+key+" "+"is "+(i-1));
		}
	}
	
	System.out.println("total words present in string "+tot);
  }
}

Solution 8 - Java

I have some big code but working nicely....

   class strDemo
   { 
       public static void main(String args[])
       {
       String s1=new String("The Ghost of The Arabean Sea");
           String s2=new String ("The");
           String s6=new String ("ehT");
           StringBuffer s3;
           StringBuffer s4=new StringBuffer(s1);
           StringBuffer s5=new StringBuffer(s2);
           char c1[]=new char[30];
           char c2[]=new char[5];
           char c3[]=new char[5];
           s1.getChars(0,28,c1,0);
           s2.getChars(0,3,c2,0);
           s6.getChars(0,3,c3,0); s3=s4.reverse();      
           int pf=0,pl=0;
           char c5[]=new char[30];
           s3.getChars(0,28,c5,0);
           for(int i=0;i<(s1.length()-s2.length());i++)
           {
               int j=0;
               if(pf<=1)
               {
                  while (c1[i+j]==c2[j] && j<=s2.length())
                  {           
                    j++;
                    System.out.println(s2.length()+" "+j);
                    if(j>=s2.length())
                    {
                       System.out.println("first match of(The) :->"+i);
                   
                     }
                     pf=pf+1;         
                  }   
             }                
       }       
         for(int i=0;i<(s3.length()-s6.length()+1);i++)
        {
            int j=0;
            if(pl<=1)
            {
             while (c5[i+j]==c3[j] && j<=s6.length())
             {
                 j++;
                 System.out.println(s6.length()+" "+j);
                 if(j>=s6.length())
                 {
                         System.out.println((s3.length()-i-3));
                         pl=pl+1;
                  
                 }   
                }                 
              }  
           }  
         }
       }

Solution 9 - Java

//finding a particular word any where inthe string and printing its index and occurence  
class IndOc
{
	public static void main(String[] args) 
	{
		String s="this is hyderabad city and this is";
		System.out.println("the given string is ");
		System.out.println("----------"+s);
		char ch[]=s.toCharArray();
		System.out.println(" ----word is found at ");
		int j=0,noc=0;
		for(int i=0;i<ch.length;i++)
		{
			j=i;
			
			if(ch[i]=='i' && ch[j+1]=='s')
			{
				System.out.println(" index "+i);
			noc++;	
			}
          
		}
		System.out.println("----- no of occurences are "+noc);
		
	}
}

Solution 10 - Java

	String match = "hello";
	String text = "0123456789hello0123456789hello";

	int j = 0;
	String indxOfmatch = "";
			
	for (int i = -1; i < text.length()+1; i++) {
		j =  text.indexOf("hello", i);
		if (i>=j && j > -1) {
			indxOfmatch += text.indexOf("hello", i)+" ";
		}
	}
	System.out.println(indxOfmatch);

Solution 11 - Java

If you're going to scan for 'n' matches of the search string, I'd recommend using regular expressions. They have a steep learning curve, but they'll save you hours when it comes to complex searches.

Solution 12 - Java

for multiple occurrence and the character found in string??yes or no

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class SubStringtest {

	public static void main(String[] args)throws Exception {
	BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	 System.out.println("enter the string");
	String str=br.readLine();
	System.out.println("enter the character which you want");
	CharSequence ch=br.readLine();   
	boolean bool=str.contains(ch);
	System.out.println("the character found is " +bool);
	int position=str.indexOf(ch.toString());
	
	while(position>=0){
		System.out.println("the index no of character is " +position); 
		position=str.indexOf(ch.toString(),position+1);
	}
	

	}

}

Solution 13 - Java

public int NumberWordsInText(String FullText_, String WordToFind_, int[] positions_)
   {
    int iii1=0;
    int iii2=0;
    int iii3=0;
    while((iii1=(FullText_.indexOf(WordToFind_,iii1)+1))>0){iii2=iii2+1;}
    // iii2 is the number of the occurences
    if(iii2>0) {
        positions_ = new int[iii2];
        while ((iii1 = (FullText_.indexOf(WordToFind_, iii1) + 1)) > 0) {
            positions_[iii3] = iii1-1;
            iii3 = iii3 + 1;
            System.out.println("position=" + positions_[iii3 - 1]);
        }
    }
    return iii2;
}

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