Search for string in 2 dimensional string array

You already have the loop for that. So you can do :

if(sTabell[i][0].equals("Sandefjord")){System.out.println("Sandefjord found at "+i);}

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after the sort() of course.

The problem with a binary search is that the array/collection must be sorted first. If, as you are, you are sorting by row, then the search token would have to be a whole row, not just a string in a row. What is your requirement – to find the indices of a complete string, or a partial string?
An iterative approach is probably right. Try

    public static int[] findMatch(String[][] table, String pattern) {
        int[] result = { -1, -1 };
        for (int row = 0; result[0] < 0 && row < table.length; row++) { 
            for (int col = 0; result[1] < 0 && col < table[row].length; col++) {
                if (table[row][col].matches(pattern)) {
                    result[0] = row;
                    result[1] = col;
                }
            }
        }
        return result;
    }

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Thanks CEHJ
The main idea was to create the fastest serch, and binarySearch seems to be a quick one (hence the sorting).
Your suggestion is a sequential search so sorting gives no benefit.

When you say the search token needs to be the whole row, how will that look like in a binary search (to take advantage of the sorting).
The row in question holds "Sandefjord" and "ei" (column 0 an 1)

Appreciate your patience :)

not a real world problem yet,  but what would you suggest for eg 5000 rows with names like this read from a database ?

You would run that search on the database

Multidimensional arrays - apart from maybe holding co-ordinates for graphic applications or some maths operations - seem to me to be a bit of a throwback to the days when less sophisticated tools were available. So although I don't have any more interesting insight than that, I am making this comment because, as has been said, it's not completely clear what it is you are trying to achieve - with database involvement or without. Why are you holding such a small amount of related data in this form ? I'm not getting involved in the database conversation, because I don't tread in that territory, but in OOP terms I would venture to say that a more conventional approach would be to hold this dimensioned data as objects with multiple fields of a class. What would an entry like :

Sandefjord  ei

be doing floating around on its own ? Maybe I've missed something, or you have another particular aim in mind that I can't see.

@both:
The above is more like a test on how to do it.
I.e. how to create and navigate in a 2dim string table/data structure and how to do a binary search.
What would be the most efficient data structure to do this in Java (i.e. not involving a database)?

Well, as I've indicated above, a binary search is not really practicable. You would have to know the whole row to be able to call a binary search on it. If you knew the whole row, then you'd presumably also know what its position was…

OK CEHJ
Thanks for the update. Appreciated :)
Case closed for now.

In case you inadvertently 'left early' again, I thought you might like to see my take on the OOP paradigm that I commented on, (money where my mouth was stuff) which holds individual pairs of your String array variables in single Objects (Twin, for want of a better class name), and maintains the array of those Twins in a separate class object called Creator (again for want of a better name).

Once the twins are created and stored in a HashSet, (so you can't have two identical entries), they can be sorted (in my approach, I decided to use an array). The first sort() will produce a first-field sort, and the second routine will do a subsort on the second field.

import java.util.Arrays;
import java.util.Comparator;
import java.util.Iterator ;
import java.util.HashSet ;


public class Creator {

    public static void main(String [] args) {
    
             int place = 0;
    
             HashSet<Twin> ts = new HashSet<Twin>() ;
           
             Twin twin = new Twin("Ola","var");ts.add(twin);
             twin = new Twin("fra","Sandefjord");ts.add(twin);
             twin = new Twin("Han","var");ts.add(twin);
             twin = new Twin("lettmatros","ombord!");ts.add(twin);
             twin = new Twin("Ola","kunne");ts.add(twin);
             twin = new Twin("engelsk","ei");ts.add(twin);
             twin = new Twin("klarte","han");ts.add(twin);
             twin = new Twin("Ola","seg");ts.add(twin);
             
             Iterator it = ts.iterator();
             
             Twin[] twa = new Twin[ts.size()];
             
             while(it.hasNext()){twa[place++]=(Twin)it.next();}
            
                    
            Arrays.sort(twa, new Comparator<Object>() {      
        
                public int compare(Object a, Object b) {
              
                    return (((Twin)a).first.compareToIgnoreCase(((Twin)b).first));
                    
                }
            });
            
            /*** Include the routine below if a subsort (i.e. on the second field) is required.
            */
            Arrays.sort(twa, new Comparator<Object>() {      
        
                public int compare(Object a, Object b) {
              
                    if(((Twin)a).first.equals(((Twin)b).first)){
                        return (((Twin)a).second.compareToIgnoreCase(((Twin)b).second));
                    }
                    else{return ((Twin)a).first.compareToIgnoreCase(((Twin)b).first) ;}
                }
            });
            

            for (int i = 0; i < twa.length; i++) {
            
            System.out.println(twa[i].first+" "+twa[i].second) ;
            
            } 
              
      }   
}



 class Twin {


 public String first ;
 public String second ;


 public Twin(String f, String s){

    this.first = f ;
    this.second = s ;

 }

 public Twin(){}

 }

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I've put both classes in the same file, to make it easier.
 

Thanks a lot krakatoa !
Very useful sample :)

I have to admit that I in the end ended up with a HashMap structure to solve my issue (as there was a few other additional things to include in the code )
Thanks !

ok.