CODEWITHSUNDEEP
androidarrayslistview
user1592708
user1592708

Reputation: 21

Onlistitemclick not Working

I am loading an array into a listview and when I click on the list item, I want to display the item in a textview within the same activity. The list is currently loading, but the program crashes when an item is clicked. Any thoughts why?

import java.util.ArrayList;
import android.app.ListActivity;
import android.content.res.Resources;
import android.os.Bundle;
import android.view.View;
import android.widget.ArrayAdapter;
import android.widget.ListView;
import android.widget.TextView;

public class WordList extends ListActivity {
    /**
     * @see android.app.Activity#onCreate(Bundle)
     * 
    */
    //public final static String TERM_EXTRA="com.myapp._word";
    //public final static String DEF_EXTRA="com.myapp._pic";

    TextView selection;
    ArrayList<String> words=new ArrayList<String>();


    @Override
    public void onCreate(Bundle icicle) {
        super.onCreate(icicle);
        setContentView(R.layout.main);
        selection=(TextView)findViewById(R.id.selection);
        Resources res=getResources();
        String[] words = res.getStringArray(R.array.words);
        setListAdapter(new ArrayAdapter<String>(this,android.R.layout.simple_list_item_1,words));
    }

    public void onListItemClick(ListView parent, View v, int position,long id) {
        selection.setText(words.get(position).toString());
    }
}

Upvotes: 2

Views: 638

Answers (1)

Serdar Samancıoğlu
Serdar Samancıoğlu

Reputation: 740

Because words.get(position).toString() returns null if you did not append any object to "words" which is an ArrayList as declared this way:

ArrayList<String> words=new ArrayList<String>();

"String[] words" is not what you are using in onListItemClick, but "ArrayList<String> words" is.

Edit:

Comment the arraylist, and define "String[] words" outside of onCreate.

//ArrayList<String> words=new ArrayList<String>();
String[] words;

Then in your onListItemClick;

selection.setText(words[position]);

And you are good to go.

Upvotes: 3

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