I've got two list but one itemclick method

The problem is that I have two listView but one itemClick method, if I put a switch, the position on one list is the same than the other list... And by example if I want to open a popup on an item in one list, the item in the same position in the other list gonna do the same thing, and I really don't want, can you help me?

Thanks

There is the code :

public void onItemClick(AdapterView<?> parent, View v, int position, long id) {
    Object str = parent.getId();
    if(str.equals(adapter_todo)){
        switch(position){
        case 0 :
            new AlertDialog.Builder(this).setTitle("test").setMessage("blah blah").setNeutralButton("close", new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    // TODO Auto-generated method stub
                }
            }).show();
        case 1 :
        case 2 :
        case 3 :
        case 4 :
        case 5 :
        case 6 :
        case 7 :
        case 8 :
        case 9 :
        case 10 :
        }
    }
    else if(str.equals(adapter_not_todo)){
        switch(position){
            case 0 : 
                new AlertDialog.Builder(this).setTitle("test 2").setMessage("blah blah blah").setNeutralButton("close", new DialogInterface.OnClickListener() {
                    @Override
                    pub开发者_如何学Pythonlic void onClick(DialogInterface dialog, int which) {
                        // TODO Auto-generated method stub
                    }
                }).show();
            case 1 :
            case 2 :
            case 3 :
            case 4 :
            case 5 :
            case 6 :
            case 7 :
            case 8 :
            case 9 :
            case 10 :
        }
    }
}

In onItemClick, call parent.getId() to tell which list view was clicked.