Question

Good day, i was having a little trouble with the following code below and i got a solution from a ticked answer in this link. I would like to know why.

Note: i had a similar type of layout as in the question, but with one TextView only.

solution link here

my code below:

    public void onItemClick(AdapterView<?> av, View v, int position, long id) {
        blueadapter.cancelDiscovery();

    String info = ((TextView) v).getText().toString(); //classcastException here
    String Bluetooth_address = info.substring(info.length()-17);
 }

so from his solution, if i change this to this

String info = ((TextView) v.findViewById(R.id.search_device_id)).getText().toString();

it solves my ClassCastException problem.

Now please can someone kindly explain to me or point me to the right direction, what he meant by in the answer “You receive the whole LinearLayout as the parameter v. You should try v.findViewById() and then use this textview.” its nice it solves my problem, but i would like to understand why i had to do that?.. i have dealt with ListViews before, but i have not come across this, so its kind of strange for me. Thank you.

Answer 1

The View v that you get is the entire Layout for the object that has the TextView (and probably other views) inside it.

It is not the TextView itself, you have to find the TextView that is inside of it.