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POCan my spinner's entries depend on another spinner's position?
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copied!<p>I have had no formal training in programming at all yet (I'm a freshman in college, though, and going to major in it), and am trying to learn Android after only knowing Visual Basic, so I need a lot of help. Here's the code I have so far:</p> <pre><code>@Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); Spinner spinner1 = (Spinner) findViewById(R.id.spinner1); ArrayAdapter<CharSequence> adapter1 = ArrayAdapter.createFromResource(this, R.array.Books, android.R.layout.simple_spinner_item); adapter1.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item); spinner1.setAdapter(adapter1); spinner1.setOnItemSelectedListener( new OnItemSelectedListener() { public void onItemSelected(AdapterView<?> parent, View view, int position, long id) { // Here's what I need help with. I basically want it to say: switch(position) { case 0: // Here I want it to set spinner2's entries to R.array.Five case 1: //Here I want to set spinner2's entries to R.array.Three //etc... } public void onNothingSelected(AdapterView<?> parent) { } }); Spinner spinner2 = (Spinner) findViewById(R.id.spinner2); ArrayAdapter<CharSequence> adapter2 = ArrayAdapter.createFromResource(this, R.array.Five, android.R.layout.simple_spinner_item); adapter2.setDropDownViewResource(android.R.layout.simple_spinner_dropdown_item); spinner2.setAdapter(adapter2); spinner2.setOnItemSelectedListener( new OnItemSelectedListener() { public void onItemSelected(AdapterView<?> parent, View view, int position, long id) { //(something similar) } public void onNothingSelected(AdapterView<?> parent) { } }); } </code></pre> <p>Is it possible to change the entries array of one spinner from another spinner's onItemSelected method? What code do I need to add where? Trying to say something about spinner2 or adapter2 in spinner1's OnItemSelectedListener says it's a misplaced construct. Which makes sense, but I don't know how to get around that.</p>

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