StackOverflow2013

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PO
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5569397
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2011-04-06T15:57:22.087
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2011-04-06T15:57:22.087
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Why don't you display a dialog with the two options before loading your new Activity? And then, according the user's selection display the corresponding xml? I would do it this way: <ol> <li>You have your mainActivity running</li> <li>When the user clicks your button, you show an AlertDialog with the "positive" button representing the first xml file and the "negative" button representing your second xml.</li> <li>Upon click on any of the buttons, you show your second Activity by using different Intent-actions and checking which intent was used in the <code>onCreate</code> method of your second Activity.</li> </ol> For your button: <pre><code>Button button = (Button) findViewById(R.id.button); button.setOnClickListener(new OnClickListener() { @Override public void onClick(View v) { DialogInterface.OnClickListener listener = new DialogInterface.OnClickListener() { @Override public void onClick(DialogInterface dialog, int which) { Intent act2 = null; switch (which) { case DialogInterface.BUTTON_POSITIVE: // layout 1 act2 = new Intent("package.example.act2_layout1"); startActivity(act2); break; case DialogInterface.BUTTON_NEGATIVE: act2 = new Intent("package.example.act2_layout2"); startActivity(act2); } } }; AlertDialog.Builder builder = new AlertDialog.Builder(mainActivity.this); builder.setTitle("Make your choice") .setPositiveButton("Layout 1", listener) .setNegativeButton("Layout 2", listener); } }); </code></pre> And then in the <code>onCreate</code> method: <pre><code>Intent starter = getIntent(); if (starter.getAction().equals("package.example.act2_layout1")) { // use layout 1 } else if (starter.getAction().equals("package.example.act2_layout2")) { // use layout 2 } </code></pre>
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1. POAndroid context menu displayed in onCreate
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1. PTAnswer
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1. USMichael Rose
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