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copied!<p>The answer by Christian is excellent. However, the method for getting the sender's address did not work for me. The Long.parseLong statement doesn't do anything except possibly throw an exception and new String(...) ?.</p> <p>On my device the cursor count is 2 or more. The first typically has a "type" of 137 and the others have a "type" of 151. I cannot find where this is documented, but one can deduce 137 is "from" and 151 is "to". Thus, if I run the method as is, I do not get an exception, and it returns the last row, which is a recipient and only one of several in many cases.</p> <p>Also AFAICT the selection is not necessary as all the rows have the same msg_id. However, it doesn't hurt.</p> <p>This is what works for me to get the sender's address:</p> <pre><code>public static String getMMSAddress(Context context, String id) { String addrSelection = "type=137 AND msg_id=" + id; String uriStr = MessageFormat.format("content://mms/{0}/addr", id); Uri uriAddress = Uri.parse(uriStr); String[] columns = { "address" }; Cursor cursor = context.getContentResolver().query(uriAddress, columns, addrSelection, null, null); String address = ""; String val; if (cursor.moveToFirst()) { do { val = cursor.getString(cursor.getColumnIndex("address")); if (val != null) { address = val; // Use the first one found if more than one break; } } while (cursor.moveToNext()); } if (cursor != null) { cursor.close(); } // return address.replaceAll("[^0-9]", ""); return address; } </code></pre> <p>I didn't care about whether it is all numeric, but I included a way to eliminate everything but numerals as a comment if that is desired. It can easily be modified to return all the recipients, as well.</p> <p>I assume it worked for him. It looks like it would give the right answer if the exception occurred on the first row.</p>

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