选择与电话号码联系而不是阅读电话号码时出现问题

时间:2011-11-20 14:48:20

标签: android contact

调用联系人选择器显示所有联系人已完成(如此在SO上多次说明):

Intent intent = new Intent( Intent.ACTION_PICK, Contacts.CONTENT_URI );
startActivityForResult( intent, REQ_CODE );

我在onActivityResult中获取了联系人姓名及其所有电话号码,并附有以下片段:

public void onActivityResult( int requestCode, int resultCode, Intent intent )
{
    Uri contactUri = intent.getData();
    ContentResolver resolver = getContentResolver();
    long contactId = -1;

    // get display name from the contact
    Cursor cursor = resolver.query( contactUri,
                                    new String[] { Contacts._ID, Contacts.DISPLAY_NAME }, 
                                    null, null, null );
    if( cursor.moveToFirst() )
    {
        contactId = cursor.getLong( 0 );
        Log.i( "tag", "ContactID = " + Long.toString( contactId ) );
        Log.i( "tag", "DisplayName = " + cursor.getString( 1 ) );
    }

    // get all phone numbers with type from the contact
    cursor = resolver.query( Phone.CONTENT_URI,
                             new String[] { Phone.TYPE, Phone.NUMBER }, 
                             Phone.CONTACT_ID + "=" + contactId, null, null );
    while( cursor.moveToNext() )
    {
        Log.i( "tag", "PhoneNumber = T:" + Integer.toString( cursor.getInt( 0 ) ) + " / N:" + cursor.getString( 1 ) );
    }

拨打联系人选择器并仅显示带有电话号码的联系人可以这样做(也可以在SO上找到):

Intent intent = new Intent( Intent.ACTION_PICK );
intent.setType( ContactsContract.CommonDataKinds.Phone.CONTENT_TYPE );
startActivityForResult( intent, REQ_CODE );

如果我这样做,我只会在联系人选择器中看到那些至少有一个电话号码的联系人,这正是我所需要的。不幸的是,通过上面的代码片段,我只得到显示名称,但不再是任何电话号码。

是否有人知道我需要更改哪些内容才能获取电话号码?

提前致谢

1 个答案:

答案 0 :(得分:2)

更改Phone._ID的where子句中的Phone.Contact_Id,如下所示:

   cursor = resolver.query( Phone.CONTENT_URI,
                             new String[] { Phone.TYPE, Phone.NUMBER }, 
                             Phone._ID + "=" + contactId, null, null );
    while( cursor.moveToNext() )
    {
        Log.i( "tag", "PhoneNumber = T:" + Integer.toString( cursor.getInt( 0 ) ) + " / N:" + cursor.getString( 1 ) );
    }

question中的详细信息。

希望有所帮助:)