1

我是安卓新手。所以我请求你对我有耐心。我正在尝试编写一个应用程序,在按下按钮时会打开联系人 API。然后用户选择联系人。然后我想获得选择的联系人onAcitvityResult(int reqcode, int rescode, Intent data)。有没有办法在不获取 Uri 并通过整个数据库查询它的情况下做到这一点?这是我的第二个活动开始的代码:

public void onClick(View v) {

      Log.d(TAG, "contact button clicked");
      Toast.makeText(ctx, "Contact button clicked",
      Toast.LENGTH_SHORT).show();

      Intent contacts = new Intent(Intent.ACTION_GET_CONTENT,Contacts.CONTENT_URI);
      contacts.setType(ContactsContract.Contacts.CONTENT_ITEM_TYPE);

      startActivityForResult(contacts,1);
}

我正确使用Contacts.CONTENT_URIandIntent.ACTION_GET_CONTENT吗?

我使用的意图是否正确?我觉得我应该在此处包含电子邮件详细信息。

这就是我处理活动结果的地方:

protected void onActivityResult(int requestCode, int resultCode, Intent data){

    super.onActivityResult(requestCode, resultCode, data);
    String email=""; long id;
    if (requestCode == 1) {
        if (resultCode == Activity.RESULT_OK) {

            // get the contact ID
            Uri contacturi= data.getData();
            Cursor c= getContentResolver().query(contacturi,null,null,null,null);
            id = c.getLong(c.getColumnIndex(ContactsContract.Contacts._ID));
            c.close();

            // get the data package containing the email address for the contact
            c=getContentResolver().query(ContactsContract.Data.CONTENT_URI,
                    new String[]{Email.DATA1}, 
                    ContactsContract.Data.CONTACT_ID + "=? AND " + Email.MIMETYPE + "=?",
                    new String[]{String.valueOf(id), Email.CONTENT_ITEM_TYPE}, null);

            email=c.getString(c.getColumnIndex(Email.DATA1));
            Log.d(TAG, "email is" + email);
        }
    }
    else {
        Log.d(TAG, "requestCode is not 1");
    }

    EditText ctext= (EditText) findViewById(R.id.contacttxt);
    ctext.append(email);
    Log.d(TAG, "onActivityResult() ends");
}

我得到一个运行时异常说failure delivering result
我应该包括一些setResult()方法吗?
先感谢您。

4

1 回答 1

0

I got it :) You have to add c.moveToFirst(); two times, after the two initializations as the 2 cursors. But I am not getting why. Could anyone explain it?

于 2011-01-04T21:01:41.430 回答