如何为数组中的每个ID细节带来值？

Question

我正在为我的问题寻找解决方案。

我有以下代码：

while($row = mysql_fetch_array($resultQuery)){
    $response = array(
        'name' => $row['name'],
        'data' => $row['data']          
    );
    $responses[] = $response;
}
echo json_encode($responses);

代码将带来这个结果：

[{"name":"Name001","data":"1"},
{"name":"Name001","data":"2"},
{"name":"Name001","data":"3"},
{"name":"Name002","data":"4"},
{"name":"Name002","data":"5"},
{"name":"Name002","data":"6"}]

但我希望得到以下结果：

[{"name":"Name001","data":[1,2,3]},
{"name":"Name002","data":[4,5,5]}

每个＆＃34;姓名＆＃34;有自己的ID，我想在＆＃34; while＆＃34;把每个id都有自己的数据。

感谢您的帮助。

1。更新：

    $id_name = "xx";

             while($row = mysql_fetch_array($resultQuery)){

    if($id !== $row['id']){
                $response = array(
                    'name' => $row['name'],
                    'data' => array($row['data'])         
                );
                $responses[] = $response;
             }
 $id_name = $row['id'];
            }
            echo json_encode($responses);

我首先要检查ID，如果不同会给我带来：

[{"name":"Name001","data":["1"]},{"name":"Name002","data":["4"]}]

Answer 1

while($row = mysql_fetch_array($resultQuery))
    $tmp[$row['name']][] = $row['data']; 

foreach($tmp as $key => $item)
    $responses[] = array('name' => $key, 'data' => $item); 

echo json_encode($responses);

结果

[{"name":"Name001","data":["1","2","3"]},{"name":"Name002","data":["4","5","6"]}]

Answer 2

我在这里做了一个测试，希望你喜欢。

$arr = array();

$arr[] = array(
  'name' => 'val',
  'data' => array(1, 2, 3)
);

$arr[] = array(
  'name' => 'val',
  'data' => array(1, 2, 3)
);

echo json_encode($arr);

输出：

[{"name":"val","data":[1,2,3]},{"name":"val","data":[1,2,3]}]