MySQL选择连接表作为结果数组

我需要编写一个mysql查询，它将从Task表中选择父数据，在子SubTask表上执行左连接并将子数据追加到Task数组下的数组像这样......MySQL选择连接表作为结果数组

任务

id| details  | created 
=============================== 
1 | This is.... | 2017-08-19 
2 | This is.... | 2017-08-20 
3 | This is.... | 2017-08-21

子课题负责人

id | task_id | details  | created 
============================================ 
123 | 1  | This is.... | 2017-08-19 
234 | 1  | This is.... | 2017-08-20 
345 | 2  | This is.... | 2017-08-21 
456 | 2  | This is.... | 2017-08-21 
567 | 3  | This is.... | 2017-08-21 


[ 
     [ 
      'id'=>1, 
      'details'=>'This is from the task table', 
      'subTasks'=>[ 
       [ 
        'id'=>123, 
        'details'=>'This is from the sub task table', 
       ], 
       [ 
        'id'=>234, 
        'details'=>'This is from the sub task table', 
       ] 
      ] 
     ], 
     [ 
      'id'=>2, 
      'details'=>'This is from the task table', 
      'subTasks'=>[ 
       [ 
        'id'=>345, 
        'details'=>'This is from the sub task table', 
       ], 
       [ 
        'id'=>456, 
        'details'=>'This is from the sub task table', 
       ] 
      ] 
     ] 
     [ 
      'id'=>3, 
      'details'=>'This is from the task table', 
      'subTasks'=>[ 
       [ 
        'id'=>567, 
        'details'=>'This is from the sub task table', 
       ] 
      ] 
     ] 
    ]

这里是我的查询：

$sql = " SELECT t.*, 
         (select st.* from SubTask) as SubTasks 
      FROM  Task t 
      LEFT JOIN SubTask st 
       ON  st.task_id = t.id 
      ";

我知道这个查询是不正确的，我只是不知道如何写它，否则。

来源

2017-05-31 LargeTuna

显示您的餐桌结构 –

这是非常困难的帮助您没有显示您的表与一些信息。 – jjoselon

添加了最基本的表详细信息 – LargeTuna

你并不需要一个子查询select，只是：

SELECT t.id AS task_id, t.details AS task_details, st.id AS subtask_id, st.details AS subtask_details 
FROM Task t 
LEFT JOIN SubTask st ON st.task_id = t.id

然后你的PHP代码可以收集所有的子任务信息到一个数组。

$tasks = []; 
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { 
    $task_id = $row['task_id']; 
    if (!isset($tasks[$task_id])) { 
     $tasks[$task_id] = ['id' => $task_id, 'details' => $row['task_details'], 'subtasks' => []]; 
    } 
    if ($row['subtask_id'] !== null) { 
     $tasks[$task_id]['subtasks'][] = ['id' => $row['subtask_id'], 'details' => $row['subtask_details']]; 
    } 
}

来源

2017-05-31 19:54:34 Barmar

这很好，谢谢！ – LargeTuna

MySQL选择连接表作为结果数组

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