Question

我有两个与topic_id相关的表格（讲座和主题）。我需要作为相关主题的孩子参加讲座。所需的json_encode（$ result）应为：

[
    {"id":"2","name":"topic 3",
      "lectures": [
        {"id":"9", "topic_id":"2","name":"lecture 1"},
        {"id":"10","topic_id":"2","name":"lecture 2"},
        {"id":"11","topic_id":"2","name":"lecture 3"}
      ]
    },
    {"id":"3","name":"topic 4",
      "lectures": [
        {"id":"12","topic_id":"3","name":"lecture 1"},
        {"id":"13","topic_id":"3","name":"lecture 2"},
        {"id":"14","topic_id":"3","name":"lecture 3"}
      ]
    }
]

一种可能的解决方案是像这样重新生成数组

$topics = $db->query("select * FROM topic")->fetchAll(PDO::FETCH_ASSOC);
$lectures = $db->query("SELECT * FROM lecture")->fetchAll(PDO::FETCH_ASSOC);

foreach($topics AS $topic) {
    $result[$topic["id"]] = $topic;
    $result[$topic["id"]]["lectures"] = array();
}
foreach($lectures AS $lecture) {
    $result[$lecture["topic_id"]]["lectures"][] = $lecture;
}

echo json_encode($result);

结果是：

[
    {"2": {"id":"2","name":"topic 3",
       "lectures": [    
        {"id":"9", "topic_id":"2","name":"lecture 1"},
    {"id":"10","topic_id":"2","name":"lecture 2"},
    {"id":"11","topic_id":"2","name":"lecture 3"}
   ]
    },
    {"3": {"id":"3","name":"topic 4",
       "lectures": [
    // ...
]

这仍然不是我们要求的。我需要删除最顶层的id（用作键）可以通过重新生成结果数组在服务器或客户端完成，只保留值。（不那么优雅）的解决方案可能是：

$result2 = array();
foreach($result AS $res) {
    $result2[] = $res;
}
echo json_encode($result2);

这让我获得了理想的结果，但解决方案效率很低。

任何有关更好的方法的建议都将受到赞赏。建议可能包括更有效的方式：

通过改进MySQL查询来完成一些工作。
php中的数组操作
通过操作客户端（Javascript，jQuery或Underscore便捷方法）获得所需结果

由于

Answer 1

你的问题有点含糊，所以我猜这里。我认为你只需要一个MySQL声明。

SELECT lecture.id lid,lecture.name lname,topic.id tid,topid.name tname 
FROM lecture 
LEFT JOIN topic ON lecture.id = topic.id
ORDER BY lid,tid

您的结果应该是正确的顺序。

将MySQL一对多结果集转换为JSON

1 个答案: