Question

我可能做错了，但我真的需要帮助掌握如何通过嵌套数组工作。我有两个表User和Reviews。

这是我正在使用的SELECT语句：

SELECT * 
FROM `users` 
LEFT JOIN `reviews` 
ON users.userId = reviews.user_id 
WHERE installation_id = $installation_id

这就是我接收数组的方式：

Array
(
    [0] => Array
        (
            [userId] => 60
            [usersEmail] => gggg@gmail.com
            [password] => 0Vg+sLdlALchd39l+3K3QXFZlvh79bwqXFp/J3nIR+o=
            [usersName] => Lauren 
            [role] => scout
            [id] => 1
            [user_id] => 60
            [client_id] => 62
            [comments] => This person was extremely professional and went above and beyond. Her pricing was reasonable.
            [stars] => 4
        )

    [1] => Array
    (
        [userId] => 60
        [usersEmail] => gggg@gmail.com
        [password] => 0Vg+sLdlALchd39l+3K3QXFZlvh79bwqXFp/J3nIR+o=
        [usersName] => Lauren Rothlisberger
        [role] => scout
        [id] => 2
        [user_id] => 60
        [client_id] => 1
        [comments] => She was pretty good. I have a couple small complaints but overall decent experience. 
        [stars] => 3
    )

[2] => Array
    (
        [userId] => 63
        [usersEmail] => proth@mail.com
        [password] => 
        [usersName] => Paul Rothlisberger
        [role] =>
        [id] => 
        [user_id] => 
        [client_id] => 
        [comments] => 
        [stars] => 
    )

）

关于这一点对我来说没有意义。我认为这更像是

[1] => Array
    (
        [userId] => 60
        [usersEmail] => gggg@gmail.com
        [password] => 0Vg+sLdlALchd39l+3K3QXFZlvh79bwqXFp/J3nIR+o=
        [usersName] => Lauren Rothlisberger
        [role] => scout
        [id] => 2
        [user_id] => 60
        [client_id] => 1
                Array [0](
                      [comments]  
                      [stars] ) 
    )

如何获得该阵列？此外，他们如何迭代用户然后向用户展示他们所有的评论和明星。

我也无法弄清楚如何找到UserId = 60的星星，然后给出该用户的平均星星。

感谢您提供给我的任何帮助。我显然缺少一些基础知识。请不要指向我通过它的文档，但不能应用它。

Answer 1

因为您已经注意到所返回的每条记录都是完整记录，当涉及多对一关系时，您会获得每条记录的一条记录。

当我遇到这个时，我应用以下短循环来将结果打包成更类似于第二种格式的内容，但它包括每次重复用户级别字段。通过这种方式，我可以轻松计算每个用户的子项数，通过在输出端使用嵌套的foreach进行中间排序或其他操作。

$results = array();
$results_raw = db.select(' ... the select statement ');
if (is_array($results_raw)) foreach ($results_raw as $result) {
    $results[$result['userId']] = $result;
}

使用LEFT JOIN TABLE提取结果的嵌套数组

1 个答案: