Question

如何在 ONE 查询中获取一对多关系数据？

示例：一本书有很多作者。

结果返回应该是这样的：

array(
   'book_title' => 'HTML for Dummies',
   'book_desc'  => '...',
   'authors' => array(
       [0] => array(
           'name' => 'Someone',
           'dob'  => '...'
       ),
       [1] => array(
           'name' => 'Someone',
           'dob'  => '...'
       )
    )
)

我尝试使用子查询来选择结果但不是运气：

SELECT *, (
    SELECT * 
    FROM authors
    WHERE book_id = b.id
) AS authors
FROM book b;

Mysql错误＆＃34;操作数应包含1列＆＃34;这意味着我只能选择一列。

您可能会建议我使用加入，但它如何归档返回结果格式，就像我向您展示的那样？

Answer 1

SELECT 
 B.id AS book_id, B.name AS book_name,
 A.name AS author_name ,A.ID AS author_id 
FROM book B
LEFT JOIN 
 author A ON B.id = A.book_id;

参考：http://sqlfiddle.com/#!9/633cfa1/2

您的输出将是数字数组。

Array
(
    [0] => Array
        (
            [bookId] => 1
            [bookName] => Book 1
            [authorId] => 1
            [authorName] => Author 1
        )

    [1] => Array
        (
            [bookId] => 1
            [bookName] => Book 1
            [authorId] => 2
            [authorName] => Author 2
        )

    [2] => Array
        (
            [bookId] => 2
            [bookName] => Book 2
            [authorId] => 3
            [authorName] => Author 3
        )

    [3] => Array
        (
            [bookId] => 3
            [bookName] => Book 3
            [authorId] => 4
            [authorName] => Author 4
        )

)

将此数组转换为多维数组

Array
(
    [0] => Array
        (
            [bookId] => 1
            [bookName] => Book 1
            [authorName] => Array
                (
                    [0] => Author 1
                    [1] => Author 2
                )

        )

    [1] => Array
        (
            [bookId] => 2
            [bookName] => Book 2
            [authorName] => Array
                (
                    [2] => Author 3
                )

        )

    [2] => Array
        (
            [bookId] => 3
            [bookName] => Book 3
            [authorName] => Array
                (
                    [3] => Author 4
                )

        )

)

参考：https://3v4l.org/q1dli

Answer 2

如果您想快速而简单地使用，请使用带有适当分隔符的GROUP_CONCAT。然后在PHP应用程序（或您正在使用的任何语言）中分割生成的连接字符串字段，以快速获取所需的内部数组。

这样做的好处是输出中每本书只有一行。

Answer 3

SELECT b.*, a.*
FROM book b
INNER JOIN authors a ON (a.book_id = b.id);

Answer 4

从mysql中获取数据可能并不容易。 Mysql将以行方式返回。所以书籍或作者都会在结果集中重复。并且你必须相应地解析mysql结果以准备你的数组。

请尝试以下操作：

1）所有以作者为逗号分隔ID的书籍。

"SELECT B.*,group_concat(a.id) as authors 
 From book b inner join authors as a on b.author_id=a.id 
 group by b.id;"

2）所有作者都有书的详细信息：

"SELECT A.*,group_concat(B.id) as books 
 From book B inner join authors as A on B.author_id=A.id 
 group by A.id;"

MySQL在一个查询中有一对多的关系

4 个答案: