Question

我们需要几个表来根据user_id提取数据。一切都很好，除了现在我们已经分离出地址，每个用户可以有多个物理地址位置（最多3个）。我们需要将这些地址中的每一个标识为a1，a2，a3，知道a2和a3是可选的并且可以为null。

这是地址表架构：

    $sql[] = "CREATE TABLE {$modules->tables->members_addresses} (
                id bigint(20) NOT NULL AUTO_INCREMENT PRIMARY KEY,
                user_id bigint(20) NOT NULL,
                mailing tinyint(1) DEFAULT 0 NOT NULL,
                public tinyint(1) DEFAULT 0 NOT NULL,
                address varchar(255) NOT NULL,
                city varchar(255) NOT NULL,
                region_id bigint(20) NOT NULL,
                country_id bigint(20) NOT NULL,
                postalcode varchar(200) NOT NULL,
                date_created datetime DEFAULT '0000-00-00 00:00:00' NOT NULL,
                date_updated datetime DEFAULT '0000-00-00 00:00:00' NOT NULL
        ) {$modules->tables->charset_collate};";

这是我到目前为止的查询SQL。它的问题是我为每个位置得到一个单独的结果（即不是将a1-a3包装成一个查询结果，而是给我单独的查询结果）：

SELECT p.*, concat( u1.meta_value, ' ', u2.meta_value ) as fullname, u1.meta_value as first_name, u2.meta_value as last_name, a.address, a.city, a.postalcode, case when p.certifications = '' then roles.name else concat( roles.name, ',', p.certifications ) end as certification, case when r.name is null then c.name else concat( r.name, ', ', c.name ) end as location, case when r.name is null then '' else r.name end as region, case when c.name is null then '' else c.name end as country
FROM modules_profiles p
LEFT JOIN moonlight_usermeta u1 ON p.user_id = u1.user_id AND u1.meta_key = 'first_name' 
LEFT JOIN moonlight_usermeta u2 ON p.user_id = u2.user_id AND u2.meta_key = 'last_name'
LEFT JOIN modules_members_addresses a ON p.user_id = a.user_id LEFT JOIN modules_roles roles ON roles.slug = p.certification 
LEFT JOIN modules_regions r ON r.id = a.region_id LEFT JOIN modules_countries c ON c.id = a.country_id 
WHERE p.user_id IN ( 1 )
ORDER BY p.user_id

分离出a1，a2和a3的SQL是什么？顺便说一下，我需要a1.address，a1.city，a1.region，a1.country，a1.postalcode（并重复a2和a3）。

Answer 1

您可以在查询中多次加入同一个表

SELECT u.id, u.name, a1.*, a2.*, a3.*
FROM users AS u
LEFT OUTER JOIN addresses AS a1 ON a1.userId = u.id
LEFT OUTER JOIN addresses AS a2 ON a2.userId = u.id AND a2.id < a1.id
LEFT OUTER JOIN addresses AS a3 ON a3.userId = u.id AND a3.id < a2.id
WHERE u.id = 1 AND a1.id = (SELECT MIN(id) FROM addresses WHERE userId = 1)

a1.id = (SELECT MIN(id) FROM addresses WHERE userId = 1)是为了让您获得第一个＆＃39;先解决。使用ID进行排序通常是一个坏主意，因此您可能希望实现一个序列＆＃39;或者＆＃39;偏好＆＃39;地址字段。

MySQL - 如何将多个表行分离为一个类似查询结果的行

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