2

我正在制作用户列表。我想返回匹配特定字段（匹配电子邮件、匹配电子邮件和类型、匹配名称和类型或匹配所有字段）的用户（和项目）。我想减少重复的用户，所以如果它与电子邮件和类型匹配，就不要返回电子邮件。

示例数据：

+----+---------------------+--------+---------+
| id | email               | type   | name    |
+----+---------------------+--------+---------+
| 1  | rod@example.com     | user   | Rod     |
| 2  | jane@example.com    | user   | Jane    |
| 3  | rod2@example.com    | user   | Rod     |
| 4  | rod3@example.com    | admin  | Rod     |
| 5  | helen@example.com   | admin  | helen   |
+----+---------------------+--------+---------+

所以如果 $email = 'rod@example.com', $type = 'user', $name = 'Rod' 我想返回以下结果。我想返回以下内容：

+----------------+----+
| matches        | id |
+----------------+----+
| 3 (all)        | 1  |
| 2 (name, type) | 3  |
| 1 (name)       | 4  |
+----------------+----+

我正在考虑为每个匹配集单独的 sql 查询，1 个用于电子邮件，另一个用于电子邮件和类型，另一个用于名称和类型，1 个用于所有。最后使用 php.ini 过滤结果。

$users = array();

// match email
$sql = 'SELECT u.*, GROUP_CONCAT(i.name) as items
        FROM users u
        INNER JOIN items i
            ON u.id = i.user_id
        WHERE u.email = '.$email.'
        GROUP BY u.id';
$users[] = mysqli_query($sql);

// match email & type
$sql = 'SELECT u.*, GROUP_CONCAT(i.name) as items
        FROM users u
        INNER JOIN items i
            ON u.id = i.user_id
        WHERE u.email = '.$email.' AND u.type = '.$type.'
        GROUP BY u.id';
$users[] = mysqli_query($sql);

// match name & type
$sql = 'SELECT u.*, GROUP_CONCAT(i.name) as items
        FROM users u
        INNER JOIN items i
            ON u.id = i.user_id
        WHERE u.name = '.$name.' AND u.type = '.$type.'
        GROUP BY u.id';
$users[] = mysqli_query($sql);

// match email, name & type
$sql = 'SELECT u.*, GROUP_CONCAT(i.name) as items
        FROM users u
        INNER JOIN items i
            ON u.id = i.user_id
        WHERE u.email = '.$email.' AND u.name = '.$name.' AND u.type = '.$type.'
        GROUP BY u.id';
$users[] = mysqli_query($sql);

我觉得必须有一种方法可以创建一个查询来替换它们并避免使用 php 进行任何过滤。任何人都可以提供任何建议吗？

4

3 回答 3

1

我会这样做：

SELECT id,
  email = 'rod@example.com' EmailMatch,
  type = 'user' TypeMatch,
  name = 'Rod' NameMatch
FROM t
HAVING EmailMatch + TypeMatch + NameMatch > 0
ORDER BY EmailMatch + TypeMatch + NameMatch DESC

输出：

| ID | EMAILMATCH | TYPEMATCH | NAMEMATCH |
|----|------------|-----------|-----------|
|  1 |          1 |         1 |         1 |
|  3 |          0 |         1 |         1 |
|  2 |          0 |         1 |         0 |
|  4 |          0 |         0 |         1 |

在这里拉小提琴。

您还可以更改id表中的实际字段并避免以后的连接，因为您将直接从同一个查询中检索数据。

如果您确实需要实际查看每行的总匹配数，那么您可以添加另一行，在其中复制所有条件并添加它们或将您的查询包装在另一个 select 语句中。无论哪种方式都不容易阅读或效率低下。所以我建议如果您确实需要查看每行的匹配数，然后在 PHP 中添加匹配列。

于 2013-09-06T17:27:44.273 回答
0

这可能是可能的查询：

select (CASE 

WHEN CONCAT(users.name, '@')=LEFT(users.email, LENGTH(users.name)+1)
and users.type='user' THEN "3 (all)"

WHEN users.type='user' and CONCAT(users.name, '@')!=LEFT(users.email, 
LENGTH(users.name)+1) THEN "2 (name, type)"

ELSE "1 (name)" END)matches, users.id from users 

inner join items on users.id=items.user_id and 

users.name in (select users.name from users 

group by users.name having count(users.name)>1);

输出：

+----------------+----+
| matches        | id |
+----------------+----+
| 3 (all)        | 1  |
| 2 (name, type) | 3  |
| 1 (name)       | 4  |
+----------------+----+
于 2013-09-06T20:16:41.870 回答
0

我会用 CASE WHEN 来执行（对不起，如果语法不是 100% 正确）：

$sql = 'SELECT u.id, 
  CASE WHEN (u.name = ' .$name. ' AND u.type = '.$type.' AND u.email = '.$email.') THEN "3 (all)"
    ELSE WHEN (u.name = ' .$name. ' AND u.type = '.$type.') THEN "2 (name, type)"
    ELSE WHEN (u.name = ' .$name. ' AND u.emal = '.$email.') THEN "2 (name, email)"
    ELSE WHEN (u.type = ' .$type. ' AND u.email = '.$email.') THEN "2 (email,type)"
    ELSE WHEN (u.name = ' .$name. ') THEN "1 (name)"
    ELSE WHEN (u.type = ' .$type. ') THEN "1 (type)"
    ELSE WHEN (u.email = ' .$email. ') THEN "1 (email)"
    ELSE "0" END AS matches
  FROM users u
  HAVING matches != "0"';
于 2013-09-06T17:23:06.470 回答