Question

我正在制作用户列表。我想返回与特定字段匹配的用户（和项目）（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）。我想减少重复的用户，所以如果它与电子邮件和电子邮件匹配，请不要返回电子邮件。类型。

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。我想回复以下内容：

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

我在想每个匹配集的单独sql查询，1个用于电子邮件，另一个用于电子邮件＆amp;类型，另一个名称＆amp;类型，1为所有。然后使用php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来创建一个查询来替换它们并避免使用php进行任何过滤。任何人都可以提出任何建议吗？

Answer 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 |

小提琴here。

您还可以更改表中实际字段的id，并避免以后加入，因为您将直接从同一个查询中检索数据。

如果确实需要实际查看每行的总匹配数量，那么您可以添加另一行，在该行中复制所有条件并添加它们或将查询包装在另一个select语句中。无论哪种方式都不容易阅读或低效。所以我建议如果你确实需要查看每行的匹配数，那么在PHP中添加匹配列。

Answer 2

我会用CASE WHEN执行该操作（抱歉，如果语法不是100％正确的话）：

$sql = 'SELECT u.id, 
  CASE WHEN (u.name = ' .$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"';

Answer 3

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);

<强>输出：

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

SELECT匹配1到多个字段的用户并删除重复项

3 个答案: