Question

我正在使用PHP + MySQL进行过时的搜索系统，现在...所有者要求我引入两个按钮以将过滤器应用于主查询。第一个按钮应仅查找具有相同first_name和last_name的用户，第二个按钮应仅查找具有相同last_name的用户。现在，ATM查询使用不同的联接（内部和左侧确实从其他表中获取其他信息）

$query = "
  SELECT s1.*,s2.first_name,s2.last_name,s2.email,s2.id as subscriber_id,s2.phone,s2.info,a.id as agency_id,a.info as agency_info,a.invoice as agency_invoice
  FROM subscription as s1
  LEFT JOIN agency as a
  ON s1.agency=a.id
  INNER JOIN subscriber as s2
  ON s1.subscriber=s2.id
  WHERE $matches AND s1.deleted=0
  ORDER BY s1.id DESC
  LIMIT $maxid,$limit
";

一切正常，但是如果我通过相同的full_name向子用户添加子句，则每个全名仅返回一个结果...

所以，如果我有这样的桌子

+------------+-----------+
| first_name | last_name |
+------------+-----------+
| Foo        | Bar       |
| Foo        | Bar       |
| Bar        | Foo       |
| Bar        | Foo       |
| John       | Doe       |
+------------+-----------+

返回

+------------+-----------+
| first_name | last_name |
+------------+-----------+
| Foo        | Bar       |
| Bar        | Foo       |
+------------+-----------+

现在...我添加到查询中的子句是这些（count和groupby）

$query = "
  SELECT s1.*,s2.first_name,s2.last_name,s2.email,s2.id as subscriber_id,s2.phone,s2.info,a.id as agency_id,a.info as agency_info,a.invoice as agency_invoice, Count(s2.*) AS Cnt
  FROM subscription as s1
  LEFT JOIN agency as a
  ON s1.agency=a.id
  INNER JOIN subscriber as s2
  ON s1.subscriber=s2.id
  WHERE $matches AND s1.deleted=0
  GROUP BY s2.first_name, s2.last_name HAVING Cnt>1
  ORDER BY s1.id DESC
  LIMIT $maxid,$limit
";

但是正如我已经说过的那样...它不起作用，我也不知道为什么...我在哪里错了？是否取决于加入？有没有找到解决方案的方法，或者使用该查询我需要找到其他解决方案？请帮助我，我尝试了其他解决方案，但是没有办法！：D

在内部联接查询中查找具有相同名称的用户PHP + MySQL

0 个答案: