Question

我有这个查询工作正常。没有错误或者是什么。但它似乎不适用于最后一部分，并且不会填充contact_data cd中的最后三个字段。

 select c.*,cd.* FROM
 (select * from
 (select con.id_contact as v,substr(concat(ifnull(con.firstname,''),ifnull(con.lastname,'')),1,4) as s0, null as s1, null as s2
 from contact con where id_user=123 and firstname != '' and firstname != ' ' and firstname is not null) as tbl235768 where 1=1 AND v IN
 (select v from
   (select * from
      (select con.id_contact as v,substr(concat(ifnull(con.firstname,''),ifnull(con.lastname,'')),1,4) as s0, null as s1, null as s2
      from contact con where id_user=123 and email = '') as tbl235770 where 1=1) as tblAnd) AND v IN
      (select v from
          (select * from (select con.id_contact as v,substr(concat(ifnull(con.firstname,''),ifnull(con.lastname,'')),1,4) as s0, null as s1, null as s2
          from contact con where id_user=123 and mobile != '' and mobile != ' ' and mobile is not null) as tbl235772 where 1=1) as tblAnd)) as tblBucket
          left join contact c on c.id_contact=v left join
          (select id_contact, group_concat(name) as custom_names,group_concat(value) as custom_values FROM contact_data where
          id_user=20 group by id_contact) as cd on cd.id_contact=c.id_contact group by (c.id_contact)

但是，如果我只运行以下

的最后一部分

 select id_contact, group_concat(name) as custom_names,group_concat(value) as custom_values 
 FROM contact_data where id_user=798 group by id_contact

它给了我想要的结果。我的查询有什么问题？任何帮助都会非常感激，谢谢。

编辑：我在得到几个答案后正在编辑我的问题。

我删除了所有嵌套部件，但仍然没有运气。

select c.*,cd.* FROM (select * from (select con.id_contact as v,substr(concat(ifnull(con.firstname,''),ifnull(con.lastname,'')),1,4) as s0,  null as s1, null as s2 from contact con where id_user=10879) as tbl235785
where 1=1) as tblBucket  left join contact c on c.id_contact=v left join  (select  id_contact, group_concat(name) as custom_names,group_concat(value) as     custom_values 
FROM contact_data where id_user=798 group by id_contact) as cd on cd.id_contact=c.id_contact group by (c.id_contact)

Answer 1

这是你的sql更简单：

所有1 = 1和SELECT * FROM（子查询）都可以消失 - 他们什么都不做。然后逻辑变得清晰 - 你只需要在where语句而不是子查询中添加一些附加条款。我认为当它清理完毕后你可以看到 - 在第一个选择中你有id_user=123，在第二个选择中你有id_user=20。

可能希望那些是一样的？你甚至不需要第二个加入选择中的where子句。由于它已加入，您应该只在外部查询中加入此字段的值。

select c.*,cd.* 
FROM (select con.id_contact as v,substr(concat(ifnull(con.firstname,''),ifnull(con.lastname,'')),1,4) as s0, null as s1, null as s2
        from contact con 
        where id_user=123 and 
              firstname != '' and firstname != ' ' and firstname is not null
          and 
            email = ''
              OR
            (
               mobile != '' and mobile != ' ' and mobile is not null
            )  
      ) as tbl235768 
left join contact c on c.id_contact=v 
left join (select id_contact, group_concat(name) as custom_names,group_concat(value) as custom_values
           FROM contact_data 
           where id_user=20 
           group by id_contact
          ) as cd on cd.id_contact=c.id_contact 
group by (c.id_contact)

Answer 2

在一个地方，你在id_user = 10879匹配，然后在id_user = 798匹配的不同子查询中，然后你加入两个，因为id_user字段都在查看不同的数字，你得不到你的匹配加入。

SQL查询未显示所需结果

2 个答案: