Question

MySQL的。链接两个表，table1是WORKERS，table2是FIRM。表1约有50,000条记录，表2约有15,000条记录。使用链接表的内部联接。不适用于未知输入电子邮件。我来自公司或员工通过电子邮件提供。我通过电子邮件的价值搜索公司及其条件...... 当知道emailW或emailF（变量＆＃34; $ W_EMAIL＆＃34;）时，查找requestX的值（来自FIRM或工人的电子邮件。对于某些输入值 - 程序没有响应..）

TABLE_1

id     | emailW
------ | ------

100    | "bbbb@bb.com"
100    | "ccc@bb.com"
100    | "bbbb@bb.com"

TABLE_2

id     | requestF      | emailF
------ | ------------- | ------
100    | "service xx"  | "aaa@bb.com"
200    | "service xx"  | "bbb1@bb.com"
300    | "service zz"  | "bbb2@bb.com"

如果找到bbbb@bb.com ...没问题（工人中存在电子邮件）
找到aaa@bb.com ...没问题（电子邮件存在于FIRM中）
如果找到xxx@yyy.info ...没有回复......没有错误......＆＃34;白屏＆＃34;！（如果电子邮件不存在！）

我的代码：

SELECT table1.id,
       table1.emailW, 
       table2.id,
       table2.requestF,
       table2.emailF            
  FROM table1
 INNER JOIN table2 ON table1.id = table2.id
 WHERE table1.emailW='$W_EMAIL'
    OR table2.emailF='$W_EMAIL'

Answer 1

select case when t.obs > 0 then 'Ok' else 'Not Ok' end as 'OKorNotOk'
from
(
select s.*,count(*) Obs
from
(
SELECT table_1.id,
       table_1.emailW, 
       table_2.id t2id,
       table_2.requestF,
       table_2.emailF            
  FROM table_1
 left join table_2 ON table_1.id = table_2.id
 union 
 SELECT table_1.id,
       table_1.emailW, 
       table_2.id ,
       table_2.requestF,
       table_2.emailF            
  FROM table_1
 right join table_2 ON table_1.id = table_2.id
 ) s
 where s.emailw = 'aaa@bb.com' or s.emailf = 'aaa@bb.com'
 ) t

结果

+-----------+
| OKorNotOk |
+-----------+
| Ok        |
+-----------+
1 row in set (0.00 sec)

或者

select   t.emailtofind,
            s.id,s.requestf, 
            case when s.srce = 't1' then 'y' else '' end as 'ok found in t1',
            case when s.srce = 't2' then 'y' else '' end as 'ok found in t2',
            case when s.email is null then 'Not Found' else 'Ok' end as Message
from
(
select 't1' as srce,t1.ID,'' as requestf , t1.emailw email from table_1 t1
union
select 't2',t2.ID,t2.requestf , t2.emailf from table_2 t2
) s
right join (select cast('aaa@bb.com' as char(20)) as emailtofind) t on t.emailtofind = cast(s.email as char(20))

结果

+-------------+------+------------+----------------+----------------+---------+
| emailtofind | id   | requestf   | ok found in t1 | ok found in t2 | Message |
+-------------+------+------------+----------------+----------------+---------+
| aaa@bb.com  |  100 | service xx |                | y              | Ok      |
+-------------+------+------------+----------------+----------------+---------+
1 row in set (0.00 sec)

mysql内连接table1和表2

1 个答案: