根据条件将两个MySql查询组合成一个查询

我使用的是

5.7.20-enterprise-commercial

，有一个MySql查询，如下所示：

SELECT name FROM jobTable where jobId IN (SELECT dataId from logTable
WHERE eValue ='1098hbfce');

如果上述查询返回空值，则我希望运行以下查询：

SELECT name FROM jobTable where procId IN (
SELECT dataId from logTable WHERE eValue ='1098hbfce');

第一个查询和第二个查询之间的唯一区别是在第一个查询中使用了

jobId

。 并且，在第二个查询中使用了

procId

另外，如果第一个查询返回

non-null

值，我不想运行第二个查询

我正在使用以下方法，但它似乎无法正常工作

IF (SELECT name FROM jobTable where jobId IN (SELECT dataId from logTable
   WHERE eValue ='1098hbfce')) IS NOT NULL 
   EXECUTE (SELECT name FROM jobTable where procId IN (
   SELECT dataId from logTable WHERE eValue ='1098hbfce'));

要求是在SQL查询本身中执行，而不创建存储过程。

---

使用

CASE

表达式，

jobTable

和

logTable

之间没有关系：

SELECT name 
FROM jobTable 
where (CASE 
       WHEN jobId IN (SELECT dataId from logTable WHERE eValue ='1098hbfce')
       THEN 1
       WHEN procId IN (SELECT dataId from logTable WHERE eValue ='1098hbfce') 
       THEN 1 
      END) = 1 ;

---

使用

CASE

表达式：

SELECT name 
FROM jobTable 
where (CASE 
       WHEN jobId IN (SELECT dataId from logTable WHERE eValue ='1098hbfce')
       THEN 1
       WHEN procId IN (SELECT dataId from logTable WHERE eValue ='1098hbfce') 
       THEN 1 
      END) = 1 ;

结合

drop table if exists t,log_table;
create table t(name varchar(10),jobid int,procid int);
create table log_table(dataid int,evalue varchar(10));

insert into t values ('aaa',null,null),('bbb',null,1),('ccc',1,null);
insert into log_table values(1,'1098hbfce');

select *,
        coalesce(jobid,procid),
        coalesce(procid,jobid)
        
from t
join log_table lt on 
        dataid = coalesce(jobid,procid) or dataid = coalesce(procid,jobid)  ;

+------+-------+--------+--------+-----------+------------------------+------------------------+
| name | jobid | procid | dataid | evalue    | coalesce(jobid,procid) | coalesce(procid,jobid) |
+------+-------+--------+--------+-----------+------------------------+------------------------+
| bbb  |  NULL |      1 |      1 | 1098hbfce |                      1 |                      1 |
| ccc  |     1 |   NULL |      1 | 1098hbfce |                      1 |                      1 |
+------+-------+--------+--------+-----------+------------------------+------------------------+
2 rows in set (0.001 sec)

        

结合

drop table if exists t,log_table;
create table t(name varchar(10),jobid int,procid int);
create table log_table(dataid int,evalue varchar(10));

insert into t values ('aaa',null,null),('bbb',null,1),('ccc',1,null);
insert into log_table values(1,'1098hbfce');

select *,
        coalesce(jobid,procid),
        coalesce(procid,jobid)
        
from t
join log_table lt on 
        dataid = coalesce(jobid,procid) or dataid = coalesce(procid,jobid)  ;

+------+-------+--------+--------+-----------+------------------------+------------------------+
| name | jobid | procid | dataid | evalue    | coalesce(jobid,procid) | coalesce(procid,jobid) |
+------+-------+--------+--------+-----------+------------------------+------------------------+
| bbb  |  NULL |      1 |      1 | 1098hbfce |                      1 |                      1 |
| ccc  |     1 |   NULL |      1 | 1098hbfce |                      1 |                      1 |
+------+-------+--------+--------+-----------+------------------------+------------------------+
2 rows in set (0.001 sec)

        

---

看起来很熟悉你昨天问这个了吗？看起来很熟悉你昨天问这个了吗？