Question

以下是表格：

info
+----------------------+
|userid|job_id|position|
+----------------------+
|1     |1     |2       |
|2     |2     |1       |
+----------------------+

job
+----------------+
|job_id|job_title|
+----------------+
|1     |army     |
|2     |grocer   |
+----------------+

army
+-----------------------+
|position|title   |income|
+------------------------+
|1       |private |$125  |
|2       |corporal|$175  |
+------------------------+

grocer
+-----------------------+
|position|title  |income|
+-----------------------+
|1       |bagger |$100  |
|2       |cashier|$120  |
+-----------------------+

我想：

+----------------------------+
|id|job_title|position|income|
+----------------------------+
|1 |army     |corporal|$175  |
+----------------------------+

每当 id 1 登录并

时

+----------------------------+
|id|job_title|position|income|
+----------------------------+
|2 |grocer   |bagger  |$100  |
+----------------------------+

每当 id 2 被记录在

中时

所以，

if job_title=army

then select table army and info.position=army.position

if job_title=grocer

then select table grocer and  info.position=grocer.position

我无法弄清楚这一点。我试过了

SELECT * FROM info,job,army,grocer 
WHERE userid='$userid' and passid='$passid' and info.job_id=job.job_id;

适用于job_title但不适用于title或income。所以我试过

SELECT * FROM info,job,army,grocer 
WHERE userid='$userid' and passid='$passid' and info.job_id=job.job_id and info.position=army.position;

适用于 id 1 但不适用于 id 2 。我需要像

这样的东西

SELECT * FROM info,job,army,grocer 
WHERE userid='$userid' and passid='$passid' and info.job_id=job.job_id and info.position=_.position;

其中_是军队或杂货店，基于用户ID 和position

Answer 1

试试这个：

--drop table #info
SELECT  *
INTO #info
FROM    
(
    select 1 as userid , 1 as job_id, 2 as position
        UNION
    select 2 as userid , 2 as job_id, 1 as position
) a

--drop table #job
SELECT  *
into #job
FROM    
(
SELECT 1 as job_id , 'army' as job_title
    union
SELECT 2 as job_id , 'grocer' as grocer
) a

--drop table #army
SELECT  *
into #army
FROM    
(
select 1 as position , 'private' as title, 125 as income
    union
select 2 as position , 'corporal' as title, 175 as income
) a

--drop table #grocer
SELECT  *
into #grocer
FROM    
(
select 1 as position , 'bagger' as title, 100 as income
    union
select 2 as position , 'cashier' as title, 120 as income
) a


--for whenever id 1 is logged in:
SELECT  a.userid as Id, b.job_title, c.title as position, c.income
FROM    #info a
    inner join #job b
        on a.job_id = b.job_id
    inner join #army c
        on a.position = c.position
where a.userid = 1

--for whenever id 2 is logged in:
SELECT  a.userid as Id, b.job_title, c.title as position, d.income
FROM    #info a
    inner join #job b
        on a.job_id = b.job_id
    inner join #army c
        on a.position = c.position
    inner join #grocer d
        on c.position = d.position
where a.userid = 2

根据不同表中的值选择表

1 个答案: