我有两张桌子:
表A
+-------+--------+
| Data | Acc_No |
+-------+--------+
| Unix | 10 |
| Linux | 20 |
+-------+--------+
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表B
+-----+----------+--------+
| Obj | Type | Amount |
+-----+----------+--------+
| 10 | rev | 100.00 |
| 10 | revision | 200.00 |
+-----+----------+--------+
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我希望输出像这样
+-------+--------+----------+--------+-----------+----------+
| Data | Acc_No | Type | Amount | Type_1 | Amount_1 |
+-------+--------+----------+--------+-----------+----------+
| Unix | 10 | rev | 100 | revision | 200 |
+-------+--------+----------+--------+-----------+----------+
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我尝试使用简单的连接.这是查询:
SELECT a.Data,a.Acc_No, b.Type, b.Amount, bb.Type AS "Type_1", bb.Amount AS "Amount_1"
FROM TableA a,TableB b, TableB bb
WHERE a.Acc_No = b.Obj AND
b.Obj = bb.Obj AND
bb.Obj = a.Acc_No AND
a.Acc_No =10;
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但我得到了这个输出.
+------+--------+----------+--------+----------+----------+
| Data | Acc_No | Type | Amount | Type_1 | Amount_1 |
+------+--------+----------+--------+----------+----------+
| Unix | 10 | rev | 100.00 | rev | 100.00 |
| Unix | 10 | revision | 200.00 | rev | 100.00 |
| Unix | 10 | rev | 100.00 | revision | 200.00 |
| Unix | 10 | revision | 200.00 | revision | 200.00 |
+------+--------+----------+--------+----------+----------+
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我试着在这个网站上搜索答案,我甚至用Google搜索,但我找不到合适的答案.
您当前的查询已关闭,但我建议对其进行一些小的更改以获得结果.如果要使用JOIN"旋转"数据,则需要区分要TableB在每个后续连接中返回的值.
例如,当您想要返回时type=rev,您需要为该值包含特定的过滤器.然后你会做同样的事情type=revision.我还建议使用a LEFT JOIN加入,TableB如果你没有两个type值,Acc_no那么你仍然会返回数据.
select
a.data,
a.acc_no,
b.type,
b.amount,
bb.type as type_1,
bb.amount as amount_1
from tablea a
left join tableb b
on a.acc_no = b.obj
and b.type = 'rev'
left join tableb bb
on a.acc_no = bb.obj
and bb.type = 'revision';
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您也可以使用某些条件聚合获得此结果,然后您不必TableB多次加入:
select
a.data,
a.acc_no,
max(case when b.type = 'rev' then b.type end) as type,
max(case when b.type = 'rev' then b.Amount end) as Amount,
max(case when b.type = 'revision' then b.type end) as type_1,
max(case when b.type = 'revision' then b.Amount end) as Amount_1
from tablea a
left join tableb b
on a.acc_no = b.obj
group by a.data, a.acc_no
order by a.acc_no;
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