[英]SQL - SELECT from a table if a certain condition is met, else SELECT from another
如何在SQL中執行此操作:
SELECT CLIENT, PAYMENT_CODE FROM PAYMENTS_TABLE
If PAYMENT_CODE = 1, SELECT other data from table_1
Else, SELECT other data from table_2
我想要的結果是:
| Client | Payment Code = 1 | Address from table_1 | Phone from table_1 | ...
要么
| Client | Payment Code 1 | Address from table_2 | Phone from table_2 | ...
提前致謝!
盡管Rajesh的解決方案很好,但是您也可以使用兩個LEFT JOIN來完成;
SELECT pt.CLIENT, pt.PAYMENT_CODE,
COALESCE(t1.address, t2.address) address,
COALESCE(t1.phone, t2.Phone) phone
FROM PAYMENTS_TABLE pt
LEFT JOIN Table_1 t1
ON pt.client_code = t1.client_code AND pt.PAYMENT_CODE = 1
LEFT JOIN Table_2 t2
ON pt.client_code = t2.client_code AND pt.PAYMENT_CODE <> 1
編輯:根據對另一個答案的評論,添加了client_code
作為鏈接條件。
您需要使用UNION
,當payment_code = 1時,一個SELECT
查詢可從表_1獲取詳細信息,而另一個SELECT
,如果payment_code不等於1,則可從表_2獲取詳細信息
SELECT CLIENT, PAYMENT_CODE, T1.Address, T1.Phone FROM PAYMENTS_TABLE
JOIN Table_1 T1
ON -- your condition
WHERE PAYMENT_CODE =1
UNION
SELECT CLIENT, PAYMENT_CODE, T2.Address, T2.Phone FROM PAYMENTS_TABLE
JOIN Table_2 T2
ON -- your condition
WHERE PAYMENT_CODE <> 1
您沒有告訴我們table_1和table_2如何鏈接到payments_table,但是這樣的方法應該起作用:
select p.client,
p.payment_code,
case
when payment_code = 1 then t1.address
else t2.address
end as address,
case
when payment_code = 1 then t1.phone
else t2.phone
end as phon
from payments_table p
left join table_1 t1 on p.some_column = t1.some_column
left join table_2 t2 on p.some_column = t2.come_column
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