从不同的表嵌套选择

Question

我想做一个嵌套的选择，但不同的表。这里是我的查询：

SELECT contact_profile.name, main_app.fk_lkp_app, main_app.id as main_id, 
(
    -- if the main_app.fk_lkp_app value is 1 then do this 
    SELECT SUM(translation_app.amount) 
    FROM translation_app 
    WHERE translation_app.fk_main_app = main_app.id 
    AND translation_app.status = 2 
    AND main_app.srf_number is not null 
    AND main_app.fk_invoice is not null 

    -- if the main_app.fk_lkp_app value is 2 then do this 
    SELECT SUM(interpretation_app.amount) 
    FROM interpretation_app 
    WHERE interpretation_app.fk_main_app = main_app.id 
    AND interpretation_app.status =2 
    AND main_app.srf_number is not null 
    AND main_app.fk_invoice is not null 

    -- if the main_app.fk_lkp_app value is 3 then do this 
    SELECT SUM(course_app.amount) 
    FROM course_app 
    WHERE course_app.fk_main_app = main_app.id 
    AND course_app.status =2 
    AND main_app.srf_number is not null 
    AND main_app.fk_invoice is not null 
) as amount 
FROM contact_profile 
LEFT JOIN main_app ON main_app.fk_contact_profile = contact_profile.id 
WHERE main_app.fk_lkp_app in (1,2,3) 
AND main_app.srf_number is not null 
AND main_app.fk_invoice is not null 
GROUP BY contact_profile.name 
ORDER BY amount DESC 

正如你所看到的，“量”字段基于main_app.fk_lkp_app值不同的表中选择。问题是如何做这个查询的最佳方法？我坚持使用“main_app.fk_lkp_app”值参数部分。

我甚至使用CASE尝试作为建议，但它不断给我的错误代码＃1064

SELECT contact_profile.name, main_app.fk_lkp_app, main_app.id as main_id, 
(
CASE 
    WHEN main_app.fk_lkp_app = '1' 
    THEN ( 
     SELECT SUM(translation_app.amount) 
     FROM translation_app 
     WHERE translation_app.fk_main_app = main_app.id 
    ) 
    WHEN main_app.fk_lkp_app = '2' 
    THEN (
     SELECT SUM(interpretation_app.amount) 
     FROM interpretation_app 
     WHERE interpretation_app.fk_main_app = main_app.id 
    ) 
    WHEN main_app.fk_lkp_app = '3' 
    THEN (
     SELECT SUM(course_app.amount) 
     FROM course_app 
     WHERE course_app.fk_main_app = main_app.id 
    ) 
    ELSE 0 
END CASE 
) as amount 
FROM contact_profile 
LEFT JOIN main_app ON main_app.fk_contact_profile = contact_profile.id 
WHERE main_app.fk_lkp_app in (1,2,3) 
AND main_app.srf_number is not null 
AND main_app.fk_invoice is not null 
GROUP BY contact_profile.name 
ORDER BY amount DESC 

奇怪的是，如果我不使用的情况下，仅使用1从3变体中（例如：我只从translation_app表中选择）查询正在工作。

Answer 1

你可以使用CASE表达

http://dev.mysql.com/doc/refman/5.7/en/control-flow-functions.html#operator_case

检查main_app.fk_lkp的价值，并执行基于该value.I查询无法测试，但这样的事情应该工作

SELECT contact_profile.name, main_app.fk_lkp_app, main_app.id as main_id, 
CASE 
    WHEN main_app.fk_lkp_app = 1 THEN (/* your query here */) 
    WHEN main_app.fk_lkp_app = 2 THEN (/* your query here */) 
    WHEN main_app.fk_lkp_app = 2 THEN (/* your query here */) 
    ELSE 0 
END AS amount ....