Question

SELECT *, j.company_id as companyid, j.id as jid, p.id as pid, f.id as fave_id, f.user_id as user_id
FROM people p
LEFT JOIN job j 
ON p.job_id = j.id
LEFT JOIN favourites f
ON p.id = f.people_id
WHERE p.company_id = 1
ORDER BY p.id ASC

我有上面的MySQL查询工作正常，最后两个选项是我需要解决的问题。

默认情况下，每行输出null，否则会输出fave_id为46和user_id的整数为6

我真正需要弄清楚的是，如果fave_id和user_id在我的头脑中都不是空的话，是否可以从查询的SELECT部分输出这样的事情

SELECT *, j.company_id as companyid, j.id as jid, p.id as pid, f.id as fave_id, f.user_id as user_id, [if fave_id && user_id then favorite = true] 
FROM people p
LEFT JOIN job j 
ON p.job_id = j.id
LEFT JOIN favourites f
ON p.id = f.people_id
WHERE p.company_id = 1
ORDER BY p.id ASC

Answer 1

CASE语句可以检查值并返回所需的值：

SELECT *, j.company_id as companyid, j.id as jid, p.id as pid, f.id as fave_id, f.user_id as user_id, 
    (CASE WHEN fave_id IS NOT NULL AND user_id IS NOT NULL THEN 1 ELSE 0 END) AS favorite 
FROM people p
LEFT JOIN job j 
ON p.job_id = j.id
LEFT JOIN favourites f
ON p.id = f.people_id
WHERE p.company_id = 1
ORDER BY p.id ASC

Answer 2

使用and：

fave_id is not null and user_id is not null

MySQL SELECT查询选择中的where子句

2 个答案: