如果所有记录都匹配，如何从连接中获取结果

Question

我在编写SQL查询时遇到问题。我有两个名为'loans'和'loan_emilists'

的表

我想从第一张表中获取贷款ID，即贷款，其中loan.emilists表状态全部为1。

例如： -

Loan id = 1因为在第二个表中，即'loan_emilists'loor_id = 1，所有状态值都为1。

MySQL查询： -

  

SELECT DISTINCT（loans.id）FROM loans LEFT JOIN loan_emilists ON   loans.id = loan_emilists.loan_id WHERE loan_emilists.status = 1

但它给了我1和2贷款.id。

Answer 1

你可以使用not exists（）：

SELECT s.id FROM loans s
WHERE NOT EXISTS(select 1 from loan_emilists t
                 where s.id = t.loan_id 
                   and t.status = 0)

如果贷款中的每个id在loan_emilists中总是有一行，如果你不需要加入，然后使用not exists（）：

SELECT DISTINCT s.id FROM loans s
INNER JOIN loan_emilists s1
 ON(s1.loan_id = s.id and s1.status = 1)
WHERE NOT EXISTS(select 1 from loan_emilists t
                 where s.id = t.loan_id 
                   and t.status = 0)

也可以使用左连接：

SELECT DISTINCT s.id FROM loans s
INNER JOIN loan_emilists s1
 ON(s1.loan_id = s.id and s1.status = 1)
LEFT JOIN loan_emilists t
 ON(t.loan_id = s.id AND t.status = 0)
WHERE t.loan_id is null