Mysql子查询问题

Question

• 贷款(列:id，用户)
• 返回(列:id、installment、scheduleDate)

如果一笔贷款要分12期还清，则有12 strip 贷款id的还款记录。

我想找出哪些用户的 scheduleDate 的所有返回之和大于 2000。

SELECT user 
FROM {loans} ka 
WHERE 
(
   SELECT MAX(inst) FROM 
   (
        SELECT SUM(installment) AS inst 
        FROM {payback} 
        WHERE id IN 
        (
            SELECT id 
            FROM {loans} 
            WHERE user = ka.user
        )  
   GROUP BY scheduleDate
   ) as t1 LIMIT 0,1
) > 2000

我收到错误: “where 子句”中的未知列“ka.user”

Answer 1

这是否按要求工作？它假定 payback.id 是引用 loads.id 的外键。

SELECT
  user, scheduleDate, sumInstallments
FROM
(
    SELECT
      l.user, pb.scheduleDate, SUM(pb.installment) AS sumInstallments
    FROM
      payback AS pb
    JOIN
      loans AS l
    ON
      l.id = pb.id
    GROUP BY
      l.user, pb.scheduleDate
) AS tempId
WHERE
  sumInstallments > 2000