sql-server - 基于JOIN和聚合条件的SQL DELETE

Question

我有两个包含类别和日期的表。

表 1:

cat date
A   20160102
A   20160103
A   20160104
B   20170202
B   20170203
B   20170204

表 2:

cat date
A   20160103
A   20160104
A   20160105
B   20170203
B   20170206
B   20170207

我现在想要从表 1 中删除日期等于或晚于每个类别表 2 的最早日期的所有行.

类别 A 的最早日期是20160103。类别 B 的最早日期是20170203。因此，条目 ('A','20160103')、('A','20160104') 和 ('B','20170204' ) 应从表 1 中删除。

为了进行测试，我尝试创建一个 SELECT 语句来选择我要删除的值。目前我想出了这个:

SELECT
    t1.id
   ,t1.holiday
   ,MIN(t2.holiday)
FROM
    table1 t1
    INNER JOIN table2 t2
        ON t2.id = t1.id
GROUP BY
    t1.id
   ,t1.holiday

(对我来说)下一个逻辑步骤是添加以下 WHERE 子句

SELECT
    t1.id
   ,t1.holiday
   ,MIN(t2.holiday)
FROM
    table1 t1
    INNER JOIN table2 t2
        ON t2.id = t1.id
WHERE
    t1.holiday >= MIN(t2.holiday)
GROUP BY
    t1.id
   ,t1.holiday

但是，这会产生以下错误:

An aggregate may not appear in the WHERE clause unless it is in a subquery contained in a HAVING clause or a select list, and the column being aggregated is an outer reference.

如何解决这个问题？

Answer 1

这回答了第一个问题(关于从表 1 中删除)

DELETE t1
FROM Table1 AS t1
     INNER JOIN (
         SELECT
            cat,
            min_date=MIN([date])
         FROM
            Table2
         GROUP BY
            cat
    ) AS t2 ON
        t2.cat=t1.cat
WHERE
    t1.[date]>=t2.min_date