sql - 如何返回所有行的值，其中至少有一行满足条件？

Question

我正在尝试返回 ID 的所有行，其中一个或多个 Num_Occurrence 行 >=10。

这是原始数据的示例:

+------+-----------+----------------+
| ID   | YearMonth | Num_Occurrence |
+------+-----------+----------------+
| 1234 | 201601    | 7              |
+------+-----------+----------------+
| 1234 | 201602    | 8              |
+------+-----------+----------------+
| 1234 | 201603    | 12             |
+------+-----------+----------------+
| 1234 | 201604    | 9              |
+------+-----------+----------------+
| 9898 | 201601    | 9              |
+------+-----------+----------------+
| 9898 | 201602    | 8              |
+------+-----------+----------------+
| 9898 | 201603    | 9              |
+------+-----------+----------------+
| 9898 | 201604    | 6              |
+------+-----------+----------------+

这是期望的输出:

+------+-----------+----------------+
| ID   | YearMonth | Num_Occurrence |
+------+-----------+----------------+
| 1234 | 201601    | 7              |
+------+-----------+----------------+
| 1234 | 201602    | 8              |
+------+-----------+----------------+
| 1234 | 201603    | 12             |
+------+-----------+----------------+
| 1234 | 201604    | 9              |
+------+-----------+----------------+

我知道以下内容不起作用:

SELECT *
FROM tbl
WHERE Num_Occurrence >= 10

因为那只会返回这一行:

+------+-----------+----------------+
| ID   | YearMonth | Num_Occurrence |
+------+-----------+----------------+
| 1234 | 201603    | 12             |
+------+-----------+----------------+

如前所述，我需要返回 Num_Occurrence >=10 的任何 ID 的所有行。

谢谢!!

Answer 1

SELECT * FROM [tbl] t1
WHERE EXISTS (SELECT * FROM [tbl] t2
              WHERE t2.ID = t1.id
              AND t2.Num_Occurrence >= 10);

此处的“EXISTS”子句使用子查询来查找 Num_Occurrence >= 10 的所有行，然后将其与完整表进行比较以获取具有匹配 ID 的所有行。