在至少一行滿足條件的情況下，如何返回所有行的值？

Question

我正在嘗試返回ID的所有行，其中一個或多個Num_Occurrence行大於等於10。

這是原始數據的示例：

+------+-----------+----------------+
| 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              |
+------+-----------+----------------+

這是所需的輸出：

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

我了解以下內容不起作用：

SELECT *
FROM tbl
WHERE Num_Occurrence >= 10

因為那只會返回以下行：

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

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

謝謝！！

Answer 1

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

這里的“ EXISTS”子句使用子查詢來查找Num_Occurrence> = 10的所有行，然后將其與完整表進行比較以獲取具有匹配ID的所有行。

Answer 2

您可以這樣做：

select t.*
from tbl t
where exists (select 1
              from tbl t2
              where t2.id = t.id and t2.id >= 10
             );