Question

很抱歉我的描述不好，但我想知道最后一个吃了每个水果的人的最新日期。我不认为我正在将它分组，或者我完全错过了一些东西。有人可以帮我弄这个吗？提前谢谢。

T1
+--------+--------+--------+
| person | fruit  |  date  |
+--------+--------+--------+
|  Bob   | apple  | 1-1-13 |
|  Bob   | apple  | 1-4-13 |
|  Bob   | apple  | 1-6-13 |
|  Bob   | orange | 1-2-13 |
|  John  | apple  | 1-3-13 |
|  John  | orange | 1-2-13 |
|  John  | orange | 1-4-13 |
|  John  | orange | 1-9-13 |
+--------+--------+--------+

Results I'm looking for
+--------+--------+--------+
| person | fruit  |  date  |
+--------+--------+--------+
|  Bob   | apple  | 1-6-13 |
|  Bob   | orange | 1-2-13 |
|  John  | apple  | 1-3-13 |
|  John  | orange | 1-9-13 |
+--------+--------+--------+

我在阅读类似的帖子时尝试了什么。

SELECT * FROM T1 
  JOIN (
    SELECT person,max(date) AS latest 
      FROM T1 GROUP BY person
    ) T2
  ON T1.person = T2.person AND T1.date = T2.latest

Resulting in
+--------+--------+--------+
| person | fruit  |  date  |
+--------+--------+--------+
|  Bob   | apple  | 1-6-13 |
|  John  | orange | 1-9-13 |
+--------+--------+--------+

如何将水果柱分开？

Answer 1

此处无需JOIN。简单聚合GROUP BY。

SELECT person, fruit, DATE_FORMAT(MAX(date), '%c-%e-%y') AS max_date
FROM T1
GROUP BY person, fruit

<强>结果

| PERSON |  FRUIT | MAX_DATE |
------------------------------
|    Bob |  apple |   1-6-13 |
|    Bob | orange |   1-2-13 |
|   John |  apple |   1-3-13 |
|   John | orange |   1-9-13 |

获取mysql行的最新日期

1 个答案:

See it in action