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Mysql查询(相同)不同的结果集返回不同的结果顺序

In lại Tác giả: Walker 123 更新时间:2023-11-29 23:33:35 28 4
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我对 Mysql 世界很天真......:)我试图使用连接从表中查询,

我遇到结果集问题...表结构如下

下面...

VIDEO_XXXXX

| Field | Type | Null | Key | Default | Extra |
| VIDEO_ID | int(255)
| TAG_NAME | varchar(1000)
| SHOP_ID | varchar(100) FK
| LOVES | int(255)
| VIDEO_URL | varchar(1000)
| UPLOADED_DATE |


PLACE_TABLE...

SHOP_ID | varchar(100) PK

查询看起来像..

   SELECT V.SHOP_ID
FROM VIDEO_xxxx V
JOIN STUB_xxxx_TABLE S ON S.SHOP_ID=V.SHOP_ID
WHERE S.COUPON_STATUS='REDEEMED'
GROUP BY V.SHOP_ID
ORDER BY COUNT(S.COUPON_GENERATED_ID) DESC;

结果集如下...

+-----------+
| SHOP_ID |
+-----------+
| 001 |
| 004 |
| 005 |
| 003 |
| 002 |

第二个查询是......

   SELECT V.SHOP_ID
FROM SECOND_TABLE V
JOIN STUB_xxxx_TABLE S ON S.SHOP_ID=V.SHOP_ID
WHERE S.COUPON_STATUS='REDEEMED'
GROUP BY V.SHOP_ID
ORDER BY COUNT(S.COUPON_GENERATED_ID) DESC;

结果集是...

+--------+
| SHOP_ID|
+--------+
|001 |
|004 |
|003 |
|002 |
005 |

我想知道为什么 shop_id 的顺序会打乱....:( :(任何帮助将不胜感激...:),

câu trả lời hay nhất

您可以将表达式 COUNT(S.COUPON_GENERATED_ID) 添加到 where 子句:

   SELECT V.SHOP_ID,COUNT(S.COUPON_GENERATED_ID)
FROM SECOND_TABLE V
JOIN STUB_xxxx_TABLE S ON S.SHOP_ID=V.SHOP_ID
WHERE S.COUPON_STATUS='REDEEMED'
GROUP BY V.SHOP_ID
ORDER BY COUNT(S.COUPON_GENERATED_ID) DESC;

如果您不希望结果中出现此额外值,您可以使用子查询:

 SELECT SHOP_ID
TỪ (
SELECT V.SHOP_ID,COUNT(S.COUPON_GENERATED_ID) AS CNT
FROM SECOND_TABLE V
JOIN STUB_xxxx_TABLE S ON S.SHOP_ID=V.SHOP_ID
WHERE S.COUPON_STATUS='REDEEMED'
GROUP BY V.SHOP_ID
)
ORDER BY CNT DESC

关于Mysql查询(相同)不同的结果集返回不同的结果顺序,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26465722/

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