好的,这是我的表:
product_id version_id update_id patch_id
1 1 0 0
1 1 1 0
1 1 1 1
1 1 2 0
1 1 2 1
2 1 0 0
2 2 0 0
2 3 0 0
2 3 0 1
3 1 0 0
3 1 0 1
现在我想选择产品的最新版本,因此具有最高update_id和版本的版本PATCH_ID。
例如,
的最新版本我在使用GROUP BY和HAVING尝试各种各样的东西,尝试了子查询,但我仍然无法找到实现这一目标的方法。
有人可以帮我找到合适的查询,还是应该考虑为此编写一个php函数?
修改
其他一些信息: - 列一起是主键(有更多列,但对于这个问题,它们并不重要) - 没有列是自动增量
这是表格:
CREATE TABLE IF NOT EXISTS `db`.`patch` (
`product_id` INT NOT NULL ,
`version_id` INT NOT NULL ,
`update_id` INT NOT NULL ,
`patch_id` INT NOT NULL
PRIMARY KEY (`product_id`, `version_id`, `update_id`, `patch_id`) ,
INDEX `fk_patch_update1` (`product_id` ASC, `version_id` ASC, `update_id` ASC) )
修改2
标记为重复,但不是:另一个问题是查找高于三个不同列中任何一列的值的记录。
在这个问题中,我们寻找按product_id分组的最高版本号。
编辑3
rgz的回答再次告诉我,这是重复的。首先:这个问题比较老了。其次,我不认为答案是一样的。
rgz建议使用以下查询:
SELECT product_id, GREATEST(version_id, update_id, patch_id) AS latest_version FROM patch.
GREATEST(1,2,3)返回3,对吗? Wat,如果我们有这些值:
product_id version_id update_id patch_id
1 1 0 0
1 1 2 8
1 3 0 0
据我了解,此查询将返回:
product_id latest_version
1 1
1 8
1 3
但它应该回归:
product_id version_id update_id patch_id
1 3 0 0
我认为GREATEST无法提供帮助。如果你认为它会,请证明我错了。
答案 0 :(得分:9)
这是独特标识有用的一个例子。
想象一下,您有一个autoincrememnting ID字段,然后您可以使用相关的子查询找到每个产品所需的ID ...
SELECT
*
FROM
yourTable
WHERE
id = (
SELECT id
FROM yourTable AS lookup
WHERE lookup.product_id = yourTable.product_id
ORDER BY version_id DESC, update_id DESC, patch_id DESC
LIMIT 1
)
没有唯一标识符的等效需要多个相关的子查询...
SELECT
*
FROM
yourTable
WHERE
version_id = (
SELECT MAX(version_id)
FROM yourTable AS lookup
WHERE lookup.product_id = yourTable.product_id
)
AND update_id = (
SELECT MAX(update_id)
FROM yourTable AS lookup
WHERE lookup.product_id = yourTable.product_id
AND lookup.version_id = yourTable.version_id
)
AND patch_id = (
SELECT MAX(patch_id)
FROM yourTable AS lookup
WHERE lookup.product_id = yourTable.product_id
AND lookup.version_id = yourTable.version_id
AND lookup.update_id = yourTable.update_id
)
这将明显慢于具有唯一标识符列的表。
另一种选择(没有唯一标识符)是在不同的聚合级别上自我加入。
SELECT
yourTable.*
FROM
(SELECT product_id, MAX(version_id) AS max_version_id FROM yourTable GROUP BY product_id) AS version
INNER JOIN
(SELECT product_id, version_id, MAX(update_id) AS max_update_id FROM yourTable GROUP BY product_id, version_id) AS update
ON update.product_id = version.product_id
AND update.version_id = version.max_version_id
INNER JOIN
(SELECT product_id, version_id, updatE_id, MAX(patch_id) AS max_patch_id FROM yourTable GROUP BY product_id, version_id) AS patch
ON patch.product_id = update.product_id
AND patch.version_id = update.version_id
AND patch.update_id = update.max_update_id
INNER JOIN
yourTable
ON yourTable.product_id = patch.product_id
AND yourTable.version_id = patch.version_id
AND yourTable.update_id = patch.update_id
AND yourTable.patch_id = patch.max_patch_id
答案 1 :(得分:3)
尝试:
Select product_id, version_id, update_id, patch_id
from (select * from MyTable
order by product_id, version_id desc, update_id desc, patch_id desc) as v
group by product_id
对不起,这是@RolandoMySQLDBA。请原谅入侵,但我会根据示例数据发布您的查询及其输出:
mysql> Select product_id, version_id, update_id, patch_id
-> from (select * from prod
-> order by product_id, version_id desc, update_id desc, patch_id desc) as v
-> group by product_id ;
+------------+------------+-----------+----------+
| product_id | version_id | update_id | patch_id |
+------------+------------+-----------+----------+
| 1 | 1 | 2 | 1 |
| 2 | 3 | 0 | 1 |
| 3 | 1 | 0 | 1 |
+------------+------------+-----------+----------+
3 rows in set (0.00 sec)
您的查询如图所示。它也比我的回答简单。好节目!!!
答案 2 :(得分:1)
我没有测试它,但我确信你可以做这样的事情
SELECT val.product_id, MAX(val.myVersion)
FROM (SELECT product_id, ((version_id * 10 + update_id) * 10 + patch_id) as myVersion
FROM myTable) as val
GROUP by product_id;
对不起,这是@RolandoMySQLDBA。请原谅入侵,但我会根据示例数据发布您的查询及其输出:
mysql> SELECT val.product_id, MAX(val.myVersion)
-> FROM (SELECT product_id, ((version_id * 10 + update_id) * 10 + patch_id) as myVersion
-> FROM prod) as val
-> GROUP by product_id;
+------------+--------------------+
| product_id | MAX(val.myVersion) |
+------------+--------------------+
| 1 | 121 |
| 2 | 301 |
| 3 | 101 |
+------------+--------------------+
3 rows in set (0.01 sec)
虽然显示不同,但您的查询在逻辑上适用于原始问题。来自我的+1!
答案 3 :(得分:0)
您可以按max(update_id * maxpatchidvalue + patch_id)
进行分组,以获得您想要的内容。
答案 4 :(得分:0)
试试这个。
SELECT
version_id,
max_update.update_id,
max_patch.patch_id
FROM
TABLE
OUTER APPLY
(
SELECT TOP 1
update_id
FROM
TABLE t1
WHERE
TABLE.version_id = t1.version_id
AND TABLE.product_id = t1.product_id
ORDER BY update_id DESC
) max_update
OUTER APPLY
(
SELECT TOP 1
patch_id
FROM
TABLE t1
WHERE
TABLE.version_id = t1.version_id
AND TABLE.product_id = t1.product_id
AND max_update.update_id = t1.update_id
ORDER BY patch_id DESC
) max_patch
答案 5 :(得分:0)
使用问题中的输入数据和没有键的表格,这是一个相当恶心的方法:
首先,您需要的查询是:
select A.* from prod A LEFT JOIN
(select product_id,max(tagnum) maxtag from
(select *,(version_id*10000+update_id*100+patch_id) tagnum from prod) AA
group by product_id) B
USING (product_id)
WHERE (version_id*10000+update_id*100+patch_id) = B.maxtag;
以下是一些使用所有整数,没有索引和示例数据的示例代码:
DROP TABLE IF EXISTS prod;
CREATE TABLE prod
(
product_id INT,
version_id INT,
update_id INT,
patch_id INT
) ENGINE=MyISAM;
INSERT INTO prod VALUES
( 1, 1, 0, 0 ),
( 1, 1, 1, 0 ),
( 1, 1, 1, 1 ),
( 1, 1, 2, 0 ),
( 1, 1, 2, 1 ),
( 2, 1, 0, 0 ),
( 2, 2, 0, 0 ),
( 2, 3, 0, 0 ),
( 2, 3, 0, 1 ),
( 3, 1, 0, 0 ),
( 3, 1, 0, 1 );
select * from prod;
select *,(version_id*10000+update_id*100+patch_id) tagnum from prod;
select A.* from prod A LEFT JOIN
(select product_id,max(tagnum) maxtag from
(select *,(version_id*10000+update_id*100+patch_id) tagnum from prod) AA
group by product_id) B
USING (product_id)
WHERE (version_id*10000+update_id*100+patch_id) = B.maxtag;
结果如下:
mysql> DROP TABLE IF EXISTS prod;
Query OK, 0 rows affected (0.00 sec)
mysql> CREATE TABLE prod
-> (
-> product_id INT,
-> version_id INT,
-> update_id INT,
-> patch_id INT
-> ) ENGINE=MyISAM;
Query OK, 0 rows affected (0.04 sec)
mysql> INSERT INTO prod VALUES
-> ( 1, 1, 0, 0 ),
-> ( 1, 1, 1, 0 ),
-> ( 1, 1, 1, 1 ),
-> ( 1, 1, 2, 0 ),
-> ( 1, 1, 2, 1 ),
-> ( 2, 1, 0, 0 ),
-> ( 2, 2, 0, 0 ),
-> ( 2, 3, 0, 0 ),
-> ( 2, 3, 0, 1 ),
-> ( 3, 1, 0, 0 ),
-> ( 3, 1, 0, 1 );
Query OK, 11 rows affected (0.00 sec)
Records: 11 Duplicates: 0 Warnings: 0
mysql> select * from prod;
+------------+------------+-----------+----------+
| product_id | version_id | update_id | patch_id |
+------------+------------+-----------+----------+
| 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 |
| 1 | 1 | 1 | 1 |
| 1 | 1 | 2 | 0 |
| 1 | 1 | 2 | 1 |
| 2 | 1 | 0 | 0 |
| 2 | 2 | 0 | 0 |
| 2 | 3 | 0 | 0 |
| 2 | 3 | 0 | 1 |
| 3 | 1 | 0 | 0 |
| 3 | 1 | 0 | 1 |
+------------+------------+-----------+----------+
11 rows in set (0.00 sec)
mysql> select *,(version_id*10000+update_id*100+patch_id) tagnum from prod;
+------------+------------+-----------+----------+--------+
| product_id | version_id | update_id | patch_id | tagnum |
+------------+------------+-----------+----------+--------+
| 1 | 1 | 0 | 0 | 10000 |
| 1 | 1 | 1 | 0 | 10100 |
| 1 | 1 | 1 | 1 | 10101 |
| 1 | 1 | 2 | 0 | 10200 |
| 1 | 1 | 2 | 1 | 10201 |
| 2 | 1 | 0 | 0 | 10000 |
| 2 | 2 | 0 | 0 | 20000 |
| 2 | 3 | 0 | 0 | 30000 |
| 2 | 3 | 0 | 1 | 30001 |
| 3 | 1 | 0 | 0 | 10000 |
| 3 | 1 | 0 | 1 | 10001 |
+------------+------------+-----------+----------+--------+
11 rows in set (0.01 sec)
mysql> select A.* from prod A LEFT JOIN
-> (select product_id,max(tagnum) maxtag from
-> (select *,(version_id*10000+update_id*100+patch_id) tagnum from prod) AA
-> group by product_id) B
-> USING (product_id)
-> WHERE (version_id*10000+update_id*100+patch_id) = B.maxtag;
+------------+------------+-----------+----------+
| product_id | version_id | update_id | patch_id |
+------------+------------+-----------+----------+
| 1 | 1 | 2 | 1 |
| 2 | 3 | 0 | 1 |
| 3 | 1 | 0 | 1 |
+------------+------------+-----------+----------+
3 rows in set (0.00 sec)
mysql>
试一试!!!
CAVEAT
我的回答将支持每种产品
答案 6 :(得分:-1)
MySQL具有GREATEST
功能,因此您应该像这样使用它:
SELECT product_id, GREATEST(version_id, update_id, patch_id) AS latest_version FROM patch.
我将这个问题标记为重复,因为这两个问题基本上都是同一个问题,选择几列的最大值。在这里,您可以在SELECT子句中使用它,在WHERE子句中使用它的另一个问题中。在任何一种情况下,GREATEST都可以解决问题。