在MySQL中有一个很好的方法来复制SQL Server函数ROW_NUMBER()?
例如:
SELECT
col1, col2,
ROW_NUMBER() OVER (PARTITION BY col1, col2 ORDER BY col3 DESC) AS intRow
FROM Table1
然后,例如,我可以添加一个条件,将intRow限制为1,以获得每个(col1, col2)对的col3最高的单行。
在MySQL中有一个很好的方法来复制SQL Server函数ROW_NUMBER()?
例如:
SELECT
col1, col2,
ROW_NUMBER() OVER (PARTITION BY col1, col2 ORDER BY col3 DESC) AS intRow
FROM Table1
然后,例如,我可以添加一个条件,将intRow限制为1,以获得每个(col1, col2)对的col3最高的单行。
当前回答
我总是遵循这个模式。给定这个表格:
+------+------+
| i | j |
+------+------+
| 1 | 11 |
| 1 | 12 |
| 1 | 13 |
| 2 | 21 |
| 2 | 22 |
| 2 | 23 |
| 3 | 31 |
| 3 | 32 |
| 3 | 33 |
| 4 | 14 |
+------+------+
你可以得到这样的结果:
+------+------+------------+
| i | j | row_number |
+------+------+------------+
| 1 | 11 | 1 |
| 1 | 12 | 2 |
| 1 | 13 | 3 |
| 2 | 21 | 1 |
| 2 | 22 | 2 |
| 2 | 23 | 3 |
| 3 | 31 | 1 |
| 3 | 32 | 2 |
| 3 | 33 | 3 |
| 4 | 14 | 1 |
+------+------+------------+
通过运行这个不需要定义任何变量的查询:
SELECT a.i, a.j, count(*) as row_number FROM test a
JOIN test b ON a.i = b.i AND a.j >= b.j
GROUP BY a.i, a.j
其他回答
在MySQL中没有像rownum, row_num()这样的函数,但方法如下:
select
@s:=@s+1 serial_no,
tbl.*
from my_table tbl, (select @s:=0) as s;
我也会投票给Mosty Mostacho的解决方案,对他的查询代码进行了轻微的修改:
SELECT a.i, a.j, (
SELECT count(*) from test b where a.j >= b.j AND a.i = b.i
) AS row_number FROM test a
会得到相同的结果:
+------+------+------------+
| i | j | row_number |
+------+------+------------+
| 1 | 11 | 1 |
| 1 | 12 | 2 |
| 1 | 13 | 3 |
| 2 | 21 | 1 |
| 2 | 22 | 2 |
| 2 | 23 | 3 |
| 3 | 31 | 1 |
| 3 | 32 | 2 |
| 3 | 33 | 3 |
| 4 | 14 | 1 |
+------+------+------------+
对于表格:
+------+------+
| i | j |
+------+------+
| 1 | 11 |
| 1 | 12 |
| 1 | 13 |
| 2 | 21 |
| 2 | 22 |
| 2 | 23 |
| 3 | 31 |
| 3 | 32 |
| 3 | 33 |
| 4 | 14 |
+------+------+
唯一的区别是查询不使用JOIN和GROUP BY,而是依赖于嵌套选择。
当我们有一个以上的列时,这个工作完美地为我创建RowNumber。这里是两列。
SELECT @row_num := IF(@prev_value= concat(`Fk_Business_Unit_Code`,`NetIQ_Job_Code`), @row_num+1, 1) AS RowNumber,
`Fk_Business_Unit_Code`,
`NetIQ_Job_Code`,
`Supervisor_Name`,
@prev_value := concat(`Fk_Business_Unit_Code`,`NetIQ_Job_Code`)
FROM (SELECT DISTINCT `Fk_Business_Unit_Code`,`NetIQ_Job_Code`,`Supervisor_Name`
FROM Employee
ORDER BY `Fk_Business_Unit_Code`, `NetIQ_Job_Code`, `Supervisor_Name` DESC) z,
(SELECT @row_num := 1) x,
(SELECT @prev_value := '') y
ORDER BY `Fk_Business_Unit_Code`, `NetIQ_Job_Code`,`Supervisor_Name` DESC
我发现最好的解决方案是使用这样的子查询:
SELECT
col1, col2,
(
SELECT COUNT(*)
FROM Table1
WHERE col1 = t1.col1
AND col2 = t1.col2
AND col3 > t1.col3
) AS intRow
FROM Table1 t1
分区BY列只是用'='进行比较,并用and分隔。ORDER BY列将与'<'或'>'进行比较,并以or分隔。
我发现这是非常灵活的,即使它有点昂贵。
这也可以是一个解决方案:
SET @row_number = 0;
SELECT
(@row_number:=@row_number + 1) AS num, firstName, lastName
FROM
employees