Question

我有以下数据。

ID        UserID        Grade
1           1             A
2           1             A
3           1             a
4           1             a
5           1             b
6           1             C
7           1             c
8           1             b
9           2             b
10          2             C
11          1             b
12          2             A

我想要下面的结果。

RunningNumber UserID         Result         Count
1                1             a              4
2                1             b              1
3                1             c              2
4                1             b              2
5                2             b              1
6                2             c              1
7                2             a              1

结果将计算列级别（不区分大小写），并且必须按用户ID分组。

Answer 1

使用如下

 SELECT UserID, Grade, COUNT(ID) AS Count 
     FROM Table1
 GROUP BY UserID,Grade;

更新

SELECT ROW_NUMBER() 
    OVER (ORDER BY Grade) AS RunningNumber , 
   UserID, Grade, COUNT(ID) AS Count 
     FROM Table1
 GROUP BY UserID,Grade;

Answer 2

只需使用Group by＆amp;数

SELECT  ID ,UserID,Grade 
COUNT(ID) FROM Tabel
GROUP BY Grade,UserID;

Answer 3

您可以DENSE_RANK使用RunningNumber，每个ID使用COUNT(*)OVER，ROW_NUMBER仅使用第一行：

WITH CTE AS
(
    SELECT 
       RunningNumber = DENSE_RANK() OVER (ORDER BY UserID, Grade),
       UserID, Result = Grade,        
       [Count] = COUNT(*) OVER (PARTITION BY UserID, Grade),
       RowNum  = ROW_NUMBER()  OVER (PARTITION BY UserID, Grade ORDER BY ID)
    FROM dbo.TableName
)
SELECT RunningNumber, UserID, Result,  [Count]
FROM CTE
WHERE RowNum = 1

Demo

Answer 4

$query_not="SELECT count(status) AS sum FROM `user_leave_details`WHERE     (status='2' Or status='3') AND user_id_no='$user_id_no'";
$result=mysqli_query($bd,$query_not);
while($arr=mysqli_fetch_array($result))
{   
$sum=$arr['sum'];
}

connect.php
<?php
$mysql_hostname = "localhost";
$mysql_user = "root";
$mysql_password = "";
$mysql_database = "";
$bd=mysqli_connect($mysql_hostname,$mysql_user,$mysql_password,$mysql_database);
?>

从数据库计算值

4 个答案: