Question

我将值“星期日”“星期六”“假日”“离开”“付费休假”放在名为“离开”的同一列中。但是，我将名为“WorkedDays”的列需要表格中的工作日总和。

$sqls = mysqli_query($conn, "SELECT EmployeeNoA, 
                                    EmployeeNameA, 
                                    SUM(AttenInDays), 
                                    COUNT(Leaves=='Sunday'), 
                                    COUNT(Leaves=='Saturday'), 
                                    COUNT(Leaves=='Holiday'), 
                                    COUNT(Leaves=='Leave'), 
                                    COUNT(Leaves=='Paid Leave') 
                            FROM attend 
                            WHERE (SignInDate BETWEEN '$StartDate' AND '$EndDate') 
                            GROUP BY EmployeeNoA, EmployeeNameA");

我需要输出为，

-------------------------------------------------------------------------------------------
Emp No.  | Emp. Name | sum(Worked Days) | count (Saturday)|count (Sunday)|count (Holiday)
-------------------------------------------------------------------------------------------
xxx001   |xxx        |    20            |     4           |     4        |  2          
-------------------------------------------------------------------------------------------
xxx002   |yyy        |    22            |     4           |     4        |  0          
-------------------------------------------------------------------------------------------
xxx003   |zzz        |    21            |     4           |     4        |  1          
-------------------------------------------------------------------------------------------

这些计数（星期六），计数（星期日），计数（假日）是单列，字符串值为星期六，星期日，假日，休假，带薪休假，病假等....我需要的计数使用工作日总和的单独字符串（表中为整数）

我是php的新手。所以请建议使用这些类型的查询需要什么语法。

Answer 1

您需要使用SUM()代替COUNT()作为col = value（另请注意，只有一个=）返回true或false ，1或0。

SELECT 
  EmployeeNoA, 
  EmployeeNameA, 
  SUM(AttenInDays), 
  SUM(Leaves='Sunday'), 
  SUM(Leaves='Saturday'), 
  SUM(Leaves='Holiday'), 
  SUM(Leaves='Leave'), 
  SUM(Leaves='Paid Leave') 
FROM attend 
WHERE (SignInDate BETWEEN '$StartDate' AND '$EndDate') 
GROUP BY EmployeeNoA, EmployeeNameA

Answer 2

检查聚合子句中的值的想法是返回一个值，如果它是真的。通常，该值为1，然后将它们相加以提供计数。

如果IF语句被明确编码为返回1或0，则会更清楚。

SELECT EmployeeNoA, 
        EmployeeNameA, 
        SUM(AttenInDays), 
        SUM(IF(Leaves='Sunday', 1, 0)), 
        SUM(IF(Leaves=Leaves='Saturday', 1, 0)), 
        SUM(IF(Leaves=Leaves='Holiday', 1, 0)), 
        SUM(IF(Leaves=Leaves='Leave', 1, 0)), 
        SUM(IF(Leaves=Leaves='Paid Leave', 1, 0)) ,
        SUM(WorkedDays)
FROM attend 
WHERE (SignInDate BETWEEN '$StartDate' AND '$EndDate') 
GROUP BY EmployeeNoA, EmployeeNameA

然而，这可以缩写为仅评估为1或0的简单比较： -

SELECT EmployeeNoA, 
        EmployeeNameA, 
        SUM(AttenInDays), 
        SUM(Leaves='Sunday'), 
        SUM(Leaves='Saturday'), 
        SUM(Leaves='Holiday'), 
        SUM(Leaves='Leave'), 
        SUM(Leaves='Paid Leave') ,
        SUM(WorkedDays)
FROM attend 
WHERE (SignInDate BETWEEN '$StartDate' AND '$EndDate') 
GROUP BY EmployeeNoA, EmployeeNameA

如何使用相邻列中的和值计算同一列中的多个字符串值

2 个答案: