Question

如何显示缺席或结果

请帮我解决此问题，谢谢

现在显示如此结果

________________________________________
 Name   TeacherNo   Attendance
________________________________________
 XYZ       993             5 Days Present

 Abc       991             7 Days Present

 123       955             2 Days Present

________________________________________

我想要这样

________________________________________________________________
Name    TeacherNo      Present Day          Absent Days
_________________________________________________________________
XYZ       993             5 Days Present        3 Days Present

Abc       991             7 Days Present        0 Days Present

123       955             2 Days Present        4 Days Present  

________________________________________________________________

这是代码

 $result1 = mysql_query("select name ,id ,`teacherno` as teacherno, CONCAT(count(`teacherno`),' Present days') as Attendance
   from tattendance 
   where `Attendance` = 'present' AND date BETWEEN '2013-04-01' AND '2013-04-07' group by `teacherno` order by attendance desc");

Answer 1

您是否在寻找类似以下内容的信息：

SELECT name, teacherno, 
    CONCAT(COALESCE(SUM(CASE WHEN Attendance = 'present' THEN 1 END),0),' Present days') as Present,
    CONCAT(COALESCE(SUM(CASE WHEN Attendance = 'absent' THEN 1 END),0),' Absent days') as Absent
FROM tattendance  
GROUP BY teacherno

这会SUM与CASE一起使用，根据出勤列添加日期。我还包括COALESCE以返回0而不是NULL。

SQL Fiddle Demo

如何显示缺席或现在的结果

1 个答案: