如何解决按学生数分组的日期和的MySQL连接即使行连接结果为0,也要返回值
如何使没有结果(匹配项)的学生在此查询中也显示0 TotalHours。
添加了我在dbfiddle及以下版本中所需要的真实示例:
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我想在查询中显示记录,以便它显示ID 2593,它没有任何出勤记录,因此我希望它在totalhours列中返回0。
select students.StudentNumber,IFNULL(round(sum(TimestampDiff(minute,Time_Start,Time_End)) / 60,2),0) as TotalHours from students left join attendance on students.StudentNumber = attendance.StudentNumber where time_start between '2019-01-01' and '2020-09-12' Group by students.StudentNumber
https://www.db-fiddle.com/f/iZDMnePyCvzoW82eScXoRF/0
解决方法
您可以将time where子句放在子选择项中以便出勤,然后获得结果。
您应仅将SELECT * FROM attendance WHERE time_start BETWEEN '2019-01-01' AND '2020-09-12'
替换为实际需要的列
查询#1
SELECT
students.StudentNumber,IFNULL(ROUND(SUM(TIMESTAMPDIFF(MINUTE,Time_Start,Time_End)) / 60,2),0) AS TotalHours
FROM
students
LEFT JOIN
attendance ON students.StudentNumber = attendance.StudentNumber
WHERE
time_start BETWEEN '2019-01-01' AND '2020-09-12' or time_start IS NULL
GROUP BY students.StudentNumber;
| StudentNumber | TotalHours |
| ------------- | ---------- |
| 2591 | 7.48 |
| 2592 | 6.50 |
| 2593 | 0.00 |
查询#2
SELECT
s.StudentNumber,0) AS TotalHours
FROM
students s
LEFT JOIN
( SELECT * FROM attendance WHERE
time_start BETWEEN '2019-01-01' AND '2020-09-12') a ON s.StudentNumber = a.StudentNumber
GROUP BY s.StudentNumber;
| StudentNumber | TotalHours |
| ------------- | ---------- |
| 2591 | 7.48 |
| 2592 | 6.50 |
| 2593 | 0.00 |
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