我需要使用 MySQL 根据列值进行聚合的方法 [英] I need the way to aggregate based on column value using MySQL

问题描述

我正在 Laravel 中学习图表，我需要为那些迟到或准时来的学生绘制每日学生出勤的折线图.我尝试编写 MYSQL 查询，但它不起作用

I am learning Charts in Laravel, i need to draw Line graph for daily Student Attendance for those students come late or on time. I tried to write MYSQL query but it doesn't work

我尝试在同一个表上进行子查询以获取日常学生的数据，我还需要 7 个日期，但不是完整日期，例如日期存储在 db 中为 09/08/2019，但我需要将其作为日期为 08.

I tried subquery on same table to get data for daily students and i also need 7 dates only not full date, like date is stored in db as 09/08/2019 but i need it as 08 as date.

SELECT Date, COUNT(*) AS TimeStudent
   FROM attendance WHERE `Attendance`='OnTime' AND (SELECT COUNT(*) AS 
   LateStudent FROM attendance 
   WHERE `Attendance`='Late'
   GROUP BY `Date`
   ORDER BY LateStudent DESC) 
   GROUP BY `Date`
   ORDER BY TimeStudent DESC 

但是我得到了

[Err] 1241 - 操作数应包含 1 列

[Err] 1241 - Operand should contain 1 column(s)

，因为我不能在子查询中再次获取日期，而在 where 子句之后使用它.任何人请帮助我.

, because i can't use to fetch Date again in subquery while use it after where clause. Any one help me plz.

推荐答案

这里是一种基于列值聚合的方法.

Here is a way to aggregate based on column value.

此查询将为您提供特定日期准时和迟到学生的计数.

This query will give you count of on time and late student for a particular date.

SELECT 
    `Date`,
    DATE_FORMAT(`Date`, '%d') AS Month_Date,  -- You can modify it as per your requirement
    SUM(IF(`Attendance` = 'OnTime', 1, 0)) AS OnTime_Count,
    SUM(IF(`Attendance` = 'Late', 1, 0)) AS Late_Count
FROM attendance
WHERE `Date` >= CURRENT_DATE - INTERVAL 7 DAY
GROUP BY `Date`;

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