Mysql使用JOIN

Question

我通过选择过去六周内完成评估的所有员工数据来编辑查询。从逻辑上讲，如果每个员工在两周内完成评估，那么每个员工应该两次，但此查询显示单个记录。

 select 
   AssessmentEmployee.
   EmployeeName,
   AVG(AssessmentListing.AssessmentScore),
   DATE_FORMAT((STR_TO_DATE(`AssessmentSubmittedDatetime`, '%d-%b-%Y %I:%i %p')) , '%Y-%m-%v') as _month 
 from AssessmentEmployee 
 LEFT JOIN AssessmentListing 
 ON  AssessmentEmployee.AssessmentID=AssessmentListing.AssessmentID 
 WHERE (STR_TO_DATE(`AssessmentSubmittedDatetime`, '%d-%b-%Y %I:%i %p') >= DATE_FORMAT(NOW() - INTERVAL 6 Week, '%Y' )) 
 group by AssessmentEmployee.EmployeeName 

我有以下表格，我正在使用。

AssessmentEmployee
 ID
 AssessmentID
 EmployeeName

其他表格是评估列表

ID
AssessmentID
AssessmentSubmittedDateTime
AssessmentScore

我想让那些在过去的性生活周中得分/完成评估的员工及其平均分数。

评估列表数据列的示例

ID  AssessmentID     AssessmentSubmittedDatetime     AssessmentScore 
1     040416024720     04-Apr-2016 02:48 PM             50

Answer 1

你正在做一个平均值，所以如果他们两次得分，它将平均得分为2分。

Answer 2

请尝试以下查询：

select 
   AssessmentEmployee.
   EmployeeName,
   AVG(AssessmentListing.AssessmentScore),
   DATE_FORMAT((STR_TO_DATE(`AssessmentSubmittedDatetime`, '%d-%b-%Y %I:%i %p')) , '%Y-%v') as year_week
 from AssessmentEmployee 
 LEFT JOIN AssessmentListing 
 ON  AssessmentEmployee.AssessmentID=AssessmentListing.AssessmentID 
 WHERE UNIX_TIMESTAMP(DATE_FORMAT(STR_TO_DATE(`AssessmentSubmittedDatetime`,'%d-%b-%Y %I:%i %p'),'%Y-%m-%d')) >= UNIX_TIMESTAMP(CURDATE() - INTERVAL 6 WEEK)
 group by AssessmentEmployee.EmployeeName, year_week;

您不应将日期/时间存储为字符串。否则，在处理这些繁琐的工作时，请接受它们。

Answer 3

SELECT *, SUM(AssessmentScore) as total, SUM(AssessmentScore)/6 as avg
FROM `assessmentlisting` 
WHERE STR_TO_DATE(`AssessmentSubmittedDatetime`, '%d-%b-%Y %I:%i %p') > DATE_FORMAT(NOW() - INTERVAL 6 Week, '%Y-%m-%d %I:%i %p' ) 
GROUP BY assessmentlisting.AssessmentID

希望它能奏效。