Mysql 计算行数并仅从表中获取最后一行

我有一个名为

employeexam

的表，其结构和数据如下：

--------------------------------------------------------
| id | course_id   | employee_id | degree | date
--------------------------------------------------------
| 1  |  1          | 3           | 8      | 2013-01-14
| 2  |  2          | 4           | 15     | 2013-01-14
| 3  |  2          | 4           | 17     | 2013-01-15
--------------------------------------------------------

预期结果将是：

---------------------------------------------------------------------------
| id | course_id   | employee_id | degree | date        | numOfTakingExams 
---------------------------------------------------------------------------
| 1  |  1          | 3           | 8      | 2013-01-14  | 1
| 3  |  2          | 4           | 17     | 2013-01-15  | 2
---------------------------------------------------------------------------

我的MySQL查询：

SELECT DISTINCT(employeexam.employee_id) as employeeid, 
                employeexam.*, 
                exam.numOfTakingExams 
     FROM employeexam
     JOIN (
            SELECT employee_id , COUNT(employee_id ) as numOfTakingExams 
            FROM employeexam  
            GROUP BY employee_id
          ) exam
     ON exam.employee_id  = employeexam.employee_id 
     ORDER BY employeexam.id DESC   

这将正确输出

numoffakingexamies

值，但我无法仅选择他上次参加考试的数据。有什么帮助吗

SELECT  a.*, b.numOfTakingExams
FROM    employeeExam a
        INNER JOIN
        (
            SELECT  employee_id,
                    MAX(date) max_Date,
                    COUNT(*) numOfTakingExams 
            FROM    employeeExam
            GROUP BY course_ID, employee_id
        ) b ON  a.employee_id = b.employee_id AND
                a.date = b.max_Date

您还可以通过最大

ID

获取最新记录，如果将其设置为

AUTO_INCREMENT

，下面的查询将产生与上面查询相同的结果

SELECT  a.*, b.numOfTakingExams
FROM    employeeExam a
        INNER JOIN
        (
            SELECT  employee_id,
                    MAX(id) max_Date,
                    COUNT(*) numOfTakingExams 
            FROM    employeeExam
            GROUP BY course_ID, employee_id
        ) b ON  a.employee_id = b.employee_id AND
                a.id = b.max_Date

尝试此查询-

SELECT
  t1.id, t1.course_id, t1.employee_id, t1.degree, t1.date, t2.numOfTakingExams
FROM
  mployeexam t1
  JOIN (
        SELECT employee_id, MAX(date) date, COUNT(*) numOfTakingExams
        FROM mployeexam
        GROUP BY employee_id
    ) t2
    ON t1.employee_id = t2.employee_id AND t1.date = t2.date

---

这将正确处理在一个日期参加多个考试的情况。

您是否尝试了与自身的连接？第一步是在包含“最后一次考试”的ID上选择，第二步是加入所需的内容。大致如下：

select A.* FROM
employeexam A INNER JOIN (
    SELECT EMPLOYEE_ID, MAX(DATE)
    FROM EMPLOYEEXAM
    GROUP BY EMPLOYEE_ID
) B
ON A.EMPLOYEE_ID = B.EMPLOYEE_ID AND
A.DATE = B.DATE

---

当然，假设每个员工id的日期是唯一的。

这是有效的，谢谢。我对您的进行了编辑，以输出员工曾经输入的课程。@Yasmeen，您还可以在子查询中包含

course\u ID

，

SELECT x.*
     , y.ttl
  FROM employeexam x 
  JOIN 
     ( SELECT course_id
            , employee_id
            , MAX(date) max_date
            , COUNT(*) ttl 
         FROM employeexam 
        GROUP 
           BY course_id
            ,employee_id
     ) y 
    ON y.course_id = x.course_id 
   AND y.employee_id = x.employee_id 
   AND y.max_date = x.date;

SELECT x.*
     , y.ttl
  FROM employeexam x 
  JOIN 
     ( SELECT course_id
            , employee_id
            , MAX(date) max_date
            , COUNT(*) ttl 
         FROM employeexam 
        GROUP 
           BY course_id
            ,employee_id
     ) y 
    ON y.course_id = x.course_id 
   AND y.employee_id = x.employee_id 
   AND y.max_date = x.date;