2

我需要将员工姓名分组在一起。我尝试使用“Order By”，但它不适用于空单元格。

在此处输入图像描述

顶部的表格是我当前的表格，我想要底部表格的结果。这是我当前的选择查询。

SELECT 
    DAYNAME(calendar.DATE) AS DAY,
    DATE_FORMAT(calendar.DATE, '%d %b %Y') AS Date,
    t2.Department,
    t2.Name,
    TIME(t2.time_enter) AS 'Time In',
    TIME(t2.time_exit) AS 'Time Out'
FROM
    calendar
        LEFT JOIN
    (SELECT 
        employee_hr_id,
            time_enter,
            time_exit,
            department.NAME As Department,
            CONCAT(employee.F_NAME, ' ', employee.L_NAME) As Name
    from
        entry_log
    Inner Join employee ON entry_log.EMPLOYEE_HR_ID = employee.HR_ID
    Inner Join department ON employee.DEPARTMENT_ID = department.ID
    where
        CONCAT(employee.F_NAME, ' ', employee.L_NAME) LIKE @NAME) t2 ON date(t2.time_enter) = calendar.date
where
    calendar.DATE >= @TIME_ENTER
        AND calendar.DATE <= @TIME_EXIT
4

2 回答 2

2

您应该使用以下子查询，而不仅仅是 CALENDAR 表。

select calendar.DATE,employee.HR_ID,employee.DEPARTMENT_ID 
  from calendar,employee 
  where
    calendar.DATE >= @TIME_ENTER
        AND calendar.DATE <= @TIME_EXIT

在这种情况下，您将获得 DATE 和 EMPLOYEE 的所有可能组合。然后DEPARTMENT_ID, HR_ID, DATE从这张桌子订购。

因此，您的查询应如下所示：

SELECT 
    DAYNAME(NEW_calendar.DATE) AS DAY,
    DATE_FORMAT(NEW_calendar.DATE, '%d %b %Y') AS Date,
    t2.Department,
    t2.Name,
    TIME(t2.time_enter) AS 'Time In',
    TIME(t2.time_exit) AS 'Time Out'
FROM
    (
       select calendar.DATE as DATE,
              employee.HR_ID as HR_ID,
              employee.DEPARTMENT_ID as DEPARTMENT_ID
            from calendar,employee 
            where
              calendar.DATE >= @TIME_ENTER
                AND calendar.DATE <= @TIME_EXIT
     ) NEW_calendar
        LEFT JOIN
    (SELECT 
        employee_hr_id,
            time_enter,
            time_exit,
            department.NAME As Department,
            CONCAT(employee.F_NAME, ' ', employee.L_NAME) As Name
    from
        entry_log
    Inner Join employee ON entry_log.EMPLOYEE_HR_ID = employee.HR_ID
    Inner Join department ON employee.DEPARTMENT_ID = department.ID
    where
        CONCAT(employee.F_NAME, ' ', employee.L_NAME) LIKE @NAME) t2 

        ON date(t2.time_enter) = NEW_calendar.date
           and (t2.EMPLOYEE_HR_ID=NEW_calendar.HR_ID)
where
    calendar.DATE >= @TIME_ENTER
        AND calendar.DATE <= @TIME_EXIT

order by NEW_calendar.DEPARTMENT_ID,NEW_calendar.HR_ID,NEW_calendar.Date
于 2013-04-25T08:35:35.463 回答
1

空行中没有任何内容表明您将与他们关联的员工。事实上，它们甚至根本不应该重复。可行的方法是连接三个表（或子查询）：一个用于日期，第二个用于所有（相关）员工，第三个用于实际分配。然后您可以使用所有员工的第二个表对行进行排序，即使您选择不打印姓名。前两个表之间的完全交叉连接为您提供了日期和员工之间的所有组合，然后可以将带有日志条目的那些组合保留下来。

我在 SQLFiddle 创建了一个示例。它使用员工姓名而不是 ID 来匹配事物，并将其t2作为表而不是更复杂的子查询，但这个想法应该是干净的。这是该小提琴的查询：

SELECT …
FROM
    (calendar,
     (SELECT DISTINCT Name FROM t2) AS employees)
    LEFT JOIN t2 ON DATE(t2.time_enter) = calendar.date
                AND t2.Name = employees.Name
ORDER BY employees.Name, calendar.date
于 2013-04-25T09:10:30.820 回答