我正在为员工创建一个时钟输入/输出系统。
有一个tbl_clockins,其中包含每个时钟输入/输出时间会话的记录,其中包含有关每个会话是否已付款,员工参加该会话的时间,以及他们加班的时间等信息等。
还有另一个名为tbl_user_work_settings的表,经理可以设置员工在哪些日子度假,或者在疾病等方面起飞。
我正在创建一些报告,其中我需要每个员工的总数,例如每个员工在给定日期范围内作为假期的总天数。我有一个很长的查询,实际上获得了所有必需的信息,但它是巨大的,有点低效。有没有办法让它更小/更有效?任何帮助表示赞赏。
// get total days worked, unpaid days, bank holidays, holidays, sicknesses
// and absences within given date range for given users
$sql = "SELECT us.username, daysWorked, secondsWorked,
unpaidDays, bankHolidays, holidays, sicknesses, absences
FROM
(SELECT username FROM users WHERE clockin_valid='1') us
LEFT JOIN (
SELECT username, selectedDate, count(isUnpaid) AS unpaidDays
FROM tbl_user_work_settings
WHERE isUnpaid = '1'
AND selectedDate>='$startDate'
AND selectedDate<='$endDate'
GROUP BY username
) u ON us.username=u.username
LEFT JOIN (
SELECT username, count(isBankHoliday) AS bankHolidays
FROM tbl_user_work_settings
WHERE isBankHoliday='1'
AND selectedDate>='$startDate'
AND selectedDate<='$endDate'
GROUP BY username
) bh ON us.username=bh.username
LEFT JOIN (
SELECT username, count(isHoliday) AS holidays
FROM tbl_user_work_settings
WHERE isHoliday='1'
AND selectedDate>='$startDate'
AND selectedDate<='$endDate'
GROUP BY username
) h ON us.username=h.username
LEFT JOIN (
SELECT username, count(isSickness) AS sicknesses
FROM tbl_user_work_settings
WHERE isSickness='1'
AND selectedDate>='$startDate'
AND selectedDate<='$endDate'
GROUP BY username
) s ON us.username=s.username
LEFT JOIN (
SELECT username, count(isOtherAbsence) AS absences
FROM tbl_user_work_settings
WHERE isOtherAbsence='1'
AND selectedDate>='$startDate'
AND selectedDate<='$endDate'
GROUP BY username
) a ON us.username=a.username
LEFT JOIN (
SELECT username, count(DISTINCT DATE(in_time)) AS daysWorked,
SUM(seconds_duration) AS secondsWorked
FROM tbl_clockins
WHERE DATE(in_time)>='$startDate'
AND DATE(in_time)<='$endDate'
GROUP BY username
) dw ON us.username=dw.username";
if(count($selectedUsers)>0)
$sql .= " WHERE (us.username='"
. implode("' OR us.username='", $selectedUsers)."')";
$sql .= " ORDER BY us.username ASC";
答案 0 :(得分:2)
您只需使用SUM(condition)
表格即可使用tbl_user_work_settings
:
// get total days worked, unpaid days, bank holidays, holidays, sicknesses
// and absences within given date range for given users
$sql = "
SELECT users.username,
SUM(ws.isUnpaid ='1') AS unpaidDays,
SUM(ws.isBankHoliday ='1') AS bankHolidays,
SUM(ws.isHoliday ='1') AS holidays,
SUM(ws.isSickness ='1') AS sicknesses,
SUM(ws.isOtherAbsence='1') AS absences,
COUNT(DISTINCT DATE(cl.in_time)) AS daysWorked,
SUM(cl.seconds_duration) AS secondsWorked
FROM users
LEFT JOIN tbl_user_work_settings AS ws
ON ws.username = users.username
AND ws.selectedDate BETWEEN '$startDate' AND '$endDate'
LEFT JOIN tbl_clockins AS cl
ON cl.username = users.username
AND DATE(cl.in_time) BETWEEN '$startDate' AND '$endDate'
WHERE users.clockin_valid='1'";
if(count($selectedUsers)>0) $sql .= "
AND users.username IN ('" . implode("','", $selectedUsers) . "')";
$sql .= "
GROUP BY users.username
ORDER BY users.username ASC";
顺便说一句(也许更多是为了其他读者的利益),我真的希望通过在将PHP变量插入SQL之前正确转义PHP变量来避免SQL注入攻击。理想情况下,您根本不应该这样做,而是将这些变量作为预准备语句的参数传递给MySQL(不会对SQL进行评估):阅读有关Bobby Tables的更多信息。
另外,另外,为什么要将整数类型作为字符串处理(通过将它们括在单引号字符中)?这是不必要的,并且MySQL中的资源浪费不得不执行不必要的类型转换。实际上,如果各种isUnpaid
等列都是0
/ 1
,您可以更改上述内容以删除相等性测试,并直接使用SUM(ws.isUnpaid)
等。< / p>
答案 1 :(得分:0)
将每个将加入临时表的表... 然后在临时表的可连接字段上创建索引... 并使用临时表进行查询。
示例:
SELECT username, selectedDate, count(isUnpaid) AS unpaidDays
INTO #TempTable1
FROM tbl_user_work_settings
WHERE isUnpaid = '1'
AND selectedDate>='$startDate'
AND selectedDate<='$endDate'
GROUP BY username
create clustered index ix1 on #TempTable1 (username)