Question

我想合并两个选项以生成一个用于PHP的查询。每个用户报告时间，这个时间是正常时间（加班= 0）或加班（加班= 1）。

SELECT username, 
       userlastname,
       SUM(time0 + time1 + time2 + time3 + time4 + time5 + time6),
       userid,
       overtime 
FROM users, time 
WHERE timeyear = $y
      AND timeproid = $proid 
      AND userid = timeuserid
      AND timeweek = $week 
GROUP BY userlastname, username, userid, overtime

我有一个显示的表，每个用户有两行，一个用于加班时间，一个用于正常时间。我选择加班，因为如果超时= 0显示超时时间为0，则显示超时为SUM（time0 + time1 + time2 + time3 + time4 + time5 + time6）

有什么办法可以合并两者吗？试试这个link：它现在看起来如何。我希望6和5.5结合使11.5和5.5（加时）保持为5.5

See this

Answer 1

你可以这样做：

SELECT username,
       userlastname,
       SUM(CASE WHEN overtime = 0 THEN
              time0 + time1 + time2 + time3 + time4 + time5 + time6
           ELSE 0 END) AS normaltime,
       userid,
       SUM(overtime) AS overtime
FROM users, time 
WHERE timeyear = $y
  AND timeproid = $proid 
  AND userid = timeuserid
  AND timeweek = $week 
GROUP BY userlastname, username, userid;

这应该给你加班的“加班”总和，以及其他时间的总和为“正常时间”。如果我理解正确的话，你会把加班时间记录到加班栏中，而在其他地方记录加班时间。

相反，如果你使用加班作为“标志”，意味着“时间0现在超时”，那么你需要这样写：

SELECT username,
       userlastname,
       SUM(CASE WHEN overtime = 0 THEN
              time0 + time1 + time2 + time3 + time4 + time5 + time6
           ELSE 0 END) AS normaltime,
       userid,
       SUM(CASE WHEN overtime != 0 THEN
              time0 + time1 + time2 + time3 + time4 + time5 + time6
           ELSE 0 END) AS overtime,
FROM users, time 
WHERE timeyear = $y
  AND timeproid = $proid 
  AND userid = timeuserid
  AND timeweek = $week 
GROUP BY userlastname, username, userid;

Answer 2

您可以使用CASE WHEN / THEN END语法在查询中执行条件语句。

SELECT u.username, u.userlastname, CASE WHEN t.overtime = 1 THEN SUM(t.time0 + t.time1 + t.time2 + t.time3 + t.time4 + t.time5 + t.time6) ELSE 0 END as overtime, u.userid
FROM users u 
LEFT JOIN time t ON(u.userid=t.timeuserid)
WHERE t.timeyear = $y AND t.timeproid = $proid AND t.timeweek = $week 
GROUP BY u.userlastname, u.username, u.userid, t.overtime

Answer 3

否则联盟会满足您的需求（当然不是最佳的绩效导向答案）？

SELECT username, 
       userlastname,
       SUM(time0 + time1 + time2 + time3 + time4 + time5 + time6) as overtime,
       userid,
       overtime 
FROM users, time 
WHERE timeyear = $y
      AND timeproid = $proid 
      AND userid = timeuserid
      AND timeweek = $week 
      AND overtime=1
GROUP BY userlastname, username, userid
union
SELECT username, 
       userlastname,
       0 as overtime,
       userid,
       overtime 
FROM users, time 
WHERE timeyear = $y
      AND timeproid = $proid 
      AND userid = timeuserid
      AND timeweek = $week
      AND overtime=0 
GROUP BY userlastname, username, userid

SQL：将两个选择合并为一个查询

3 个答案: