SQL：将两个选择合并到一个查询中答案

【问题标题】：SQL: Merge two selects to one querySQL：将两个选择合并到一个查询中
【发布时间】：2012-07-18 19:06:20
【问题描述】：

我想合并两个选择以进行一个查询以在 PHP 中使用。每个用户报时间，这个时间要么是正常时间（加班=0），要么是加班（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

我显示了一个表格，其中每个用户有两行，一列用于加班时间，另一行用于正常时间。我选择加班是因为如果加班 = 0 将加班时间显示为 0 否则将加班时间显示为 SUM(time0 + time1 + time2 + time3 + time4 + time5 + time6)

有什么办法可以合并这两者吗？试试这个link：它现在的样子。我希望将 6 和 5.5 组合成 11.5，并将 5.5（加班）保留为 5.5

See this

【问题讨论】：

请提供示例数据和所需的输出。

标签： php sql join merge union

【解决方案1】：

否则联合是否满足您的需求（当然不是以性能为导向的最佳答案）？

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

【讨论】：

【解决方案2】：

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”的总和为加班，其他时间的总和为“正常时间”。如果我理解正确，您将加班时间记录在加班列中，而其他地方则记录不加班时间。

如果您将超时用作“标志”，表示“time0 现在超时”，那么您需要这样写：

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;

【讨论】：

我所拥有的是 time0...time6 和一个设置为 0 或 1 的加班列，表示 time0...6 是正常时间还是加班时间。我更新了问题谢谢。

【解决方案3】：

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

【讨论】：

请查看更新后的问题和我添加的链接，因为它仍然无法正常工作