Question

  Table - Activity(jobid,duedate,status)

状态可以是＆＃39;完成＆＃39;或＆＃34; NotComplete＆＃34;。

我需要一个SQL查询来获取

完成了多少工作
有多少工作没有完成，因为duedate大于当前日期
有多少工作没有完成，而且还没有当前日期

任何人都可以帮我这个吗？结果应该只有3行和2列。

编辑：我正在寻找最佳查询，可能是任何内置函数/关键字，它在单行查询中执行任务。不确定是否可能。

Answer 1

这样的事情：

SELECT Count(jobid) AS total, 
       'Completed' 
FROM   activity 
GROUP  BY status 
HAVING status = 'Complete' 
UNION 
SELECT Count(jobid) AS total, 
       'Completed Due in past.' 
FROM   activity 
WHERE  duedate >= Getdate() 
GROUP  BY status 
HAVING status = 'Complete' 
UNION 
SELECT Count(jobid) AS total, 
       'Completed Due in past.' 
FROM   activity 
WHERE  duedate < Getdate() 
GROUP  BY status 
HAVING status = 'Complete'

Answer 2

尝试案例：

select
  case when status='Complete' then '1'
       when status='NotComplete' and duedate >= getdate() then '2'
  else '3' end,
  count(jobid)
from Activity
group by 
  case when status='Complete' then '1'
       when status='NotComplete' and duedate >= getdate() then '2'
  else '3' end

Answer 3

试试这个： -

SELECT COUNT(JOBID) Completed Jobs, STATUS
FROM Activity
WHERE STATUS = "Complete"
GROUP BY STATUS
    UNION
SELECT COUNT(JOBID) Completed Jobs, STATUS
FROM Activity
WHERE duedate > GETDATE() AND STATUS = "NotComplete"
GROUP BY STATUS
    UNION
SELECT COUNT(JOBID) Completed Jobs, STATUS
FROM Activity
WHERE duedate < GETDATE() AND STATUS = "NotComplete"
GROUP BY STATUS;

这可能对您有所帮助。

Answer 4

Select sum( case when status = 'Complete' then 1 else 0 end ) as completedJobs, 
Sum ( case when status ='NotCompleted' and duedate > getdate() then 1 else 0 end ) as notcompletedinDuedate,
Sum ( case when status ='NotCompleted' and duedate < getdate() then 1 else 0 end ) as notcompletedJobs
From Activity

Answer 5

每个人，谢谢你的回复。我正在寻找2列的最佳查询。这是我的解决方案。如果有人有短/好版本查询，请建议。

    ;with cte as (
select 
    case 
        when stat = 'N' and duedate > GETDATE() then 'NG' 
        when stat = 'N' and duedate < GETDATE() then 'NL' 
        else stat 
    end As S, duedate from activity)
    select S,COUNT(*) from CTE group by (S)

sql查询获取dIvided组

5 个答案: