Question

这是我到目前为止所拥有的......

SELECT pr.[name], SUM(COALESCE(qm.iscompleted,0)) count_yes,
  SUM(COALESCE(qm.iscompleted,1)) count_no,
   COUNT(p.id) AS tot
FROM   (aco.practices pr
      left JOIN aco.patients_practices ppr ON (ppr.practiceid = pr.id)
      left JOIN aco.patients p ON (p.id = ppr.patientid)
      LEFT JOIN aco.qmheader qm ON (qm.patientid = p.id AND qm.practiceid = pr.id)
)
WHERE  (pr.parentaco = 30982)
GROUP BY pr.[name]
ORDER BY pr.[name]

以下是一些结果

name    count_yes   count_no    tot
name1   0            1          0
name2   0            1          0
name3   0            273        273
name4   0            114        114

在大多数情况下它是正确的，我唯一需要的是如果tot为0，则不在count_no上使用SUM。

有没有办法轻松做到这一点？

感谢。

Answer 1

将您的select条款更改为：

SELECT pr.[name], 
       SUM(COALESCE(qm.iscompleted,0)) AS count_yes,
       (case when COUNT(p.id) > 0 then 
             SUM(COALESCE(qm.iscompleted,1))
        else
             0 end) AS count_no,
       COUNT(p.id) AS tot

Answer 2

使用HAVING子句：

SELECT pr.[name], SUM(COALESCE(qm.iscompleted,0)) count_yes,
  SUM(COALESCE(qm.iscompleted,1)) count_no,
   COUNT(p.id) AS tot
FROM   (aco.practices pr
      left JOIN aco.patients_practices ppr ON (ppr.practiceid = pr.id)
      left JOIN aco.patients p ON (p.id = ppr.patientid)
      LEFT JOIN aco.qmheader qm ON (qm.patientid = p.id AND qm.practiceid = pr.id)
)
WHERE  (pr.parentaco = 30982)
GROUP BY pr.[name]
HAVING COUNT(p.id) = 0
ORDER BY pr.[name]

仅当COUNT（*）＆gt;时的SUM数据0

2 个答案: