0

我有以下加入声明：

SELECT DISTINCT dn.fname, dn.lname,w.websit,q.qual,q.year,q.postqual,p.pnumber,a.accred,n.nspecial
FROM cur_doctor_names dn
LEFT JOIN cur_website w ON dn.userid = w.userid
LEFT JOIN cur_qualification q ON dn.userid = q.userid
LEFT JOIN (Select * from cur_phone_numbers where type="primary") p On dn.userid = p.userid
Left JOIN cur_accreditation a On dn.userid = a.userid
Left JOIN cur_niche n On dn.userid = n.userid
WHERE dn.userid =69
GROUP BY p.pnumber

对于表 cur_phone_numbers (p) 、 cur_accreditation(a) 、 cur_niche (n) ，结果可以包含用户的 0 到 3 条记录，我目前得到的结果为我提供了数据的所有可能组合。我尝试在列上单独使用 group by ，但这会消除其他列的结果。我不介意用 nulls 得到结果。这是我目前得到的结果：

fname         lname        website  qual  year  postqual    pnumber  accred    nspecial
===========================================================================================
First Doctor  Last Doctor  NULL     NULL  NULL  NULL        123      ISO 9001  one
First Doctor  Last Doctor  NULL     NULL  NULL  NULL        123      ISO 9001  two
First Doctor  Last Doctor  NULL     NULL  NULL  NULL        123      ISO       one
First Doctor  Last Doctor  NULL     NULL  NULL  NULL        123      ISO       two
First Doctor  Last Doctor  NULL     NULL  NULL  NULL        1234     ISO 9001  one
First Doctor  Last Doctor  NULL     NULL  NULL  NULL        1234     ISO 9001  two
First Doctor  Last Doctor  NULL     NULL  NULL  NULL        1234     ISO       one
First Doctor  Last Doctor  NULL     NULL  NULL  NULL        1234     ISO       two
4

1 回答 1

1

您必须使用 选择其中一个值aggregate functions，例如max：

SELECT dn.fname, dn.lname,w.websit,q.qual,q.year,q.postqual,max(p.pnumber),max(a.accred),max(n.nspecial)
FROM cur_doctor_names dn
LEFT JOIN cur_website w ON dn.userid = w.userid
LEFT JOIN cur_qualification q ON dn.userid = q.userid
LEFT JOIN (Select * from cur_phone_numbers where type="primary") p On dn.userid = p.userid
Left JOIN cur_accreditation a On dn.userid = a.userid
Left JOIN cur_niche n On dn.userid = n.userid
WHERE dn.userid =69
GROUP BY p.pnumber
于 2012-10-12T11:42:53.313 回答