我对这个MYSQL查询视而不见.这不应该那么难,但我确实可以取得结果,但不是我想要的结果.非常感谢你的帮助！
医生
dctr_id | dctr_name | ...
--------------------------
60 | Bezant
造访
vist_id | dctr_id| prsnl_id | visit_date | ...
-----------------------------------------------
1 | 60 | 86 | 2018-12-31
意外
acc_id | dctr_id | prsnl_id| acc_date | ...
--------------------------------------------
51 | 60 | 86 | 2018-12-25
55 | 60 | 86 | 2018-12-20
人物
prsnl_id | prsnl_name | ...
---------------------------
79 | test_name2
86 | test_name
我得到这个结果：
dctr_id | dctr_name | visit_id | visit_date | acc_id | acc_date | prsnl_id | prsnl_name
-----------------------------------------------------------------------------------------
60 | Bezant | 1 | 2018-12-31 | 51 | 2018-12-25 | 79 | test_name2
60 | Bezant | 1 | 2018-12-31 | 51 | 2018-12-25 | 79 | test_name2
60 | Bezant | 1 | 2018-12-31 | 55 | 2018-12-20 | 86 | test_name1
SELECT DISTINCT dctr.dctr_id,dctr.dctr_name,vst.visit_id,vst.visit_date,acc.acc_id,acc.acc_date,prsnl.prsnl_id,prsnl.name
FROM doctor dctr
LEFT
JOIN visits vst
ON vst.dctr_id = dctr.dctr_id
LEFT
JOIN accidents acc
ON acc.dctr_id = dctr.dctr_id
LEFT
JOIN personell prsnl
ON prsnl.prsnl_id = vst.prsnl_id
OR prsnl.prsnl_id = acc.prsnl_id
WHERE dctr.dctr_id = 60
我想得到以下结果：
dctr_id | dctr_name | visit_id | visit_date | acc_id | acc_date | prsnl_id | prsnl_name
-----------------------------------------------------------------------------------------
60 | Bezant | 1 | 2018-12-31 | NULL | NULL | 79 | test_name2
60 | Bezant | NULL | NULL | 51 | 2018-12-25 | 79 | test_name2
60 | Bezant | NULL | NULL | 55 | 2018-12-20 | 86 | test_name1
最佳答案
为了获得访问或意外的结果,您需要将这两个表联合在一起,为该表中没有对应数据的列选择NULL值(例如,访问中的acc_id).然后可以将这些结果加入到医生和人员表中,以获取每次访问/意外的相关医生和人员信息：
原文链接：https://www.f2er.com/mysql/531900.htmlSELECT d.dctr_id,d.dctr_name,i.visit_id,i.visit_date,i.acc_id,i.acc_date,p.prsnl_id,p.prsnl_name
FROM doctor d
LEFT JOIN (SELECT dctr_id,visit_id,prsnl_id,visit_date,NULL AS acc_id,NULL AS acc_date
FROM visits
UNION
SELECT dctr_id,NULL,acc_id,acc_date
FROM accidents) i
ON i.dctr_id = d.dctr_id
LEFT JOIN personell p ON p.prsnl_id = i.prsnl_id
WHERE d.dctr_id = 60
ORDER BY i.visit_id,i.acc_id
输出：
dctr_id dctr_name visit_id visit_date acc_id acc_date prsnl_id prsnl_name
60 Bezant 1 2018-12-31 (null) (null) 79 test_name2
60 Bezant (null) (null) 51 2018-12-25 79 test_name2
60 Bezant (null) (null) 55 2018-12-20 86 test_name