找到在同一天访问过两位同一专业的不同医生的患者.
示例数据库:单击此处查看SQL Fiddle中的示例数据脚本.
CREATE VIEW DistinctVisits AS
SELECT v.vid,v.pid,d.speciality,v.date
FROM Visits v ,Doctors d
WHERE d.did=v.did
GROUP BY v.pid,v.did,v.date;
CREATE VIEW DistinctVisits2 AS
SELECT dv.pid,dv.speciality,dv.date, COUNT(dv.vid) as countv
FROM DistinctVisits dv
GROUP BY dv.pid,dv.speciality,dv.date;
SELECT dv2.pid,dv2.speciality
FROM DistinctVisits2 dv2
WHERE dv2.countv=2;
DROP VIEW DistinctVisits;
DROP VIEW DistinctVisits2;
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我如何重复相同的想法,但只是一个大问题?另外一个解决方案也会很好,但请尽量帮助我改进这个.
小智 7
您需要找到在某一天访问过两位不同医生的患者名单.
在此要求中,您的Patient表成为主表.我们先查询该表.
现在我们有患者名单.我们需要获得他们访问过的医生名单.我们不能简单地将Patients表与Doctors表一起加入,因为没有列来映射数据.我们必须使用Visits作为中间表
LEFT OUTER JOIN在患者和访问表之间添加一个并按pid列加入.
我们有病人和他们的访问清单,但现在我们需要得到医生的信息.因此,LEFT OUTER JOIN在Visits和Doctors表之间添加另一个并通过did列加入.
我们有患者和医生访问信息.但是,我们只需要患者的姓名,他们所访问的医生的专长以及他们访问的日期.因此,我们将添加列p.pname,d.speciality并v.date在SELECT子句也是在GROUP BY子句.除此之外,我们需要所有的访问次数,但有一个问题.我们只需要DISTINCT计数,换句话说,我们需要他们访问过的所有独特医生的数量.因此,如果患者在给药日两次访问同一位医生,则应将其计为1.因此,添加DISTINCT将有助于此处.另外,关键是使用正确的列名,在这种情况下,d.did代表医生.
我们拥有所需的所有数据,但我们只需要过滤同一天访问过两位不同医生的患者.要做到这一点,HAVING条款来解救我们.当您应用GROUP BY时,HAVING是合适的.我们将使用相同的COUNT(DISTINCT d.did)来检查计数是否仅匹配值2.您可以在输出中查看结果.
您不必为插入表中的每个值指定INSERT INTO语句.您可以在括号内将它们组合在一起并将它们分开commas.最后一个陈述应该结束semicolon.
查询使用LEFT OUTER JOIN.我使用这个联合来找出每个病人的所有医生就诊,即使他们从未去过医生.我只是想在我形成查询时看到输出.您可以将其更改为INNER JOIN,我认为在您的方案中更合适.
如果您不想显示访问次数,可以将其从SELECT子句中删除.
SELECT p.pname
, d.speciality
, v.date
, COUNT(DISTINCT d.did) AS visitcount
FROM Patient p
LEFT OUTER JOIN Visits v
ON v.pid = p.pid
LEFT OUTER JOIN Doctors d
ON d.did = v.did
GROUP BY p.pname
, d.speciality
, v.date
HAVING COUNT(DISTINCT d.did) = 2
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SELECT p.pname
, d.speciality
, v.date
FROM Patient p
INNER JOIN Visits v
ON v.pid = p.pid
INNER JOIN Doctors d
ON d.did = v.did
GROUP BY p.pname
, d.speciality
, v.date
HAVING COUNT(DISTINCT d.did) = 2
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PNAME SPECIALITY DATE VISITCOUNT
--------- ------------ --------- -----------
Loch Ness Assholes 17/9/2012 2
Loch Ness Orthopedist 13/1/2011 2
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create table InsuranceCompanies (
cid int,
cname varchar(20),
primary key (cid)
);
create table Patient (
pid int,
pname varchar(20),
age int,
cid int,
gender char,
primary key (pid),
constraint foreign key (cid)
references InsuranceCompanies (cid)
);
create table Doctors (
did int ,
dname varchar(20),
speciality varchar(20),
age int,
cid int,
primary key (did),
constraint foreign key (cid)
references InsuranceCompanies (cid)
);
create table Visits(
vid int,
pid int,
did int,
date varchar(20),
primary key (vid),
constraint foreign key (pid)
references Patient (pid) ,
constraint foreign key (did)
references Doctors (did)
);
INSERT INTO InsuranceCompanies(cid, cname) VALUES
( 1111, 'Harel Inc' ),
( 2222, 'Clalit Inc' );
INSERT INTO Doctors ( did, dname, speciality, age, cid) VALUES
( 100, 'Jhonny Depp', 'Heart', 42, 1111 ),
( 101, 'Tom Tolan', 'Assholes', 62, 1111 ),
( 105, 'Yom Tov', 'Assholes', 52, 1111 ),
( 102, 'Lauren Jaime', 'Throat', 27, 2222 ),
( 103, 'Gomez Flaurence', 'Legs', 37, 2222 ),
( 106, 'David Harpaz', 'Orthopedist', 37, 2222 ),
( 107, 'David Schwimmer', 'Orthopedist', 37, 2222 ),
( 108, 'Sammy Salut', 'Orthopedist', 37, 1111 );
INSERT INTO Patient ( pid, pname, age, cid,gender) VALUES
( 200, 'Jon Gilmour', 25, 2222, 'm' ),
( 206, 'Bon Gilmour', 30, 2222, 'm' ),
( 205, 'Jon Gilmour', 22, 2222, 'm' ),
( 201, 'Bon Jovy', 21, 2222, 'm' ),
( 202, 'Loch Ness', 17, 2222, 'f' ),
( 203, 'Lilach Sonin', 12, 1111, 'f' ),
( 209, 'Lilach Dba', 34, 1111, 'f' ),
( 210, 'Paulina Daf', 32, 1111, 'f' ),
( 204, 'Gerry Jalor', 23, 1111, 'm' ),
( 208, 'Jerrushalem Jalor', 23, 1111, 'm' );
INSERT INTO Visits ( vid, pid, did, date) VALUES
( 300, 204, 100, '12/12/2012' ),
( 301, 204, 101, '12/12/2012' ),
( 302, 204, 101, '02/01/2012' ),
( 303, 202, 101, '17/09/2012' ),
( 311, 202, 105, '17/09/2012' ),
( 304, 203, 102, '12/12/2011' ),
( 312, 202, 106, '13/06/2012' ),
( 314, 202, 107, '13/01/2011' ),
( 313, 202, 108, '13/01/2011' ),
( 305, 204, 102, '10/10/2011' ),
( 306, 201, 100, '12/01/2012' ),
( 316, 204, 108, '18/05/2012' ),
( 307, 202, 100, '12/07/2012' ),
( 315, 203, 108, '12/07/2012' ),
( 310, 204, 103, '10/04/2012' ),
( 308, 203, 102, '12/12/2011' ),
( 309, 200, 101, '12/12/2012' );
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