我正在尝试找到申请了 2 家以上不同公司的申请人
显示的申请人应该是：
莫妮卡 (a2)：作业 101、j02 和 j04
Jim (a3)：作业 j02、j04 和 j06
这是用于此的脚本！！
drop table Jobskills;
drop table Appskills;
drop table Applies;
drop table Applicant;
drop table Skills;
drop table Job;
drop table Company;
create table Company(compid char(5) primary key, compname varchar(20),
comptype varchar(15));
create table Job(jobid char(5) primary key, jobtitle varchar(20),
salarylow int, salaryhigh int, location char(10),
compid references Company(compid) on delete cascade);
create table Skills(skillid char(5) primary key, skillname varchar(15));
create table Jobskills(jobid references Job(jobid) on delete cascade,
skillid references Skills(skillid),
expertiseneeded int, primary key(jobid,skillid));
create table Applicant(appid char(5) primary key, name varchar(15),
age int, highdegree char(5), expected_salary int) ;
create table AppSkills(appid references Applicant(appid) on delete cascade,
skillid references Skills(skillid), expertise int,
primary key(appid, skillid));
create table Applies(jobid references Job(jobid),
appid references Applicant(appid) on delete cascade,
appdate date, decisiondate date, outcome char(10),
primary key(jobid, appid));
rem Initial Company data
insert into Company values('PWC', 'Price Waterhouse', 'consulting');
insert into Company values('MSFT', 'Microsoft', 'software');
insert into Company values('INTL', 'Intel', 'electronics');
insert into Company values('NCR', 'NCR Corp', 'server');
insert into Company values('WPAF', 'WP Air Force', 'defense');
insert into Company values('DLT', 'Deloitte', 'consulting');
rem Initial Job data
insert into Job values('101', 'Programmer', 55000, 60000, 'Redmond', 'MSFT');
insert into Job values('j02', 'Designer', 42000, 45000, 'Redmond', 'MSFT');
insert into Job values('j03', 'SAP impl', 30000, 40000, 'Chicago', 'PWC');
insert into Job values('j04', 'Proj mgmt', 35000, 55000, 'Chicago', 'PWC');
insert into Job values('j05', 'SOX', 60000, 65000, 'Detroit', 'PWC');
insert into Job values('j06', 'db admin', 45000, 50000, 'Dayton', 'NCR');
insert into Job values('j07', 'db designer', 35000, 40000, 'Dayton', 'NCR');
insert into Job values('j08', 'intern', 25000, 28000, 'Dayton', 'NCR');
insert into Job values('j09', 'engineer', 52000, 55000, 'Dayton','WPAF');
insert into Job values('j10', 'dba', 62000, 65000, 'Dayton','WPAF');
insert into Job values('j11', 'hardware dev', 50000, 65000, 'NYC','INTL');
insert into Job values('j12', 'pcb designer', 55000, 68000,'NYC','INTL');
insert into Job values('j13', 'chip designer', 40000, 55000,'Chicago','INTL');
insert into Job values('j14', 'IT', 40000, 60000, 'Dayton', 'DLT');
insert into Job values('j15', 'IT', 50000, 70000, 'Chicago', 'DLT');
rem initial Skills data
insert into Skills values('s1', 'database');
insert into Skills values('s2', 'programming');
insert into Skills values('s3', 'sox');
insert into Skills values('s4', 'project');
insert into Skills values('s5', 'hardware');
insert into Skills values('s6', 'sap');
insert into Skills values('s7', 'analysis');
rem Initial Jobskills data
insert into Jobskills values('101', 's2', 5);
insert into Jobskills values('101', 's7', 4);
insert into Jobskills values('j02', 's2', 3);
insert into Jobskills values('j02', 's7', 5);
insert into Jobskills values('j03', 's6', 5);
insert into Jobskills values('j04', 's7', 4);
insert into Jobskills values('j04', 's4', 5);
insert into Jobskills values('j04', 's2', 2);
insert into Jobskills values('j05', 's3', 5);
insert into Jobskills values('j06', 's1', 5);
insert into Jobskills values('j06', 's2', 3);
insert into Jobskills values('j07', 's1', 4);
insert into Jobskills values('j07', 's7', 3);
insert into Jobskills values('j08', 's1', 2);
insert into Jobskills values('j09', 's2', 4);
insert into Jobskills values('j09', 's4', 4);
insert into Jobskills values('j10', 's4', 3);
insert into Jobskills values('j10', 's1', 5);
insert into Jobskills values('j11', 's5', 3);
insert into Jobskills values('j11', 's4', 3);
insert into Jobskills values('j12', 's5', 5);
insert into Jobskills values('j13', 's1', 4);
insert into Jobskills values('j13', 's2', 5);
insert into Jobskills values('j14', 's7', 4);
rem initial Applicants data
insert into Applicant values('a1', 'Joe', 30, 'MS', 55000);
insert into Applicant values('a2', 'Monica', 25, 'BS', 62000);
insert into Applicant values('a3', 'Jim', 22, 'BS', 45000);
insert into Applicant values('a4', 'Monica', 25, 'BS', 34000);
rem initial Appskills data
insert into Appskills values('a1', 's1', 3);
insert into Appskills values('a1', 's2', 4);
insert into Appskills values('a1', 's4', 4);
insert into Appskills values('a1', 's6', 3);
insert into Appskills values('a1', 's7', 4);
insert into Appskills values('a2', 's2', 3);
insert into Appskills values('a2', 's3', 5);
insert into Appskills values('a2', 's6', 4);
insert into Appskills values('a3', 's4', 3);
insert into Appskills values('a3', 's1', 3);
insert into Appskills values('a3', 's2', 5);
rem Applies
insert into Applies values ('101', 'a1', '01-JAN-06', '08-JAN-06', 'hire');
insert into Applies values ('101', 'a2', '01-JAN-06', '08-JAN-06', 'hire');
insert into Applies values ('j02', 'a2', '01-JAN-06', '08-JAN-06', 'hire');
insert into Applies values ('j04', 'a2', '01-JAN-06', '08-JAN-06', 'hire');
insert into Applies values ('j02', 'a3', '01-JAN-06', '08-JAN-06', 'nohire');
insert into Applies values ('j04', 'a3', '01-JAN-06', '08-JAN-06', 'nohire');
insert into Applies values ('j06', 'a3', '01-JAN-06', '08-JAN-06', 'nohire');
这是我目前的代码...
SELECT app.name, app.appid, j.jobid, c.compname
FROM applicant app INNER JOIN applies apps ON app.appid = apps.appid INNER JOIN
job j ON j.jobid = apps.jobid INNER JOIN company c ON c.compid = j.compid
WHERE apps.appid > '2'
虽然有些不对劲，因为我得到了 Joe (a1) 而我不应该得到他。我应该只得到 Monica (a2) 和 Jim (a3)。我错过了什么？
答案 0 :(得分：1)
嗯。 . .你需要聚合。要回答您的问题，您只需要一个 client = boto3.client('dynamodb')
client.get_item(TableName='YourTable', Key={'PrimaryKeyName': {'ValueType':'Value'}})
子句，工作和公司并不重要。但是如果您愿意，您可以将它们放入分隔字符串中：
HAVING答案 1 :(得分：0)
这里在通用表表达式中，我对申请人的唯一大学名称进行了排名。然后使用exists，我只计算了拥有不止一所大学的申请人，并且只显示了这些申请人的信息。
with cte as (
SELECT app.name, app.appid, j.jobid, c.compname, dense_rank()over(partition by app.appid order by compname)companycount
FROM applicant app INNER JOIN applies apps ON app.appid = apps.appid INNER JOIN
job j ON j.jobid = apps.jobid INNER JOIN company c ON c.compid = j.compid)
select name,appid,jobid,compname from cte c
where exists (select appid from cte ct where ct.companycount>1 and c.appid=ct.appid)