在SQL中从多个表中查找数据

我试图在代顿找到需要数据库或分析技能的工作地点，然后按公司和工作id排序。

它应该是以下内容：

数据库（“s1”）：j06、j07、j08和j10

分析（“s7”）：j07和j14

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');

这是我到目前为止的代码

SELECT j.location, j.jobid, js.skillid
FROM jobskills js INNER JOIN job j ON js.jobid = j.jobid
WHERE j.location = 'Dayton'
    AND (js.skillid = 's1' OR js.skillid = 's7')
ORDER BY j.jobid

---

我只需要添加公司名称，但每次添加都会弄乱我的整个代码。

您只需在compid上使用job加入公司即可

SELECT c.compname,j.location, j.jobid, js.skillid
FROM jobskills js INNER JOIN job j ON js.jobid = j.jobid
inner join company c on c.compid=j.compid
WHERE j.location = 'Dayton'
    AND (js.skillid = 's1' OR js.skillid = 's7')
ORDER BY j.jobid

Ourput:

|COMPNAME    |LOCATION  |JOBID |SKILLID|
|------------|----------|------|-------|
|NCR Corp    |Dayton    |j06   |s1     |
|NCR Corp    |Dayton    |j07   |s1     |
|NCR Corp    |Dayton    |j07   |s7     |
|NCR Corp    |Dayton    |j08   |s1     |
|WP Air Force|Dayton    |j10   |s1     |
|Deloitte    |Dayton    |j14   |s7     | 

---

你为什么认为你会加入compID？你不认为从

jobid

加入

skillid

会更好地“弄清楚如何在这里添加技能ID”？--让我们看看，…，也许

Jobskills

表会有帮助。看一看，再试一次这一点对我来说似乎是不言而喻的，我想知道您是否真的试图自己做这件事，或者您只是希望我们能为您编写代码。这不是StackOverflow的目的。@Andreas这是我第一次学习SQL类，我已经尝试解决这个问题一段时间了，现在我决定寻求一些帮助。如果我只是询问代码，会有什么好处？我真的想弄明白这一点，但我不希望有人为我写这篇文章。“仅仅要求代码对我有什么好处？我真的想弄明白这一点，但我不希望有人为我写这篇文章”这是对你态度的赞扬。但你会惊讶于有多少人在这里发帖只是希望有人能帮他们做作业。因此，早期的反应。现在，请期待人们会问你一些引导性问题，这些问题旨在帮助你自我发现（并更好地理解）解决方案。@EdStevens谢谢你，我理解这一点，我不是在找人帮我做这件事。他的建议足以帮助我以不同的方式思考如何处理这个问题，但他的其余评论是不必要的。你能帮助我吗？我想我知道的更多，但我只需要一件事。我要更新我原来的帖子。Oracle还是MySQL？请仅标记您真正使用的DBMS。（顺便说一句：在MySQL中，所有“内联”外键都被忽略）谢谢！我不知道我能像那样做2

内部连接。这就是为什么我的代码总是乱七八糟的。