必要に応じてデータを入力し、データをどのように表示するかを教えてください。そうすれば、解決策を見つけることができます。postjob.postidで区別できますこれにより一意のデータが得られます
ライブデモ http://sqlfiddle.com/#!3/2d53c/4
create table postjob
(postid int identity(1,1),jobtitle nvarchar(50),industry nvarchar(30),
jobdescription nvarchar(150),PostingDate datetime)
insert into postjob values('Programmer','Software','User interface','2012-07-04 15:00:24.833')
insert into postjob values('Tester','Software','Automated Testing Application','2012-07-04 15:00:24.833')
insert into postjob values('Sales Executive','Marketing','Meeting with clients','2012-07-04 15:00:24.833')
insert into postjob values('Doctor','Hospitality','Treatment','2012-07-04 15:00:24.833')
insert into postjob values('Desinger','Software','User interface graphicdesinger','2012-07-04 15:00:24.833')
insert into postjob values('Desinger','Interior','Blah blah','2012-07-04 15:00:24.833')
create table Job_Location(PostigID int,Location nvarchar(50))
insert into Job_Location values ('1','India')
insert into Job_Location values ('2','Italy')
insert into Job_Location values ('3','Germany')
insert into Job_Location values ('4','France')
insert into Job_Location values ('5','India')
insert into Job_Location values ('6','USA')