Question

我有2张桌子

表1列出了测试前的所有测试题。

表1：

testid  qnid    qn
1       1       currency of iran
1       2       highest peak        
1       3       longest bridge

表2列出了所有学生在测试后的测试状态：

表2：

studentid   testid  qnid    status  

    1       1       1       unanswered      
    1       1       2       unanswered  
    2       1       1       unanswered  
    2       1       2       answered

给出的表2不完整，因为它不包含对所有qnid的响应。

结果应该是列出所有testid的表，qnid来自testnti的表1，来自表2的匹配testid，qnid for studentid = 2

即。结果：

testid  qnid    studentid    status     
1       1       2           unanswered
1       2       2           answered
1       3

表2不包含表1中testid = 1 qnid = 3的值在RESULT表中它的空格应该是空的。

我使用的查询：

select distinct table1.testid,table1.qnid,table2.status 
from table1 
  left outer join table2 
    on table1.testid = table2.testid 
where (table2.studentid = 2 
or table2.studentid =NULL)

但输出是：

testid  qnid    status
1       1       unanswered
1       1       answered
1       2       unanswered
1       2       answered
1       3       unanswered
1       3       answered

Answer 1

只需删除table2.studentid = NULL即可。

select distinct table1.testid,table1.qnid,table2.status 
from table1 
  left outer join table2 
    on table1.testid = table2.testid 
where table2.studentid = 2

Answer 2

你还需要加入问题ID ...所以你加入会是这样的：

table1 left outer join table2 on table1.testid = table2.testid AND table1.qnid = table2.qnid

Answer 3

select distinct table1.testid,table1.qnid,coalesce(table2.student_id,NULL),table2.status 
from table1 
  left outer join table2 
    on table1.testid = table2.testid 
where table2.studentid = 2

改进执行SQL / JOIN

3 个答案: