一.case when then else end條件控制查詢
1.簡單case函式用法
select id,stu_name
case when score >=90 then '優秀'
when score>=75 and score<90 then '良好'
when score>=60 and score<75 then '及格'
else '不及格' end result
from student;
select grade,count(case when *** = 1 then 1 else null end) 男生數,
count(case when *** = 2 then 1 else null end) 女生數
from student group by grade;
2.case搜尋函式
select t2.*,t1.*
from t1,t2
where (case when t2.type = 'a' and t1.name like '%海哥' then 1
when t2.type <> 'a' and t1.name not like '%海哥' then 1
else 0
end) = 1
select
case when salary <=2000 then '低收入'
when salary >2000 and salary<=4000 then '溫飽收入'
when salary >4000 and salary <8000 then '小康收入'
when salary >8000 and salary <12000 then '中產收入'
else '高等收入' end salary_level,
count(1) from employee
group by
case when salary <=2000 then '低收入'
when salary >2000 and salary<=4000 then '溫飽收入'
when salary >4000 and salary <8000 then '小康收入'
when salary >8000 and salary <12000 then '中產收入'
else '高等收入' end;
需要注意的問題，case函式只返回第乙個符合條件的值，剩下的case部分將會被自動忽略
二.遞迴查詢
1.查詢id=20178900的所有直屬子節點,所有後代(prior 後接子節點,start with 開始的地方就是根節點)
select a.id,a.mc,a.code from village a where
a.code in (select id from city s start with id='20178900' connect by s.code = prior s.id)
and a.status = '1' and mc like '%朱家莊%' order by a.time desc ,a.id desc;
2.查詢id=20178900的所有直屬父節點,所有祖宗(prior 後接父節點,start with 開始的地方就是根節點)
select a.id,a.mc,a.code from village a where
a.code in (select id from city s start with id='20178900' connect by prior s.code = s.id)
and a.status = '1' and mc like '%朱家莊%' order by a.time desc ,a.id desc;
Oracle兩種高階查詢總結（例項講解）
一.case when then else end條件控制查詢 1.簡單case函式用法 select id,stu name case when score 90 then 優秀 when score 75 and score 90 then 良好 when score 60 and score ...
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