hive學習之經典sql 50題 hive版
建表:
create table student(s_id string,s_name string,s_birth string,s_sex string) row format delimited fields terminated by '\t';
create table course(c_id string,c_name string,t_id string) row format delimited fields terminated by '\t';
create table teacher(t_id string,t_name string) row format delimited fields terminated by '\t';
create table score(s_id string,c_id string,s_score int) row format delimited fields terminated by '\t';
生成數據
vi /export/data/hivedatas/student.csv
01 趙雷 1990-01-01 男
02 錢電 1990-12-21 男
03 孫風 1990-05-20 男
04 李云 1990-08-06 男
05 周梅 1991-12-01 女
06 吳蘭 1992-03-01 女
07 鄭竹 1989-07-01 女
08 王菊 1990-01-20 女
vi /export/data/hivedatas/course.csv
01 語文 02
02 數學 01
03 英語 03
vi /export/data/hivedatas/teacher.csv
01 張三
02 李四
03 王五
vi /export/data/hivedatas/score.csv
01 01 80
01 02 90
01 03 99
02 01 70
02 02 60
02 03 80
03 01 80
03 02 80
03 03 80
04 01 50
04 02 30
04 03 20
05 01 76
05 02 87
06 01 31
06 03 34
07 02 89
07 03 98
導數據到hive
load data local inpath '/export/data/hivedatas/student.csv' into table student;
load data local inpath '/export/data/hivedatas/course.csv' into table course;
load data local inpath '/export/data/hivedatas/teacher.csv' into table teacher;
load data local inpath '/export/data/hivedatas/score.csv' into table score;
–注:–hive查詢語法
SELECT [ALL | DISTINCT] select_expr, select_expr, ...
FROM table_reference
[WHERE where_condition]
[GROUP BY col_list [HAVING condition]]
[CLUSTER BY col_list
| [DISTRIBUTE BY col_list] [SORT BY| ORDER BY col_list]
]
[LIMIT number]
– 1、查詢"01"課程比"02"課程成績高的學生的信息及課程分數:
select student.*,a.s_score as 01_score,b.s_score as 02_score
from student
join score a on student.s_id=a.s_id and a.c_id='01'
left join score b on student.s_id=b.s_id and b.c_id='02'
where a.s_score>b.s_score;
–答案2
select student.*,a.s_score as 01_score,b.s_score as 02_score
from student
join score a on a.c_id='01'
join score b on b.c_id='02'
where a.s_id=student.s_id and b.s_id=student.s_id and a.s_score>b.s_score;
– 2、查詢"01"課程比"02"課程成績低的學生的信息及課程分數:
select student.*,a.s_score as 01_score,b.s_score as 02_score
from student
join score a on student.s_id=a.s_id and a.c_id='01'
left join score b on student.s_id=b.s_id and b.c_id='02'
where a.s_score<b.s_score;
–答案2
select student.*,a.s_score as 01_score,b.s_score as 02_score
from student
join score a on a.c_id='01'
join score b on b.c_id='02'
where a.s_id=student.s_id and b.s_id=student.s_id and a.s_score<b.s_score;
– 3、查詢平均成績大于等于60分的同學的學生編號和學生姓名和平均成績:
select student.s_id,student.s_name,tmp.平均成績 from student
join (
select score.s_id,round(avg(score.s_score),1)as 平均成績
from score group by s_id)as tmp
on tmp.平均成績>=60
where student.s_id = tmp.s_id
–答案2
select student.s_id,student.s_name,round(avg (score.s_score),1) as 平均成績 from student
join score on student.s_id = score.s_id
group by student.s_id,student.s_name
having avg (score.s_score) >= 60;
– 4、查詢平均成績小于60分的同學的學生編號和學生姓名和平均成績:
– (包括有成績的和無成績的)
select student.s_id,student.s_name,tmp.avgScore from student
join (
select score.s_id,round(avg(score.s_score),1)as avgScore from score group by s_id)as tmp
on tmp.avgScore < 60
where student.s_id=tmp.s_id
union all
select s2.s_id,s2.s_name,0 as avgScore from student s2
where s2.s_id not in
(select distinct sc2.s_id from score sc2);
–答案2
select score.s_id,student.s_name,round(avg (score.s_score),1) as avgScore from student
inner join score on student.s_id=score.s_id
group by score.s_id,student.s_name
having avg (score.s_score) < 60
union all
select s2.s_id,s2.s_name,0 as avgScore from student s2
where s2.s_id not in
(select distinct sc2.s_id from score sc2);
– 5、查詢所有同學的學生編號、學生姓名、選課總數、所有課程的總成績:
select student.s_id,student.s_name,(count(score.c_id) )as total_count,sum(score.s_score)as total_score
from student
left join score on student.s_id=score.s_id
group by student.s_id,student.s_name ;
– 6、查詢"李"姓老師的數量:
select t_name,count(1) from teacher where t_name like '李%' group by t_name;
– 7、查詢學過"張三"老師授課的同學的信息:
select student.* from student
join score on student.s_id =score.s_id
join course on course.c_id=score.c_id
join teacher on course.t_id=teacher.t_id and t_name='張三';
– 8、查詢沒學過"張三"老師授課的同學的信息:
select student.* from student
left join (select s_id from score
join course on course.c_id=score.c_id
join teacher on course.t_id=teacher.t_id and t_name='張三')tmp
on student.s_id =tmp.s_id
where tmp.s_id is null;
– 9、查詢學過編號為"01"并且也學過編號為"02"的課程的同學的信息:
select * from student
join (select s_id from score where c_id =1 )tmp1
on student.s_id=tmp1.s_id
join (select s_id from score where c_id =2 )tmp2
on student.s_id=tmp2.s_id;
– 10、查詢學過編號為"01"但是沒有學過編號為"02"的課程的同學的信息:
select student.* from student
join (select s_id from score where c_id =1 )tmp1
on student.s_id=tmp1.s_id
left join (select s_id from score where c_id =2 )tmp2
on student.s_id =tmp2.s_id
where tmp2.s_id is null;
– 11、查詢沒有學全所有課程的同學的信息:
–先查詢出課程的總數量
select count(1) from course;
–再查詢所需結果
select student.* from student
left join(
select s_id
from score
group by s_id
having count(c_id)=3)tmp
on student.s_id=tmp.s_id
where tmp.s_id is null;
–方法二(一步到位):
select student.* from student
join (select count(c_id)num1 from course)tmp1
left join(
select s_id,count(c_id)num2
from score group by s_id)tmp2
on student.s_id=tmp2.s_id and tmp1.num1=tmp2.num2
where tmp2.s_id is null;
– 12、查詢至少有一門課與學號為"01"的同學所學相同的同學的信息:
select student.* from student
join (select c_id from score where score.s_id=01)tmp1
join (select s_id,c_id from score)tmp2
on tmp1.c_id =tmp2.c_id and student.s_id =tmp2.s_id
where student.s_id not in('01')
group by student.s_id,s_name,s_birth,s_sex;
– 13、查詢和"01"號的同學學習的課程完全相同的其他同學的信息:
–備注:hive不支持group_concat方法,可用 concat_ws(’|’, collect_set(str)) 實現(xiàn)
select student.*,tmp1.course_id from student
join (select s_id ,concat_ws('|', collect_set(c_id)) course_id from score
group by s_id having s_id not in (1))tmp1
on student.s_id = tmp1.s_id
join (select concat_ws('|', collect_set(c_id)) course_id2
from score where s_id=1)tmp2
on tmp1.course_id = tmp2.course_id2;
– 14、查詢沒學過"張三"老師講授的任一門課程的學生姓名:
select student.* from student
left join (select s_id from score
join (select c_id from course join teacher on course.t_id=teacher.t_id and t_name='張三')tmp2
on score.c_id=tmp2.c_id )tmp
on student.s_id = tmp.s_id
where tmp.s_id is null;
– 15、查詢兩門及其以上不及格課程的同學的學號，姓名及其平均成績:
select student.s_id,student.s_name,tmp.avg_score from student
inner join (select s_id from score
where s_score<60
group by score.s_id having count(s_id)>1)tmp2
on student.s_id = tmp2.s_id
left join (
select s_id,round(AVG (score.s_score)) avg_score
from score group by s_id)tmp
on tmp.s_id=student.s_id;
– 16、檢索"01"課程分數小于60，按分數降序排列的學生信息:
select student.*,s_score from student,score
where student.s_id=score.s_id and s_score<60 and c_id='01'
order by s_score desc;
– 17、按平均成績從高到低顯示所有學生的所有課程的成績以及平均成績:
select a.s_id,tmp1.s_score as chinese,tmp2.s_score as math,tmp3.s_score as english,
round(avg (a.s_score),2) as avgScore
from score a
left join (select s_id,s_score from score s1 where c_id='01')tmp1 on tmp1.s_id=a.s_id
left join (select s_id,s_score from score s2 where c_id='02')tmp2 on tmp2.s_id=a.s_id
left join (select s_id,s_score from score s3 where c_id='03')tmp3 on tmp3.s_id=a.s_id
group by a.s_id,tmp1.s_score,tmp2.s_score,tmp3.s_score order by avgScore desc;
– 18.查詢各科成績最高分、最低分和平均分：以如下形式顯示：課程ID，課程name，最高分，最低分，平均分，及格率，中等率，優(yōu)良率，優(yōu)秀率:
–及格為>=60，中等為：70-80，優(yōu)良為：80-90，優(yōu)秀為：>=90
select course.c_id,course.c_name,tmp.maxScore,tmp.minScore,tmp.avgScore,tmp.passRate,tmp.moderate,tmp.goodRate,tmp.excellentRates from course
join(select c_id,max(s_score) as maxScore,min(s_score)as minScore,
round(avg(s_score),2) avgScore,
round(sum(case when s_score>=60 then 1 else 0 end)/count(c_id),2)passRate,
round(sum(case when s_score>=60 and s_score<70 then 1 else 0 end)/count(c_id),2) moderate,
round(sum(case when s_score>=70 and s_score<80 then 1 else 0 end)/count(c_id),2) goodRate,
round(sum(case when s_score>=80 and s_score<90 then 1 else 0 end)/count(c_id),2) excellentRates
from score group by c_id)tmp on tmp.c_id=course.c_id;
– 19、按各科成績進行排序，并顯示排名:
– row_number() over()分組排序功能(mysql沒有該方法)
select s1.*,row_number()over(order by s1.s_score desc) Ranking
from score s1 where s1.c_id='01'order by noRanking asc
union all select s2.*,row_number()over(order by s2.s_score desc) Ranking
from score s2 where s2.c_id='02'order by noRanking asc
union all select s3.*,row_number()over(order by s3.s_score desc) Ranking
from score s3 where s3.c_id='03'order by noRanking asc;
– 20、查詢學生的總成績并進行排名:
select score.s_id,s_name,sum(s_score) sumscore,row_number()over(order by sum(s_score) desc) Ranking
from score ,student
where score.s_id=student.s_id
group by score.s_id,s_name order by sumscore desc;
后續(xù)部分參見:
https://blog.csdn.net/Thomson617/article/details/83280617
Hive下的SQL語法總結：
(1).Hive不支持join的非等值連接,不支持or
分別舉例如下及實現(xiàn)解決辦法。
不支持不等值連接
錯誤:select * from a inner join b on a.id<>b.id
替代方法:select * from a inner join b on a.id=b.id and a.id is null;
不支持or
錯誤:select * from a inner join b on a.id=b.id or a.name=b.name
替代方法:select * from a inner join b on a.id=b.id
union all
select * from a inner join b on a.name=b.name
兩個sql union all的字段名必須一樣或者列別名要一樣。
(2).分號字符:不能智能識別concat(‘;’,key)，只會將‘；’當做SQL結束符號。
?分號是SQL語句結束標記，在HiveQL中也是，但是在HiveQL中，對分號的識別沒有那么智慧，例如：
?select concat(key,concat(';',key)) from dual;
?但HiveQL在解析語句時提示：
FAILED: Parse Error: line 0:-1 mismatched input '<EOF>' expecting ) in function specification
?解決的辦法是，使用分號的八進制的ASCII碼進行轉義，那么上述語句應寫成：
?select concat(key,concat('\073',key)) from dual;
(3).不支持INSERT INTO 表 Values（）, UPDATE, DELETE等操作.這樣的話，就不要很復雜的鎖機制來讀寫數據。
INSERT INTO syntax is only available starting in version 0.8。INSERT INTO就是在表或分區(qū)中追加數據。
(4).HiveQL中String類型的字段若是空(empty)字符串, 即長度為0, 那么對它進行IS NULL的判斷結果是False，使用left join可以進行篩選行。
(5).不支持 ‘< dt <’這種格式的范圍查找，可以用dt in(”,”)或者between替代。
(6).Hive不支持將數據插入現(xiàn)有的表或分區(qū)中，僅支持覆蓋重寫整個表，示例如下：
INSERT OVERWRITE TABLE t1 SELECT * FROM t2;
(7).group by的字段,必須是select后面的字段，select后面的字段不能比group by的字段多.
如果select后面有聚合函數,則該select語句中必須有group by語句
而且group by后面不能使用別名
(8).hive的0.13版之前select , where 及 having 之后不能跟子查詢語句(一般使用left join、right join 或者inner join替代)
(9).先join(及inner join) 然后left join或right join
(10).hive不支持group_concat方法,可用 concat_ws('|', collect_set(str)) 實現(xiàn)
(11).not in 和 <> 不起作用,可用left join tmp on tableName.id = tmp.id where tmp.id is null 替代實現(xiàn)
– 21、查詢不同老師所教不同課程平均分從高到低顯示:
– 方法1
select course.c_id,course.t_id,t_name,round(avg(s_score),2)as avgscore from course
join teacher on teacher.t_id=course.t_id
join score on course.c_id=score.c_id
group by course.c_id,course.t_id,t_name order by avgscore desc;
– 方法2
select course.c_id,course.t_id,t_name,round(avg(s_score),2)as avgscore from course,teacher,score
where teacher.t_id=course.t_id and course.c_id=score.c_id
group by course.c_id,course.t_id,t_name order by avgscore desc;
– 22、查詢所有課程的成績第2名到第3名的學生信息及該課程成績:
select tmp1.* from
(select * from score where c_id='01' order by s_score desc limit 3)tmp1
order by s_score asc limit 2
union all select tmp2.* from
(select * from score where c_id='02' order by s_score desc limit 3)tmp2
order by s_score asc limit 2
union all select tmp3.* from
(select * from score where c_id='03' order by s_score desc limit 3)tmp3
order by s_score asc limit 2;
– 23、統(tǒng)計各科成績各分數段人數：課程編號,課程名稱,[100-85],[85-70],[70-60],[0-60]及所占百分比
select c.c_id,c.c_name,tmp1.s0_60, tmp1.percentum,tmp2.s60_70, tmp2.percentum,tmp3.s70_85, tmp3.percentum,tmp4.s85_100, tmp4.percentum
from course c
join(select c_id,sum(case when s_score<60 then 1 else 0 end )as s0_60,
round(100*sum(case when s_score<60 then 1 else 0 end )/count(c_id),2)as percentum
from score group by c_id)tmp1 on tmp1.c_id =c.c_id
left join(select c_id,sum(case when s_score<70 and s_score>=60 then 1 else 0 end )as s60_70,
round(100*sum(case when s_score<70 and s_score>=60 then 1 else 0 end )/count(c_id),2)as percentum
from score group by c_id)tmp2 on tmp2.c_id =c.c_id
left join(select c_id,sum(case when s_score<85 and s_score>=70 then 1 else 0 end )as s70_85,
round(100*sum(case when s_score<85 and s_score>=70 then 1 else 0 end )/count(c_id),2)as percentum
from score group by c_id)tmp3 on tmp3.c_id =c.c_id
left join(select c_id,sum(case when s_score>=85 then 1 else 0 end )as s85_100,
round(100*sum(case when s_score>=85 then 1 else 0 end )/count(c_id),2)as percentum
from score group by c_id)tmp4 on tmp4.c_id =c.c_id;
– 24、查詢學生平均成績及其名次:
select tmp.*,row_number()over(order by tmp.avgScore desc) Ranking from
(select student.s_id,
student.s_name,
round(avg(score.s_score),2) as avgScore
from student join score
on student.s_id=score.s_id
group by student.s_id,student.s_name)tmp
order by avgScore desc;
– 25、查詢各科成績前三名的記錄
–課程id為01的前三名
select score.c_id,course.c_name,student.s_name,s_score from score
join student on student.s_id=score.s_id
join course on score.c_id='01' and course.c_id=score.c_id
order by s_score desc limit 3;
–課程id為02的前三名
select score.c_id,course.c_name,student.s_name,s_score
from score
join student on student.s_id=score.s_id
join course on score.c_id='02' and course.c_id=score.c_id
order by s_score desc limit 3;
–課程id為03的前三名
select score.c_id,course.c_name,student.s_name,s_score
from score
join student on student.s_id=score.s_id
join course on score.c_id='03' and course.c_id=score.c_id
order by s_score desc limit 3;
– 26、查詢每門課程被選修的學生數:
select c.c_id,c.c_name,tmp.number from course c
join (select c_id,count(1) as number from score
where score.s_score<60 group by score.c_id)tmp
on tmp.c_id=c.c_id;
– 27、查詢出只有兩門課程的全部學生的學號和姓名:
select st.s_id,st.s_name from student st
join (select s_id from score group by s_id having count(c_id) =2)tmp
on st.s_id=tmp.s_id;
– 28、查詢男生、女生人數:
select tmp1.man,tmp2.women from
(select count(1) as man from student where s_sex='男')tmp1,
(select count(1) as women from student where s_sex='女')tmp2;
– 29、查詢名字中含有"風"字的學生信息:
select * from student where s_name like '%風%';
– 30、查詢同名同性學生名單，并統(tǒng)計同名人數:
select s1.s_id,s1.s_name,s1.s_sex,count(*) as sameName
from student s1,student s2
where s1.s_name=s2.s_name and s1.s_id<>s2.s_id and s1.s_sex=s2.s_sex
group by s1.s_id,s1.s_name,s1.s_sex;
– 31、查詢1990年出生的學生名單:
select * from student where s_birth like '1990%';
– 32、查詢每門課程的平均成績，結果按平均成績降序排列，平均成績相同時，按課程編號升序排列:
select score.c_id,c_name,round(avg(s_score),2) as avgScore from score
join course on score.c_id=course.c_id
group by score.c_id,c_name order by avgScore desc,score.c_id asc;
– 33、查詢平均成績大于等于85的所有學生的學號、姓名和平均成績:
select score.s_id,s_name,round(avg(s_score),2)as avgScore from score
join student on student.s_id=score.s_id
group by score.s_id,s_name having avg(s_score) >= 85;
– 34、查詢課程名稱為"數學"，且分數低于60的學生姓名和分數:
select s_name,s_score as mathScore from student
join (select s_id,s_score
from score,course
where score.c_id=course.c_id and c_name='數學')tmp
on tmp.s_score < 60 and student.s_id=tmp.s_id;
– 35、查詢所有學生的課程及分數情況:
select a.s_name,
SUM(case c.c_name when '語文' then b.s_score else 0 end ) as chainese,
SUM(case c.c_name when '數學' then b.s_score else 0 end ) as math,
SUM(case c.c_name when '英語' then b.s_score else 0 end ) as english,
SUM(b.s_score) as sumScore
from student a
join score b on a.s_id=b.s_id
join course c on b.c_id=c.c_id
group by s_name,a.s_id;
后續(xù)部分參見:
https://blog.csdn.net/Thomson617/article/details/83281254
Hive下的SQL經驗總結：
(1).不支持非等值連接，一般使用left join、right join 或者inner join替代。
?SQL中對兩表內聯(lián)可以寫成：
select * from dual a,dual b where a.key = b.key;
?Hive中應為:
select * from dual a join dual b on a.key = b.key;
而不是傳統(tǒng)的格式：
SELECT t1.a1 as c1, t2.b1 as c2 FROM t1, t2 WHERE t1.a2 = t2.b2
(2).分號字符:不能智能識別concat(‘;’,key)，只會將‘；’當做SQL結束符號。
?分號是SQL語句結束標記，在HiveQL中也是，但是在HiveQL中，對分號的識別沒有那么智慧，例如：
?select concat(key,concat(';',key)) from dual;
?但HiveQL在解析語句時提示：
FAILED: Parse Error: line 0:-1 mismatched input '<EOF>' expecting ) in function specification
?解決的辦法是，使用分號的八進制的ASCII碼進行轉義，那么上述語句應寫成：
?select concat(key,concat('\073',key)) from dual;
(3).不支持INSERT INTO 表 Values（）, UPDATE, DELETE等操作.這樣的話，就不要很復雜的鎖機制來讀寫數據。
INSERT INTO syntax is only available starting in version 0.8。INSERT INTO就是在表或分區(qū)中追加數據。
(4).HiveQL中String類型的字段若是空(empty)字符串, 即長度為0, 那么對它進行IS NULL的判斷結果是False，使用left join可以進行篩選行。
(5).不支持 ‘< dt <’這種格式的范圍查找，可以用dt in(”,”)或者between替代。
(6).Hive不支持將數據插入現(xiàn)有的表或分區(qū)中，僅支持覆蓋重寫整個表，示例如下：
INSERT OVERWRITE TABLE t1 SELECT * FROM t2;
(7).group by的字段,必須是select后面的字段，select后面的字段不能比group by的字段多.
如果select后面有聚合函數,則該select語句中必須有group by語句;
而且group by后面不能使用別名;
有聚合函數存在就必須有group by.
(8).select , where 及 having 之后不能跟子查詢語句(一般使用left join、right join 或者inner join替代)
(9).先join(及inner join) 然后left join或right join
(10).hive不支持group_concat方法,可用 concat_ws('|', collect_set(str)) 實現(xiàn)
(11).not in 和 <> 不起作用,可用left join tmp on tableName.id = tmp.id where tmp.id is null 替代實現(xiàn)
(12).hive 中‘不等于’不管是用！ 或者<>符號實現(xiàn)，都會將空值即null過濾掉，此時要用
where （white_level<>'3' or white_level is null）
或者 where (white_level!='3' or white_level is null ) 來保留null 的情況。
(13).union all 后面的表不加括號,不然執(zhí)行報錯;
hive也不支持頂層的union all，使用子查詢來解決;
union all 之前不能有DISTRIBUTE BY | SORT BY| ORDER BY | LIMIT 等查詢條件
– 36、查詢任何一門課程成績在70分以上的學生姓名、課程名稱和分數:
select student.s_id,s_name,c_name,s_score from student
join (select sc.* from score sc
left join(select s_id from score where s_score < 70 group by s_id)tmp
on sc.s_id=tmp.s_id where tmp.s_id is null)tmp2
on student.s_id=tmp2.s_id
join course on tmp2.c_id=course.c_id
order by s_id;
**-- 查詢全部及格的信息**
select sc.* from score sc
left join(select s_id from score where s_score < 60 group by s_id)tmp
on sc.s_id=tmp.s_id
where tmp.s_id is null;
**-- 或(效率低)**
select sc.* from score sc
where sc.s_id not in (select s_id from score where s_score < 60 group by s_id);
– 37、查詢課程不及格的學生:
select s_name,c_name as courseName,tmp.s_score
from student
join (select s_id,s_score,c_name
from score,course
where score.c_id=course.c_id and s_score < 60)tmp
on student.s_id=tmp.s_id;
–38、查詢課程編號為01且課程成績在80分以上的學生的學號和姓名:
select student.s_id,s_name,s_score as score_01
from student
join score on student.s_id=score.s_id
where c_id='01' and s_score >= 80;
– 39、求每門課程的學生人數:
select course.c_id,course.c_name,count(1)as selectNum
from course
join score on course.c_id=score.c_id
group by course.c_id,course.c_name;
– 40、查詢選修"張三"老師所授課程的學生中，成績最高的學生信息及其成績:
select student.*,tmp3.c_name,tmp3.maxScore
from (select s_id,c_name,max(s_score)as maxScore from score
join (select course.c_id,c_name from course join
(select t_id,t_name from teacher where t_name='張三')tmp
on course.t_id=tmp.t_id)tmp2
on score.c_id=tmp2.c_id group by score.s_id,c_name
order by maxScore desc limit 1)tmp3
join student
on student.s_id=tmp3.s_id;
– 41、查詢不同課程成績相同的學生的學生編號、課程編號、學生成績:
select distinct a.s_id,a.c_id,a.s_score from score a,score b
where a.c_id <> b.c_id and a.s_score=b.s_score;
– 42、查詢每門課程成績最好的前三名:
select tmp1.* from
(select *,row_number()over(order by s_score desc) ranking
from score where c_id ='01')tmp1
where tmp1.ranking <= 3
union all
select tmp2.* from
(select *,row_number()over(order by s_score desc) ranking
from score where c_id ='02')tmp2
where tmp2.ranking <= 3
union all
select tmp3.* from
(select *,row_number()over(order by s_score desc) ranking
from score where c_id ='03')tmp3
where tmp3.ranking <= 3;
– 43、統(tǒng)計每門課程的學生選修人數（超過5人的課程才統(tǒng)計）:
– 要求輸出課程號和選修人數，查詢結果按人數降序排列，若人數相同，按課程號升序排列
select distinct course.c_id,tmp.num from course
join (select c_id,count(1) as num from score group by c_id)tmp
where tmp.num>=5 order by tmp.num desc ,course.c_id asc;
– 44、檢索至少選修兩門課程的學生學號:
select s_id,count(c_id) as totalCourse
from score
group by s_id
having count(c_id) >= 2;
– 45、查詢選修了全部課程的學生信息:
select student.*
from student,
(select s_id,count(c_id) as totalCourse
from score group by s_id)tmp
where student.s_id=tmp.s_id and totalCourse=3;
–46、查詢各學生的年齡(周歲):
– 按照出生日期來算，當前月日 < 出生年月的月日則，年齡減一
select s_name,s_birth,(year(CURRENT_DATE)-year(s_birth)-
(case when month(CURRENT_DATE) > month(s_birth) then 0
when day(CURRENT_DATE) > day(s_birth) then
1 else 0 end)) as age
from student;
– 47、查詢本周過生日的學生:
–方法1
select * from student where weekofyear(CURRENT_DATE)+1 =weekofyear(s_birth);
–方法2
select s_name,s_sex,s_birth from student
where substring(s_birth,6,2)='10'
and substring(s_birth,9,2)=14;
– 48、查詢下周過生日的學生:
–方法1
select * from student where weekofyear(CURRENT_DATE)+1 =weekofyear(s_birth);
–方法2
select s_name,s_sex,s_birth from student
where substring(s_birth,6,2)='10'
and substring(s_birth,9,2)>=15
and substring(s_birth,9,2)<=21;
– 49、查詢本月過生日的學生:
–方法1
select * from student where MONTH(CURRENT_DATE)+1 =MONTH(s_birth);
–方法2
select s_name,s_sex,s_birth from student where substring(s_birth,6,2)='10';
– 50、查詢12月份過生日的學生:
select s_name,s_sex,s_birth from student where substring(s_birth,6,2)='12';
所有代碼親測有效!
如果因為hive版本及測試環(huán)境造成無法運行的還請自行修正!
hive sql中的部分方法總結:
1.case when ... then ... else ... end
2.length(string)
3.cast(string as bigint)
4.rand() 返回一個0到1范圍內的隨機數
5.ceiling(double) 向上取整
6.substr(string A, int start, int len)
7.collect_set(col)函數只接受基本數據類型，它的主要作用是將某字段的值進行去重匯總，產生array類型字段
8.concat()函數
1、功能：將多個字符串連接成一個字符串。
2、語法：concat(str1, str2,...)
返回結果為連接參數產生的字符串，如果有任何一個參數為null，則返回值為null。
9.concat_ws()函數
1、功能：和concat()一樣，將多個字符串連接成一個字符串，但是可以一次性指定分隔符～（concat_ws就是concat with separator）
2、語法：concat_ws(separator, str1, str2, ...)
說明：第一個參數指定分隔符。需要注意的是分隔符不能為null，如果為null，則返回結果為null。
10.nvl(expr1, expr2)：空值轉換函數 nvl(x,y) Returns y if x is null else return x
11.if(boolean testCondition, T valueTrue, T valueFalse)
12.row_number()over()分組排序功能,over()里頭的分組以及排序的執(zhí)行晚于 where group by order by 的執(zhí)行。
13.獲取年、月、日、小時、分鐘、秒、當年第幾周
select
year('2018-02-27 10:00:00') as year
,month('2018-02-27 10:00:00') as month
,day('2018-02-27 10:00:00') as day
,hour('2018-02-27 10:00:00') as hour
,minute('2018-02-27 10:00:00') as minute
,second('2018-02-27 10:00:00') as second
,weekofyear('2018-02-27 10:00:00') as weekofyear
獲取當前時間:
1).current_timestamp
2).unix_timestamp()
3).from_unixtime(unix_timestamp())
4).CURRENT_DATE
