hive学习笔记之六:HiveQL基础
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《hive学习笔记》系列导航
- 基本数据类型
- 复杂数据类型
- 内部表和外部表
- 分区表
- 分桶
- HiveQL基础
- 内置函数
- Sqoop
- 基础UDF
- 用户自定义聚合函数(UDAF)
- UDTF
本篇概览
- 本文是《hive学习笔记》系列的第六篇,前面的文章咱们对数据类型、表结构有了基本了解,接下来对常用的查询语句做一次集中式的学习;
- HiveQL与SQL类似, 在语法上与大部分SQL兼容, 但是并非完全兼容,例如更新、事务等都不支持,子查询和join操作也有限, 这和底层依赖Hadoop有关;
准备数据
- 本次实战要准备两个表:学生表和住址表,字段都很简单,如下图所示,学生表有个住址ID字段,是住址表里的记录的唯一ID:
2. 先创建住址表:
create table address (addressid int, province string, city string)
row format delimited
fields terminated by ",";
- 创建address.txt文件,内容如下:
1,guangdong,guangzhou
2,guangdong,shenzhen
3,shanxi,xian
4,shanxi,hanzhong
6,jiangshu,nanjing
- 加载数据到address表:
load data
local inpath "/home/hadoop/temp/202010/25/address.txt"
into table address;
- 创建学生表,其addressid字段关联了address表的addressid字段:
create table student (name string, age int, addressid int)
row format delimited
fields terminated by ",";
- 创建student.txt文件,内容如下:
tom,11,1
jerry,12,2
mike,13,3
john,14,4
mary,15,5
- 加载数据到student表:
load data
local inpath "/home/hadoop/temp/202010/25/student.txt"
into table student;
- 至此,本次操作所需数据已准备完毕,如下所示:
hive> select * from address;
OK
1 guangdong guangzhou
2 guangdong shenzhen
3 shanxi xian
4 shanxi hanzhong
6 jiangshu nanjing
Time taken: 0.043 seconds, Fetched: 5 row(s)
hive> select * from student;
OK
tom 11 1
jerry 12 2
mike 13 3
john 14 4
mary 15 5
Time taken: 0.068 seconds, Fetched: 5 row(s)
- 开始体验HiveQL
select和where
最普通的带条件查询:
hive> select * from address where city like "%a%";
OK
1 guangdong guangzhou
3 shanxi xian
4 shanxi hanzhong
6 jiangshu nanjing
Time taken: 0.128 seconds, Fetched: 4 row(s)
group by
- 按province字段分组:
select province, count(*) from address group by province;
该查询会触发MR计算,结果如下:
...
Total MapReduce CPU Time Spent: 1 seconds 910 msec
OK
guangdong 2
jiangshu 1
shanxi 2
Time taken: 17.847 seconds, Fetched: 3 row(s)
- 试试嵌套查询,内部是查出city字段带有a字母的记录,然后将这些记录按照province字段分组:
select t.province, count(*) from (
select * from address where city like "%a%"
) t
group by t.province;
结果如下:
Total MapReduce CPU Time Spent: 1 seconds 760 msec
OK
guangdong 1
jiangshu 1
shanxi 2
Time taken: 18.036 seconds, Fetched: 3 row(s)
having
- 前面的嵌套查询,结果有两个省:guangdong和shanxi,如果再加个条件:只显示城市数量大于1的省,首先想到的是再加一层嵌套:
select t1.* from (
select t.province, count(*) as cnt from (
select * from address where city like "%a%"
) t
group by t.province) t1
where t1.cnt>1;
结果如下,可见只有shanxi被显示了:
Total MapReduce CPU Time Spent: 2 seconds 250 msec
OK
shanxi 2
Time taken: 20.067 seconds, Fetched: 1 row(s)
- 对于上述SQL,可以用having语法进行分组筛选,得到同样数据:
select t.province, count(*) as cnt from (
select * from address where city like "%a%"
) t
group by t.province having cnt>1;
order by
- 对分组结果做排序:
select t.province, count(*) as cnt from (
select * from address where city like "%a%"
) t
group by t.province order by cnt;
会触发MR,结果如下:
Total MapReduce CPU Time Spent: 3 seconds 50 msec
OK
jiangshu 1
guangdong 1
shanxi 2
Time taken: 40.315 seconds, Fetched: 3 row(s)
- order by对于的实现,是在最后通过一个reducer进行全部排序,该过程可能耗时较长,针对这种情况,hive提供了sort by,功能与order by一样,但是会在每个reducer中进行排序,这样最终做排序的时候效率就会提升;
- 要注意的是:sort by解决的问题是最终结果排序的效率,因此数据量不大时,排序不是瓶颈,此时使用sort by也不会加快整体速度;
内连接(inner join)
- 内连接用join简写,与连接标准匹配的数据在两张表中都存在,才会保留:
select
s.name, s.age,
a.province, a.city
from
student s
inner join
address a
on
s.addressid=a.addressid;
结果如下:
Total MapReduce CPU Time Spent: 1 seconds 20 msec
OK
tom 11 guangdong guangzhou
jerry 12 guangdong shenzhen
mike 13 shanxi xian
john 14 shanxi hanzhong
Time taken: 17.294 seconds, Fetched: 4 row(s)
自然连接(natural join)
- 自然连接是在两张表中寻找数据类型和列明都相同的字段,并自动连接起来:
select name, age, province, city from student natural join address;
结果如下,可见不会根据student表的addressid字段值去address查找记录,而是将addrerss的记录全部连接一次:
Total MapReduce CPU Time Spent: 940 msec
OK
tom 11 guangdong guangzhou
jerry 12 guangdong guangzhou
mike 13 guangdong guangzhou
john 14 guangdong guangzhou
mary 15 guangdong guangzhou
tom 11 guangdong shenzhen
jerry 12 guangdong shenzhen
mike 13 guangdong shenzhen
john 14 guangdong shenzhen
mary 15 guangdong shenzhen
tom 11 shanxi xian
jerry 12 shanxi xian
mike 13 shanxi xian
john 14 shanxi xian
mary 15 shanxi xian
tom 11 shanxi hanzhong
jerry 12 shanxi hanzhong
mike 13 shanxi hanzhong
john 14 shanxi hanzhong
mary 15 shanxi hanzhong
tom 11 jiangshu nanjing
jerry 12 jiangshu nanjing
mike 13 jiangshu nanjing
john 14 jiangshu nanjing
mary 15 jiangshu nanjing
Time taken: 18.525 seconds, Fetched: 25 row(s)
左外连接(left outer join)
- 以连接中的左表为主:
select
s.name, s.age, s.addressid,
a.province, a.city
from
student s
left outer join
address a
on
s.addressid=a.addressid;
结果如下,可见name=mary的记录,addressid等于5,在address中不存在addressid等于5的记录,因此province和city字段都展示了NULL,而在前面使用inner join时,结果中没有这条记录:
Total MapReduce CPU Time Spent: 950 msec
OK
tom 11 1 guangdong guangzhou
jerry 12 2 guangdong shenzhen
mike 13 3 shanxi xian
john 14 4 shanxi hanzhong
mary 15 5 NULL NULL
Time taken: 18.442 seconds, Fetched: 5 row(s)
右外连接(right outer join)
和左连接类似,只不过是以右表为主,语法是right outer join:
select
s.name, s.age, s.addressid,
a.province, a.city
from
student s
right outer join
address a
on
s.addressid=a.addressid;
结果如下,可见city=nanjing的记录,在student表中没有一条记录与之关联,因此结果中展示了address的字段,而student的字段为NULL:
Total MapReduce CPU Time Spent: 970 msec
OK
tom 11 1 guangdong guangzhou
jerry 12 2 guangdong shenzhen
mike 13 3 shanxi xian
john 14 4 shanxi hanzhong
NULL NULL NULL jiangshu nanjing
Time taken: 18.294 seconds, Fetched: 5 row(s)
全外连接(full outer join)
查询结果等于左外连接和右外连接之和,语法是full outer join:
select
s.name, s.age, s.addressid,
a.province, a.city
from
student s
full outer join
address a
on
s.addressid=a.addressid;
结果如下:
Total MapReduce CPU Time Spent: 2 seconds 630 msec
OK
tom 11 1 guangdong guangzhou
jerry 12 2 guangdong shenzhen
mike 13 3 shanxi xian
john 14 4 shanxi hanzhong
mary 15 5 NULL NULL
NULL NULL NULL jiangshu nanjing
Time taken: 22.189 seconds, Fetched: 6 row(s)
- 至此,常用HiveQL体验完毕,希望能给您一些参考,接下来的章节会进一步学习HiveQL的特性;
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