目录

题目

准备数据

分析数据

总结

---

题目

写一个 SQL 查询语句，以报告在给定三个实验平台中每种实验完成的次数。请注意，每一对（实验平台、实验名称）都应包含在输出中，包括平台上实验次数是零的。

结果可以以任意顺序给出.

准备数据

Create table If Not Exists Experiments (experiment_id int, platform ENUM('Android', 'IOS', 'Web'), experiment_name ENUM('Reading', 'Sports', 'Programming'))
    Truncate table Experiments
    insert into Experiments (experiment_id, platform, experiment_name) values ('4', 'IOS', 'Programming')
    insert into Experiments (experiment_id, platform, experiment_name) values ('13', 'IOS', 'Sports')
    insert into Experiments (experiment_id, platform, experiment_name) values ('14', 'Android', 'Reading')
    insert into Experiments (experiment_id, platform, experiment_name) values ('8', 'Web', 'Reading')
    insert into Experiments (experiment_id, platform, experiment_name) values ('12', 'Web', 'Reading')
    insert into Experiments (experiment_id, platform, experiment_name) values ('18', 'Web', 'Programming')

experiments表 

分析数据

方法一

select platform,experiment_name,count(e1.platform) as num_experiments from
             (select distinct platform from experiments) p
                 cross join (select distinct experiment_name from experiments) e
left join experiments e1
using(platform, experiment_name)
group by platform,experiment_name;

方法二

with pe as
         (   select * from
             (
                 select 'IOS' platform
                 union all
                 select 'Android' platform
                 union all
                 select 'Web' platform
             )p
                 cross join
             (
                 select 'Programming' experiment_name
                 union all
                 select 'Sports' experiment_name
                 union all
                 select 'Reading' experiment_name
             )e
         )
select pe.platform,pe.experiment_name,count(w.platform) as num_experiments
from pe
         left join Experiments  as w
                   using(platform,experiment_name)
group by platform,experiment_name;

总结

• 方法一:使用distinct,但是不严谨,可能例子中含有platform或者experiment_name少于3个的情况,从而丢失情况.方法二:使用union的方法枚举
• left join Experiments表，using(platform,experiment_name)即可保全左表所有结构，这里不需要ifnull，因为count对null计数就会返回0。

using的使用情况和on一样,但是可以避免on子句中重复列名

假设有两个表employees和departments，它们都有一个名为department_id的列，你想要根据这个列将这两个表连接起来：

SELECT e.name, d.department_name
FROM employees e
JOIN departments d USING (department_id);

等同于使用on子句

SELECT e.name, d.department_name
FROM employees e
JOIN departments d ON e.department_id = d.department_id;