# Write your MySQL query statement below
select name
from customer
where id not in
(select id
from customer where referee_id =2);
or
select name from customer where id != 2 or id is null;
11
多表连接
11
select …
from 表1 as a
left join 表2 as b
on a.列名=b.列名
where b.列名 is null;
中上图黑色框里的sql解决的问题是：不在表里的数据，也就是在表A里的数据，但是不在表B里的数据。
对于这个题目“不是近视眼的学生都有谁？”，就是在“学生表”里的数据，但是不在“近视学生”表里的数据。我们选择下图黑色框里的左联结sql语句。。
-
左连接表
-
select c.Name as Customers from Customers as c left join Orders as o on c.Id = o.CustomerId where o.Id is null;
11
# Write your MySQL query statement below
select
user_id,
concat(upper(left(name,1)),lower(substr(name,2))) as name
from
Users
order by
user_id;
# Write your MySQL query statement below
select sell_date,count(distinct product) as num_sold,
group_concat(distinct product order by(product)) as products
from activities
group by sell_date
order by sell_date;
111
select *from patients where conditions like '%DIAB1%'
or conditions like ' DIAB1%';
11
# Write your MySQL query statement below
select employee_id from Employees where employee_id not in (select employee_id
from salaries) union select employee_id from salaries where employee_id not in
(select employee_id from Employees) order by employee_id;
11
# Write your MySQL query statement below
select id,
case when t.p_id is null then 'Root'
when t.id in (select p_id from tree ) then 'Inner' //判断是否右孩子结点
else 'Leaf'
end as Type //作为type列返回生成查询结果
from tree t //起别名 tree as t
11
-
要想获取第二高，需要排序，使用 order by（默认是升序 asc，即从小到大），若想降序则使用关键字 desc
-
去重，如果有多个相同的数据，使用关键字 distinct 去重
-
判断临界输出，如果不存在第二高的薪水，查询应返回 null，使用 ifNull（查询，null）方法
-
起别名，使用关键字 as …
-
因为去了重，又按顺序排序，使用 limit（）方法，查询第二大的数据，即第二高的薪水，即 limit(1,1) （因为默认从0开始，所以第一个1是查询第二大的数，第二个1是表示往后显示多少条数据，这里只需要一条）
# Write your MySQL query statement below
select ifNull((
select distinct Salary
from Employee
order by Salary desc limit 1,1),null) as SecondHighestSalary
```mysql
select FirstName, LastName, City, State
from Person left join Address
on Person.PersonId = Address.PersonId
;
11
# Write your MySQL query statement below
SELECT customer_id, COUNT(customer_id) count_no_trans #count是计算出现的次数，别名是count_no_trans（题目要求返回的）
FROM visits v
LEFT JOIN
transactions t ON v.visit_id = t.visit_id #因为起了别名可以用别名直接调用,连接的条件是主键id相等
WHERE amount IS NULL
GROUP BY customer_id; #按照id升序排序
