两种方式
1.between 在列值得某与某之间
select
device_id,
gender,
age,
university
from user_profile
WHERE
age between 20 and 23
2.用and的来连接条件范围
SELECT
device_id,
gender,
age
from user_profile
WHERE
age >= 20 and age<=23
SELECT
device_id,
gender,
age,
university
FROM
user_profile
WHERE
university NOT IN ('复旦大学')
解法一:
select device_id,gender,age,university
from user_profile
where age is not NULL;
解法二:
select device_id,gender,age,university
from user_profile
where age !='';
SELECT
device_id,
gender,
age,
university,
gpa
FROM
user_profile
WHERE
university NOT IN ('浙江大学')
SELECT
device_id,
gender,
age,
university,
gpa
FROM
user_profile
WHERE
university IN ('北京大学', '复旦大学', '山东大学')
一般形式为:
列名 [NOT ] LIKE
匹配串中可包含如下四种通配符:
_:匹配任意一个字符;
%:匹配0个或多个字符;
[ ]:匹配[ ]中的任意一个字符(若要比较的字符是连续的,则可以用连字符“-”表 达 );
[^ ]:不匹配[ ]中的任意一个字符。
# 1.查询姓张的学生
SELECT * FROM 学生表 WHERE 姓名 LIKE '张%'
# 2.查询姓张并且名字为3个字的
SELECT * FROM 学生表 WHERE 姓名 LIKE '张__'
# 3.如果把姓名列的类型改为nchar(20),在SQL Server 2012中执行没有结果。
# 原因是姓名列的类型是char(20),当姓名少于20个汉字时,系统在存储这些数据时自动在后边补空格,
# 空格作为一个字符,也参加LIKE的比较。可以用rtrim()去掉右空格。
# 4.查询学生表中姓‘张’、姓‘李’和姓‘刘’的学生的情况
SELECT * FROM 学生表 WHERE rtrim(姓名) LIKE '张__'
# 5.查询学生表中姓‘张’、姓‘李’和姓‘刘’的学生的情况。
SELECT * FROM 学生表 WHERE 姓名 LIKE '[张李刘]%'
# 6.查询学生表表中名字的第2个字为“小”或“大”的学生的姓名和学号。
SELECT 姓名,学号 FROM 学生表 WHERE 姓名 LIKE '_[小大]%'
# 7.查询学生表中所有不姓“刘”的学生
SELECT 姓名 FROM 学生 WHERE 姓名 NOT LIKE '刘%'
# 8.从学生表表中查询学号的最后一位不是2、3、5的学生信息。
SELECT * FROM 学生表 WHERE 学号 LIKE '%[^235]'
SELECT device_id,age,university FROM user_profile
WHERE university LIKE '%北京%'
# LIMIT 子句可以被用于强制 SELECT 语句返回指定的记录数。
# LIMIT 接受一个或两个数字参数。参数必须是一个整数常量。
# 如果只给定一个参数,它表示返回最大的记录行数目。
# 如果给定两个参数,第一个参数指定第一个返回记录行的偏移量,第二个参数指定返回记录行的最大数目。
# 为了检索从某一个偏移量到记录集的结束所有的记录行,可以指定第二个参数为 -1。
# 初始记录行的偏移量是 0(而不是 1)。
# 1.检索记录行 [6-7]
SELECT * FROM table LIMIT 5,2
# 2.检索前 5 个记录行[3-7]5条数据
SELECT * FROM user_profile LIMIT 5 OFFSET 2
# 3.检索记录行 11-last
SELECT * FROM table LIMIT 10,-1
# 4.检索前 5 个记录行
SELECT * FROM table LIMIT 5
# 去重两种思路
# 第一种:使用distinct(关键词distinct用于返回唯一不同的值)
select distinct university from user_profile;
# 第二种:使用group by按照university进行分组
select university from user_profile group by university;
# 方法1
select
max(gpa) as gpa
from
user_profile
where
university = '复旦大学';
# 方法2
select
gpa
from
user_profile
where
university = '复旦大学'
order by
gpa desc
limit
1
# 表头重命名,用as语法
# 浮点数的平均值可能小数点位数很多,按照示例保存一位小数,用round函数
select
count(gender) as male_num,
round(avg(gpa), 1) as avg_gpa
from user_profile where gender="male";
select
university,
avg(question_cnt) as avg_question_cnt,
avg(answer_cnt) as avg_answer_cnt
from
user_profile
group by
university
having
avg_question_cnt < 5 or avg_answer_cnt < 20
select
university,
avg(question_cnt) as avg_question_cnt
from
user_profile
group by
university
order by
avg_question_cnt
left join(左联接) ——返回包括左表中的所有记录和右表中联结字段相等的记录
right join(右联接)——返回包括右表中的所有记录和左表中联结字段相等的记录
inner join(等值连接)—— 只返回两个表中联结字段相等的行
# 1.子查询
select
device_id,
question_id,
result
from
question_practice_detail
where
device_id IN (
select
device_id
from
user_profile
where
university = '浙江大学'
)
# 2.多表连接
select
q.device_id,
question_id,
result
from
question_practice_detail q
inner join user_profile u on q.device_id = u.device_id
where
university = '浙江大学'
select
university,
count(question_id) / count(distinct qpd.device_id) as avg_answer_cnt
from
question_practice_detail as qpd
inner join user_profile as up on qpd.device_id = up.device_id
group by
university
select
university,
difficult_level,
round(
count(qpd.question_id) / count(distinct qpd.device_id),
4
) as avg_answer_cnt
from
question_practice_detail as qpd
left join user_profile as up on up.device_id = qpd.device_id
left join question_detail as qd on qd.question_id = qpd.question_id
group by
university,
difficult_level
# 第一段:select选择字段,含原生字段及构造字段,其中构造的过程中常用sum count round +-*/ 等函数及运算表达式
select
university,
difficult_level,
round(count(qpd.question_id) / count(distinct qpd.device_id),4) as avg_answer_cnt
# 第二段:from选择表格,含合并表格操作
from question_practice_detail as qpd
join user_profile as up
on up.device_id=qpd.device_id
join question_detail as qd
on qd.question_id=qpd.question_id
# 第三段:数据操作段,含 分组/排序/筛选/取指定行数等操作
group by university,difficult_level;
分别查看&结果不去重:所以直接使用两个条件的or是不行的,直接用union也不行,要用union all,分别去查满足条件1的和满足条件2的,然后合在一起不去重
# 结果不去重就用union all , 去重就用union
select
device_id,gender,age,gpa
from
user_profile
where
university = '山东大学'
union all
select
device_id,gender,age,gpa
from
user_profile
where
gender = 'male'
CASE函数
是一种多分支的函数,可以根据条件列表的值返回多个可能的结果表达式中的一个。
可用在任何允许使用表达式的地方,但不能单独作为一个语句执行。
分为:
简单CASE函数
搜索CASE函数
CASE 测试表达式
WHEN 简单表达式1 THEN 结果表达式1
WHEN 简单表达式2 THEN 结果表达式2 …
WHEN 简单表达式n THEN 结果表达式n
[ ELSE 结果表达式n+1 ]
END
#case函数
SELECT
学号,
课程号,
CASE
WHEN 成绩 >= 90 THEN '优'
WHEN 成绩 BETWEEN 80 AND 89 THEN '良'
WHEN 成绩 BETWEEN 70 AND 79 THEN '中'
WHEN 成绩 BETWEEN 60 AND 69 THEN '及格'
WHEN 成绩 < 60 THEN '不及格'
END 成绩
FROM
成绩表
WHERE
课程号 = 'M01F011'
SELECT
CASE
WHEN age < 25 OR age IS NULL THEN '25岁以下'
WHEN age >= 25 THEN '25岁及以上'
END age_cut,
COUNT(*) number
FROM
user_profile
GROUP BY
age_cut
#if()函数
SELECT
IF(age >= 25, "25岁及以上", "25岁以下") AS age_cut,
count(*) AS number
FROM
user_profile
GROUP BY
age_cut;
DAYOFWEEK(date)
返回日期date的星期索引(1=星期天,2=星期一, ……7=星期六)。这些索引值对应于ODBC标准。
select DAYOFWEEK('1998-02-03')
-> 3
WEEKDAY(date)
返回date的星期索引(0=星期一,1=星期二, ……6= 星期天)。
mysql> select WEEKDAY('1997-10-04 22:23:00');
-> 5
DAYOFMONTH(date)
返回date的月份中日期,在1到31范围内。
mysql> select DAYOFMONTH('1998-02-03');
-> 3
DAYOFYEAR(date)
返回date在一年中的日数, 在1到366范围内。
mysql> select DAYOFYEAR('1998-02-03');
-> 34
MONTH(date)
返回date的月份,范围1到12。
mysql> select MONTH('1998-02-03');
-> 2
DAYNAME(date)
返回date的星期名字。
mysql> select DAYNAME("1998-02-05");
-> 'Thursday'
MONTHNAME(date)
返回date的月份名字。
mysql> select MONTHNAME("1998-02-05");
-> 'February'
QUARTER(date)
返回date一年中的季度,范围1到4。
mysql> select QUARTER('98-04-01');
-> 2
WEEK(date)
对于星期天是一周的第一天的地方,有一个单个参数,返回date的周数,范围在0到52。
mysql> select WEEK('1998-02-20');
-> 7
WEEK(date,first)
2个参数形式WEEK()允许你指定星期是否开始于星期天或星期一。
如果第二个参数是0,星期从星期天开始,
如果第二个参数是1,从星期一开始。
mysql> select WEEK('1998-02-20',0);
-> 7
mysql> select WEEK('1998-02-20',1);
-> 8
YEAR(date)
返回date的年份,范围在1000到9999。
mysql> select YEAR('98-02-03');
-> 1998
HOUR(time)
返回time的小时,范围是0到23。
mysql> select HOUR('10:05:03');
-> 10
MINUTE(time)
返回time的分钟,范围是0到59。
mysql> select MINUTE('98-02-03 10:05:03');
-> 5
SECOND(time)
回来time的秒数,范围是0到59。
mysql> select SECOND('10:05:03');
-> 3
PERIOD_ADD(P,N)
增加N个月到阶段P(以格式YYMM或YYYYMM)。以格式YYYYMM返回值。注意阶段参数P不是日期值。
mysql> select PERIOD_ADD(9801,2);
-> 199803
PERIOD_DIFF(P1,P2)
返回在时期P1和P2之间月数,P1和P2应该以格式YYMM或YYYYMM。注意,时期参数P1和P2不是日期值。
mysql> select PERIOD_DIFF(9802,199703);
-> 11
select
day(date) as day,
count(question_id) as question_cnt
from
question_practice_detail
where
month(date) = 8
and year(date) = 2021
group by
date
<窗口函数> OVER ([PARTITION BY <用于分组的列>] ORDER BY <用于排序的列>)
lag和lead分别是向前向后的意思
参数有三个。expression:列名;offset:偏移量;default_value:超出记录窗口的默认值(默认为null,可以设置为0)
#lag和lead分别是向前向后的意思
#参数有三个。
#expression:列名;offset:偏移量;default_value:超出记录窗口的默认值(默认为null,可以设置为0)
select * from weather
#将温度向后退一行
SELECT
id,
date,
temperature,
LAG( temperature, 1, 0 ) OVER ( ORDER BY date ) AS temp
FROM
weather;
#查询出当天比前天温度高的id
SELECT
id
FROM
(SELECT id, date, temperature, LAG( temperature, 1, 0) OVER ( ORDER BY date ) AS temp FROM weather) as tmp
WHERE
temperature > temp
AND temp != 0;
#将温度向前移一行
SELECT
id,
date,
temperature,
lead( temperature, 1, 0 ) OVER ( ORDER BY date ) AS temp
FROM
weather;
#查询出当天比后一天温度高的id
SELECT
id
FROM
(SELECT id, date, temperature, lead( temperature, 1, 0) OVER ( ORDER BY date ) AS temp FROM weather) as tmp
WHERE
temperature > temp
AND temp != 0;
select count(date2) / count(date1) as avg_ret
from (
select
distinct qpd.device_id,
qpd.date as date1,
uniq_id_date.date as date2
from question_practice_detail as qpd
left join(
select distinct device_id, date
from question_practice_detail
) as uniq_id_date
on qpd.device_id=uniq_id_date.device_id
and date_add(qpd.date, interval 1 day)=uniq_id_date.date
) as id_last_next_date
select avg(if(datediff(date2, date1)=1, 1, 0)) as avg_ret
from (
select
distinct device_id,
date as date1,
lead(date) over (partition by device_id order by date) as date2
from (
select distinct device_id, date
from question_practice_detail
) as uniq_id_date
) as id_last_next_date
group by是分组函数,partition by是分析函数(然后像sum()等是聚合函数);
在执行顺序上,对于group by而言
from > where > group by > having > order by
对于partition by而言,在执行完select之后,在所得结果集之上进行partition。
在group by后的结果集上使用聚合函数,会作用在分组下的所有记录上。而如果在partition结果上聚合,千万注意聚合函数是逐条累计运行结果的!
group by将分组记录汇总成一条记录,具有去重效果
partition by显示所有数据,不会去重
#group by去重
SELECT
id,
date,
sum( temperature )
FROM
weather
GROUP BY
temperature
#partition by不去重,组内排序PARTITION BY temperature ORDER BY date
SELECT
id,
date,
temperature,
sum( temperature ) over ( PARTITION BY temperature ORDER BY date )
FROM
weather
select count(date2) / count(date1) as avg_ret
from (
select
distinct qpd.device_id,
qpd.date as date1,
uniq_id_date.date as date2
from question_practice_detail as qpd
left join(
select distinct device_id, date
from question_practice_detail
) as uniq_id_date
on qpd.device_id=uniq_id_date.device_id
and date_add(qpd.date, interval 1 day)=uniq_id_date.date
) as id_last_next_date
select avg(if(datediff(date2, date1)=1, 1, 0)) as avg_ret
from (
select
distinct device_id,
date as date1,
lead(date) over (partition by device_id order by date) as date2
from (
select distinct device_id, date
from question_practice_detail
) as uniq_id_date
) as id_last_next_date
// 得到当前时间增加1个小时的结果
select date_add(now(),interval 1 hour)
// 得到当前时间增加1天的结果
select date_add(now(),interval 1 day)
// 得到当前时间减少72个小时的结果
select date_add(now(),interval -72 hour)
// 还有一些更加高级的用法,比如增加一天一小时一分一秒
// 但是如果你不太喜欢这种用法,你仍然可以使用等同于该数值的second来进行加减
date_add(now(),interval '1 1:1:1' day_second)
select DATEDIFF(year,'2010-12-31','2019-01-01')
select DATEDIFF(Month,'2018-01-01','2019-01-01')
select DATEDIFF(Week,'2018-12-01','2018-12-31')
select DATEDIFF(day,'2018-12-01','2018-12-31')
select count(date2) / count(date1) as avg_ret
from (
select
distinct qpd.device_id,
qpd.date as date1,
uniq_id_date.date as date2
from question_practice_detail as qpd
left join(
select distinct device_id, date
from question_practice_detail
) as uniq_id_date
on qpd.device_id=uniq_id_date.device_id
and date_add(qpd.date, interval 1 day)=uniq_id_date.date
) as id_last_next_date
select avg(if(datediff(date2, date1)=1, 1, 0)) as avg_ret
from (
select
distinct device_id,
date as date1,
lead(date) over (partition by device_id order by date) as date2
from (
select distinct device_id, date
from question_practice_detail
) as uniq_id_date
) as id_last_next_date