sql的练习题
表名和字段
–1.学生表
Student(s_id,s_name,s_birth,s_sex) --学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t_id,t_name) --教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) --学生编号,课程编号,分数
测试数据
--建表
--学生表
CREATE TABLE `Student`(
`s_id` VARCHAR(20),
`s_name` VARCHAR(20) NOT NULL DEFAULT '',
`s_birth` VARCHAR(20) NOT NULL DEFAULT '',
`s_sex` VARCHAR(10) NOT NULL DEFAULT '',
PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);
练习题和sql语句
-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
select a.* ,b.s_score as 01_score,c.s_score as 02_score from
student a
join score b on a.s_id=b.s_id and b.c_id='01'
left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score
--也可以这样写
select a.*,b.s_score as 01_score,c.s_score as 02_score from student a,score b,score c
where a.s_id=b.s_id
and a.s_id=c.s_id
and b.c_id='01'
and c.c_id='02'
and b.s_score>c.s_score
-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
select a.* ,b.s_score as 01_score,c.s_score as 02_score from
student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL
join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score
-- 3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from
student b
join score a on b.s_id = a.s_id
GROUP BY b.s_id,b.s_name HAVING avg_score >=60;
-- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
-- (包括有成绩的和无成绩的)
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from
student b
left join score a on b.s_id = a.s_id
GROUP BY b.s_id,b.s_name HAVING avg_score <60
union
select a.s_id,a.s_name,0 as avg_score from
student a
where a.s_id not in (
select distinct s_id from score);
-- 5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from
student a
left join score b on a.s_id=b.s_id
GROUP BY a.s_id,a.s_name;
-- 6、查询"李"姓老师的数量
select count(t_id) from teacher where t_name like '李%';
-- 7、查询学过"张三"老师授课的同学的信息
select a.* from
student a
join score b on a.s_id=b.s_id where b.c_id in(
select c_id from course where t_id =(
select t_id from teacher where t_name = '张三'));
-- 8、查询没学过"张三"老师授课的同学的信息
select * from
student c
where c.s_id not in(
select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in(
select a.c_id from course a join teacher b on a.t_id = b.t_id where t_name ='张三'));
-- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息
select a.* from
student a,score b,score c
where a.s_id = b.s_id and a.s_id = c.s_id and b.c_id='01' and c.c_id='02';
-- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
select a.* from
student a
where a.s_id in (select s_id from score where c_id='01' ) and a.s_id not in(select s_id from score where c_id='02')
-- 11、查询没有学全所有课程的同学的信息
--@wendiepei的写法
select s.* from student s
left join Score s1 on s1.s_id=s.s_id
group by s.s_id having count(s1.c_id)<(select count(*) from course)
--@k1051785839的写法
select *
from student
where s_id not in(
select s_id from score t1
group by s_id having count(*) =(select count(distinct c_id) from course))
-- 12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息
select * from student where s_id in(
select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01')
);
-- 13、查询和"01"号的同学学习的课程完全相同的其他同学的信息
--@ouyang_1993的写法
SELECT
Student.*
FROM
Student
WHERE
s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (
#下面的语句是找到'01'同学学习的课程数
SELECT COUNT(c_id) FROM Score WHERE s_id = '01'
)
)
AND s_id NOT IN (
#下面的语句是找到学过‘01’同学没学过的课程,有哪些同学。并排除他们
SELECT s_id FROM Score
WHERE c_id IN(
#下面的语句是找到‘01’同学没学过的课程
SELECT DISTINCT c_id FROM Score
WHERE c_id NOT IN (
#下面的语句是找出‘01’同学学习的课程
SELECT c_id FROM Score WHERE s_id = '01'
)
) GROUP BY s_id
) #下面的条件是排除01同学
AND s_id NOT IN ('01')
--@k1051785839的写法
SELECT
t3.*
FROM
(
SELECT
s_id,
group_concat(c_id ORDER BY c_id) group1
FROM
score
WHERE
s_id <> '01'
GROUP BY
s_id
) t1
INNER JOIN (
SELECT
group_concat(c_id ORDER BY c_id) group2
FROM
score
WHERE
s_id = '01'
GROUP BY
s_id
) t2 ON t1.group1 = t2.group2
INNER JOIN student t3 ON t1.s_id = t3.s_id
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select a.s_name from student a where a.s_id not in (
select s_id from score where c_id =
(select c_id from course where t_id =(
select t_id from teacher where t_name = '张三')));
-- 15、查询两门及其以上不及格课程的同学的学号,姓名及其平均成绩
select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from
student a
left join score b on a.s_id = b.s_id
where a.s_id in(
select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2)
GROUP BY a.s_id,a.s_name
-- 16、检索"01"课程分数小于60,按分数降序排列的学生信息
select a.*,b.c_id,b.s_score from
student a,score b
where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC;
-- 17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文,
(select s_score from score where s_id=a.s_id and c_id='02') as 数学,
(select s_score from score where s_id=a.s_id and c_id='03') as 英语,
round(avg(s_score),2) as 平均分 from score a GROUP BY a.s_id ORDER BY 平均分 DESC;
--@喝完这杯还有一箱的写法
SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文,
MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学,
MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语,
avg(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 DESC
-- 18.查询各科成绩最高分、最低分和平均分:以如下形式显示:课程ID,课程name,最高分,最低分,平均分,及格率,中等率,优良率,优秀率
--及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name
-- 19、按各科成绩进行排序,并显示排名
-- mysql没有rank函数
select a.s_id,a.c_id,
@i:=@i +1 as i保留排名,
@k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
@score:=a.s_score as score
from (
select s_id,c_id,s_score from score GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
--@k1051785839的写法
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='01') rank
FROM score t1 where t1.c_id='01'
order by t1.s_score desc) t1)
union
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='02') rank
FROM score t1 where t1.c_id='02'
order by t1.s_score desc) t2)
union
(select * from (select
t1.c_id,
t1.s_score,
(select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='03') rank
FROM score t1 where t1.c_id='03'
order by t1.s_score desc) t3)
-- 20、查询学生的总成绩并进行排名
select a.s_id,
@i:=@i+1 as i,
@k:=(case when @score=a.sum_score then @k else @i end) as rank,
@score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
(select @k:=0,@i:=0,@score:=0)s
-- 21、查询不同老师所教不同课程平均分从高到低显示
select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
left join score b on a.c_id=b.c_id
left join teacher c on a.t_id=c.t_id
GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;
-- 22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'
ORDER BY a.s_score DESC
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'
ORDER BY a.s_score DESC
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3
UNION
select d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03'
ORDER BY a.s_score DESC
)c
left join student d on c.s_id=d.s_id
where 排名 BETWEEN 2 AND 3;
-- 23、统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比
select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a
left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)b on a.c_id=b.c_id
left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)c on a.c_id=c.c_id
left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)d on a.c_id=d.c_id
left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)e on a.c_id=e.c_id
left join course f on a.c_id = f.c_id
-- 24、查询学生平均成绩及其名次
select a.s_id,
@i:=@i+1 as '不保留空缺排名',
@k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
@avg_score:=avg_s as '平均分'
from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;
-- 25、查询各科成绩前三名的记录
-- 1.选出b表比a表成绩大的所有组
-- 2.选出比当前id成绩大的 小于三个的
select a.s_id,a.c_id,a.s_score from score a
left join score b on a.c_id = b.c_id and a.s_score<b.s_score
group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
ORDER BY a.c_id,a.s_score DESC
-- 26、查询每门课程被选修的学生数
select c_id,count(s_id) from score a GROUP BY c_id
-- 27、查询出只有两门课程的全部学生的学号和姓名
select s_id,s_name from student where s_id in(
select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);
-- 28、查询男生、女生人数
select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex
-- 29、查询名字中含有"风"字的学生信息
select * from student where s_name like '%风%';
-- 30、查询同名同性学生名单,并统计同名人数
select a.s_name,a.s_sex,count(*) from student a JOIN
student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex
-- 31、查询1990年出生的学生名单
select s_name from student where s_birth like '1990%'
-- 32、查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号升序排列
select c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC
-- 33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
select a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85
-- 34、查询课程名称为"数学",且分数低于60的学生姓名和分数
select a.s_name,b.s_score from score b join student a on a.s_id=b.s_id where b.c_id=(
select c_id from course where c_name ='数学') and b.s_score<60
-- 35、查询所有学生的课程及分数情况;
select a.s_id,a.s_name,
SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
SUM(b.s_score) as '总分'
from student a left join score b on a.s_id = b.s_id
left join course c on b.c_id = c.c_id
GROUP BY a.s_id,a.s_name
-- 36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数;
select a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id
left join student a on a.s_id=c.s_id where c.s_score>=70
-- 37、查询不及格的课程
select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id
where a.s_score<60
-- 38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名;
select a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id
where a.c_id = '01' and a.s_score>80
-- 39、求每门课程的学生人数
select count(*) from score GROUP BY c_id;
-- 40、查询选修"张三"老师所授课程的学生中,成绩最高的学生信息及其成绩
-- 查询老师id
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
-- 查询最高分(可能有相同分数)
select MAX(s_score) from score where c_id='02'
-- 查询信息
select a.*,b.s_score,b.c_id,c.c_name from student a
LEFT JOIN score b on a.s_id = b.s_id
LEFT JOIN course c on b.c_id=c.c_id
where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')
and b.s_score in (select MAX(s_score) from score where c_id='02')
-- 41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩
select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score
-- 42、查询每门功成绩最好的前两名
-- 牛逼的写法
select a.s_id,a.c_id,a.s_score from score a
where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id
-- 43、统计每门课程的学生选修人数(超过5人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,若人数相同,按课程号升序排列
select c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC
-- 44、检索至少选修两门课程的学生学号
select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2
-- 45、查询选修了全部课程的学生信息
select * from student where s_id in(
select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))
-- 46、查询各学生的年龄
-- 按照出生日期来算,当前月日 < 出生年月的月日则,年龄减一
select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') -
(case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age
from student;
-- 47、查询本周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
-- 48、查询下周过生日的学生
select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)
-- 49、查询本月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
-- 50、查询下月过生日的学生
select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)
sql的练习题的更多相关文章
- 珍藏的数据库SQL基础练习题答案
自己珍藏的数据库SQL基础练习题答案 一,基本表的定义与删除. 题1: 用SQL语句创建如下三张表:学生(Student),课程表(Course),和学生选课表(SC),这三张表的结构如表1-1到表1 ...
- Oracle SQL部分练习题
SQL练习题 注:查询列表不建议用 “*” 1.列出至少有一个雇员的所有部门: a. select * from dept where deptno in(select distinct ...
- SQL学习笔记之SQL查询练习题1
(网络搜集) 0x00 表名和字段 –1.学生表 Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别 –2.课程表 Course(c_id ...
- Mysql Sql 语句练习题 (50道)
MySql 语句练习50题 表名和字段 –1.学生表 Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别 –2.课程表 Course(c_ ...
- 数据库SQL语句练习题
一. 设有一数据库,包括四个表:学生表(Student).课程表(Course).成绩表(Score)以及教师信息表(Teacher).四个表的结构分别如表1-1的表(一)~表( ...
- SQL 经典练习题
create database 练习题gouse 练习题go create table Student( Sno char(3) primary key, Sname char(8) not null ...
- SQL查询 练习题
设有一数据库,包括四个表:学生表(Student).课程表(Course).成绩表(Score)以及教师信息表(Teacher).四个表的结构分别如表1-1的表(一)~表(四)所示,数据如表1-2的表 ...
- 20_学生选课数据库SQL语句练习题
一. 设有一数据库,包括四个表:学生表(Student).课程表(Course).成绩表(Score)以及教师信息表(Teacher).四个表的结构分别如表1-1的表(一)~表( ...
- SQL语句练习题【主供自己学习、记忆】
1.这是我在面试中遇到的一道sql题,没有答出来,o(╥﹏╥)o 这是我刚才在网上查找函数之后写的SQL语句,能得到这个结果.[谁有不同的方法,欢迎底下评论留言哈] select (DATENAME( ...
随机推荐
- python学习——函数返回值及递归
返回值 return语句是从python 函数返回一个值,在讲到定义函数的时候有讲过,每个函数都要有一个返回值.Python中的return语句有什么作用,今天小编就依目前所了解的讲解一下.pytho ...
- 关于guava实现线程池
private ListeningExecutorService executorService = MoreExecutors.listeningDecorator(Executors.newCac ...
- ibatis in语句参数传入方法
第一种:传入参数仅有数组 <select id="GetEmailList_Test" resultClass="EmailInfo_"& ...
- 关联规则之Aprior算法
关联规则挖掘在电商.零售.大气物理.生物医学已经有了广泛的应用,本篇文章将介绍一些基本知识和Aprori算法. 啤酒与尿布的故事已经成为了关联规则挖掘的经典案例,还有人专门出了一本书<啤酒与尿布 ...
- cannot be found on object of type xx.CacheExpressionRootObject
0 环境 系统环境:win10 编辑器:IDEA 1 前言->环境搭建 1-1 pom依赖 <?xml version="1.0" encoding="UTF ...
- Hadoop_在linux中安装hadopp出现的问题
问题 sudo: no valid sudoers sources found, quitting 网络解决方法 链接:sudo: no valid sudoers sources found, qu ...
- sklearn包源码分析(一)--neighbors
python如何查看内置函数的用法及其源码? 在anaconda的安装目录下,有一块会放着我们安装的所有包,在里面可以找到所有的包 找到scikit learn包,进入 这里面又有了多个子包,每个子包 ...
- C++中stoi函数
作用: 将 n 进制的字符串转化为十进制 头文件: #include <string> 用法: stoi(字符串,起始位置,n进制),将 n 进制的字符串转化为十进制 示例: stoi(s ...
- 深入JVM内核--GC算法和种类
GC的概念 Garbage Collection 垃圾收集 1960年 List 使用了GC Java中,GC的对象是堆空间和永久区 引用计数法 老牌垃圾回收算法 通过引用计算来回收垃圾 使用者 CO ...
- Qt LNK1158无法运行rc.exe解决办法
找出电脑上的rc.exe ,发现在C:\Program Files (x86)\Windows Kits\10\bin\10.0.xxxx.0\x86 路径下. 找出电脑上的rc.exe ,发现在C: ...