2019-10-17

50道SQL练习题

来自https://www.jianshu.com/p/476b52ee4f1b

sql语句的执行顺序

（9）Select
（10）distinct 字段名1,字段名2，
（7）[fun(字段名)]
（1）from 表1
（3）<join类型>join 表2
（2）on <join条件>
（4）where <where条件>
（5）group by <字段>
（6）with <cube|rollup>
（8）having <having条件>
（11）order by <排序字段>
（12）limit <起始偏移量,行数>

学生表 Student

create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));
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' , '1990-12-20' , '男');
insert into Student values('04' , '李云' , '1990-12-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-01-01' , '女');
insert into Student values('07' , '郑竹' , '1989-01-01' , '女');
insert into Student values('09' , '张三' , '2017-12-20' , '女');
insert into Student values('10' , '李四' , '2017-12-25' , '女');
insert into Student values('11' , '李四' , '2012-06-06' , '女');
insert into Student values('12' , '赵六' , '2013-06-13' , '女');
insert into Student values('13' , '孙七' , '2014-06-01' , '女');

科目表 Course

create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));
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 Teacher

create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

成绩表 SC

create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);
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
    * 
FROM
    Student s 
WHERE
    s.SId IN (
    SELECT
        sc1.SId 
    FROM
        SC sc1
        JOIN SC sc2 ON sc1.SId = sc2.SId 
        AND sc1.CId = '01' 
        AND sc2.CId = '02' 
    AND sc1.score > sc2.score 
    );
2.查询同时存在" 01 "课程和" 02 "课程的学生情况
SELECT
    * 
FROM
    Student s 
WHERE
    s.SId IN (
    SELECT
        sc1.SId 
    FROM
        SC sc1
        JOIN SC sc2 ON sc1.CId = '01' 
        AND sc2.CId = '02' 
    AND sc1.SId = sc2.SId 
    );
3.查询存在" 01 "课程但不存在" 02 "课程的学生情况(不存在时显示为 null )
SELECT
    * 
FROM
    Student s 
WHERE
    s.SId IN (
    SELECT
        sc1.SId 
    FROM
        ( SELECT * FROM SC WHERE SC.CId = '01' ) AS sc1
        LEFT JOIN ( SELECT * FROM SC WHERE SC.CId = '02' ) AS sc2 ON sc1.SId = sc2.SId 
    WHERE
    sc2.SId IS NULL 
    );
4.查询不存在" 01 "课程但存在" 02 "课程的学生情况
SELECT
    s.* 
FROM
    Student s,
    (
    SELECT
        sc2.SId 
    FROM
        ( SELECT * FROM SC WHERE SC.CId = '01' ) AS sc1
        RIGHT JOIN ( SELECT * FROM SC WHERE SC.CId = '02' ) AS sc2 ON sc1.SId = sc2.SId 
    WHERE
        sc1.SId IS NULL 
    ) AS sc2id 
WHERE
    s.SId = sc2id.SId;
5.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
    s.SId,
    s.Sname,
    avgs1.as1 
FROM
    Student s,
    ( SELECT SId, AVG( score ) AS as1 FROM SC GROUP BY SId HAVING AVG( score ) > 60 ) AS avgs1 
WHERE
    s.SId = avgs1.SId;
6.查询在 SC 表存在成绩的学生信息
SELECT DISTINCT
    s.* 
FROM
    Student s,
    SC 
WHERE
    s.SId = SC.SId;
7.查询所有同学的学生编号、学生姓名、选课总数、所有课程的成绩总和
SELECT
    s.SId,
    s.Sname,
    sccs.scc,
    sccs.scs 
FROM
    Student s
    LEFT JOIN (
    SELECT
        SC.SId,
        COUNT( SC.CId ) AS scc,
        SUM( SC.score ) AS scs 
    FROM
        SC 
    GROUP BY
    SC.SId 
    ) AS sccs ON s.SId = sccs.SId;
8.查有成绩的学生信息
SELECT
    * 
FROM
    Student s 
WHERE
    EXISTS ( SELECT SC.SId FROM SC WHERE SC.SId = s.SId );
9.查询「李」姓老师的数量
SELECT
    COUNT( * ) 
FROM
    Teacher 
WHERE
    Tname LIKE '李%';
10.查询学过「张三」老师授课的同学的信息
SELECT
    s.* 
FROM
    Teacher t,
    Course c,
    SC,
    Student s 
WHERE
    t.Tname = '张三' 
    AND t.TId = c.TId 
    AND SC.CId = c.CId 
    AND s.SId = SC.SId;
11.查询没有学全所有课程的同学的信息
SELECT
    * 
FROM
    Student s 
WHERE
    s.SId NOT IN (
    SELECT
        SC.SId 
    FROM
        SC 
    GROUP BY
        SC.SId 
    HAVING
    COUNT( SC.CId ) = ( SELECT COUNT( c.CId ) FROM Course c ) 
    );
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 DISTINCT
    s.* 
FROM
    ( SELECT * FROM SC WHERE SC.SId = '01' ) AS sc1
    JOIN ( SELECT * FROM SC WHERE SC.SId != '01' ) AS sc2
    JOIN Student s 
WHERE
    sc1.CId = sc2.CId 
    AND s.SId = sc2.SId;
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
SELECT
    SId 
FROM
    SC 
WHERE
    CId IN ( SELECT SC.CId FROM SC WHERE SC.SId = '01' ) 
GROUP BY
    SC.SId 
HAVING
    COUNT( * ) = ( SELECT COUNT( * ) FROM SC WHERE SC.SId = '01' ) 
    AND SC.SId != '01';

但是如果SId='02'增加一条CId='04',按照上面的语句也是符合条件。因为第三行的语句把这种CId='04'给过滤了。

SELECT
    * 
FROM
    Student 
WHERE
    SId IN (
    SELECT
        SId 
    FROM
        SC 
    GROUP BY
        SId 
    HAVING
        GROUP_CONCAT( CId ) = ( SELECT GROUP_CONCAT( CId ) FROM SC WHERE SId = '01' ) 
    AND SId != '01' 
    );
14.查询没学过"张三"老师讲授的任一门课程的学生姓名
SELECT
    s.SName 
FROM
    Student s 
WHERE
    s.SId NOT IN (
    SELECT
        SC.SId 
    FROM
        Teacher t,
        Course c,
        SC 
    WHERE
        t.Tname = '张三' 
        AND t.TId = c.TId 
    AND SC.CId = c.CId 
    );
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
    s.SId,
    s.Sname,
    AVG( SC.score ) 
FROM
    SC,
    Student s 
WHERE
    s.SId = SC.SId 
    AND SC.score < 60 GROUP BY SC.SId HAVING COUNT( * ) > 1;
16.检索" 01 "课程分数小于 60，按分数降序排列的学生信息
SELECT
    s.*,
    SC.score 
FROM
    SC,
    Student s 
WHERE
    SC.CId = '01' 
    AND SC.score < 60 
    AND SC.SId = s.SId 
ORDER BY
    SC.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
    SC.*,
    sc1.avgs 
FROM
    SC
    LEFT JOIN ( SELECT SId, AVG( SC.score ) AS avgs FROM SC GROUP BY SId ) AS sc1 ON SC.SId = sc1.SId 
ORDER BY
    sc1.avgs 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
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
    SC.CId,
    c.Cname,
    MAX( SC.score ) AS tops,
    MIN( SC.score ) AS lows,
    AVG( SC.score ) AS avgs,
    COUNT( * ) AS nump,
    SUM( CASE WHEN SC.score >= 60 THEN 1 ELSE 0 END ) / COUNT( * ) AS pasr,
    SUM( CASE WHEN SC.score >= 70 AND SC.score < 80 THEN 1 ELSE 0 END ) / COUNT( * ) AS medr,
    SUM( CASE WHEN SC.score >= 90 THEN 1 ELSE 0 END ) / COUNT( * ) AS excr 
FROM
    SC,
    Course c 
WHERE
    SC.CId = c.CId 
GROUP BY
    SC.CId 
ORDER BY
    COUNT( * ) DESC,
    SC.CId ASC;
19.按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺
SELECT
    sc1.CId,
    sc1.SId,
    COUNT( sc2.score ) + 1 rank 
FROM
    SC sc1
    LEFT JOIN SC sc2 ON sc1.CId = sc2.CId 
    AND sc1.score < sc2.score 
GROUP BY
    sc1.CId,
    sc1.SId 
ORDER BY
    sc1.CId,
    rank;
查询学生的总成绩，并进行排名，总分重复时不保留名次空缺
SET @rank := 0;
SELECT
    sc.SId,
    total,
    @rank := @rank + 1 AS Rank 
FROM
    ( SELECT SC.SId, SUM( SC.score ) AS total FROM SC GROUP BY SC.SId ORDER BY total DESC ) sc;
20.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]，[60-0] 及所占百分比
SELECT
    Course.CId,
    Course.Cname,
    p1.`[100-85]`,
    p1.`[100-85]%`,
    p2.`[85-70]`,
    p2.`[85-70]%`,
    p3.`[70-60]`,
    p3.`[70-60]%`,
    p4.`[60-0]`,
    p4.`[60-0]%` 
FROM
    Course
    LEFT JOIN (
    SELECT
        CId,
        SUM( CASE WHEN SC.score <= 100 AND SC.score > 85 THEN 1 ELSE 0 END ) AS `[100-85]`,
        ROUND(
            (
                SUM( CASE WHEN SC.score <= 100 AND SC.score > 85 THEN 1 ELSE 0 END ) * 100 / COUNT( * ) 
            ),
            2 
        ) `[100-85]%` 
    FROM
        SC 
    GROUP BY
        SC.CId 
    ) p1 ON Course.CId = p1.CId
    LEFT JOIN (
    SELECT
        CId,
        SUM( CASE WHEN SC.score <= 85 AND SC.score > 70 THEN 1 ELSE 0 END ) AS `[85-70]`,
        ROUND(
            (
                SUM( CASE WHEN SC.score <= 85 AND SC.score > 70 THEN 1 ELSE 0 END ) * 100 / COUNT( * ) 
            ),
            2 
        ) AS `[85-70]%` 
    FROM
        SC 
    GROUP BY
        SC.CId 
    ) AS p2 ON Course.CId = p2.CId
    LEFT JOIN (
    SELECT
        CId,
        SUM( CASE WHEN SC.score <= 70 AND SC.score > 60 THEN 1 ELSE 0 END ) AS `[70-60]`,
        ROUND(
            (
                SUM( CASE WHEN SC.score <= 70 AND SC.score > 60 THEN 1 ELSE 0 END ) * 100 / COUNT( * ) 
            ),
            2 
        ) AS `[70-60]%` 
    FROM
        SC 
    GROUP BY
        SC.CId 
    ) AS p3 ON Course.CId = p3.CId
    LEFT JOIN (
    SELECT
        CId,
        SUM( CASE WHEN SC.score <= 60 AND SC.score > 0 THEN 1 ELSE 0 END ) AS `[60-0]`,
        ROUND(
            (
                SUM( CASE WHEN SC.score <= 60 AND SC.score > 0 THEN 1 ELSE 0 END ) * 100 / COUNT( * ) 
            ),
            2 
        ) AS `[60-0]%` 
    FROM
        SC 
    GROUP BY
    SC.CId 
    ) AS p4 ON Course.CId = p4.CId;
21.查询各科成绩前三名的记录
SELECT
    sc1.* 
FROM
    SC sc1
    LEFT JOIN SC sc2 ON sc1.CId = sc2.CId 
    AND sc1.score < sc2.score 
GROUP BY
    sc1.SId,
    sc1.CId 
HAVING
    COUNT( sc2.CId ) < 3 
ORDER BY
    sc1.CId;
22.查询每门课程被选修的学生数
SELECT
    CId,
    COUNT( SId ) 
FROM
    SC 
GROUP BY
    CId;
23.查询出只选修两门课程的学生学号和姓名
SELECT
    Student.SId,
    Student.Sname 
FROM
    SC,
    Student 
WHERE
    SC.SId = Student.SId 
GROUP BY
    SC.SId 
HAVING
    COUNT( * ) = 2;
24.查询男生、女生人数
SELECT
    Ssex,
    COUNT( * ) 
FROM
    Student 
GROUP BY
    Ssex;
25.查询名字中含有「风」字的学生信息
SELECT
    * 
FROM
    Student 
WHERE
    Sname LIKE '%风%';
26.查询同名学生名单，并统计同名人数
SELECT
    Sname,
    COUNT( * ) 
FROM
    Student 
GROUP BY
    Sname 
HAVING
    COUNT( * ) > 1;
27.查询 1990 年出生的学生名单
SELECT
    * 
FROM
    Student 
WHERE
    YEAR ( Sage ) = 1990;
28.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
    SC.CId,
    Course.Cname,
    AVG( SC.score ) AS average 
FROM
    SC,
    Course 
WHERE
    SC.CId = Course.CId 
GROUP BY
    SC.CId 
ORDER BY
    average DESC,
    CId;
29.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
    Student.SId,
    Student.Sname,
    AVG( SC.score ) AS average 
FROM
    Student,
    SC 
WHERE
    Student.SId = SC.SId 
GROUP BY
    SC.SId 
HAVING
    average >= 85;
30.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数
SELECT
    Student.Sname,
    SC.score 
FROM
    Course,
    SC,
    Student 
WHERE
    Course.Cname = "数学" 
    AND Course.CId = SC.CId 
    AND SC.score < 60 
    AND Student.SId = SC.SId;
31.查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）
SELECT
    Sname,
    CId,
    score 
FROM
    Student
    LEFT JOIN SC USING ( SId );
32.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
    Student.Sname,
    Course.Cname,
    SC.score 
FROM
    SC,
    Student,
    Course 
WHERE
    SC.score > 70 
    AND Student.SId = SC.SId 
    AND SC.CId = Course.CId;
33.查询存在不及格的课程
SELECT DISTINCT
    SC.CId 
FROM
    SC 
WHERE
    SC.score < 60;
34.查询课程编号为 01 且课程成绩在 80 分及以上的学生的学号和姓名
SELECT
    Student.SId,
    Student.Sname 
FROM
    Student,
    SC 
WHERE
    Student.SId = SC.SId 
    AND CId = "01" 
    AND score >= 80;
35.求每门课程的学生人数
SELECT
    SC.CId,
    COUNT( * ) AS sn 
FROM
    SC 
GROUP BY
    CId;
36.成绩不重复，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
SELECT
    Student.*,
    SC.CId,
    SC.score 
FROM
    Teacher,
    Course,
    SC,
    Student 
WHERE
    Teacher.Tname = "张三" 
    AND Teacher.TId = Course.TId 
    AND Course.CId = SC.CId 
    AND SC.SId = Student.SId 
ORDER BY
    SC.score DESC 
    LIMIT 1;
37.成绩有重复的情况下，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
SELECT Student.*,SC.CId,SC.score 
    FROM Teacher,Course,SC,Student
    WHERE Teacher.Tname="张三"
    AND Teacher.TId=Course.TId
    AND Course.CId=SC.CId 
    AND SC.SId=Student.SId 
    AND SC.score=
    (SELECT MAX(SC.score)
    FROM Teacher,Course,SC,Student
    WHERE Teacher.Tname="张三"
    AND Teacher.TId=Course.TId
    AND Course.CId=SC.CId 
    AND SC.SId=Student.SId );
38.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
    sc1.* 
FROM
    SC sc1
    JOIN SC sc2 USING ( SId ) 
WHERE
    sc1.CId != sc2.CId 
    AND sc1.score = sc2.score 
GROUP BY
    sc1.CId;
39.查询每门课程成绩最好的前两名
SELECT
    sc1.* 
FROM
    SC sc1
    LEFT JOIN SC sc2 ON sc1.CId = sc2.CId 
    AND sc1.score < sc2.score 
GROUP BY
    sc1.SId,
    sc1.CId 
HAVING
    COUNT( sc2.CId ) < 2 
ORDER BY
    sc1.CId;
40.统计每门课程的学生选修人数（超过 5 人的课程才统计）
SELECT
    SC.CId,
    COUNT( SId ) AS cc 
FROM
    SC 
GROUP BY
    CId 
HAVING
    cc > 5;
41.检索至少选修两门课程的学生学号
SELECT
    SC.SId,
    COUNT( SId ) AS cc 
FROM
    SC 
GROUP BY
    SId 
HAVING
    cc >= 2;
42.查询选修了全部课程的学生信息
SELECT
    Student.* 
FROM
    SC,
    Student 
WHERE
    SC.SId = Student.SId 
GROUP BY
    SC.SId 
HAVING
    COUNT( * ) = ( SELECT COUNT( * ) FROM Course );
43.查询各学生的年龄，只按年份来算
SELECT
    SId,
    Sname,
    YEAR ( CURDATE( ) ) - YEAR ( Sage ) AS age 
FROM
    Student;
44.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
    SId,
    Sname,
    TIMESTAMPDIFF( YEAR, Sage, CURDATE( ) ) AS age 
FROM
    Student;
45.查询本周过生日的学生
SELECT
    * 
FROM
    Student 
WHERE
    WEEKOFYEAR( Sage ) = WEEKOFYEAR( CURDATE( ) );
查询下周过生日的学生
SELECT
    * 
FROM
    Student 
WHERE
    WEEKOFYEAR( Sage ) = WEEKOFYEAR( CURDATE( ) ) + 1;
46.查询本月过生日的学生
SELECT
    * 
FROM
    Student 
WHERE
    MONTH ( Sage ) = MONTH ( CURDATE( ) );
47.查询下月过生日的学生
SELECT
    * 
FROM
    Student 
WHERE
    MONTH ( Sage ) = MONTH ( CURDATE( ) ) + 1;
©著作权归作者所有,转载或内容合作请联系作者
  • 人面猴
    序言：七十年代末，一起剥皮案震惊了整个滨河市，随后出现的几起案子，更是在滨河造成了极大的恐慌，老刑警刘岩，带你破解...
    沈念sama阅读 214,658评论 6赞 496
  • 死咒
    序言：滨河连续发生了三起死亡事件，死亡现场离奇诡异，居然都是意外死亡，警方通过查阅死者的电脑和手机，发现死者居然都...
    沈念sama阅读 91,482评论 3赞 389
  • 救了他两次的神仙让他今天三更去死
    文/潘晓璐 我一进店门，熙熙楼的掌柜王于贵愁眉苦脸地迎上来，“玉大人，你说我怎么就摊上这事。” “怎么了？”我有些...
    开封第一讲书人阅读 160,213评论 0赞 350
  • 道士缉凶录：失踪的卖姜人
    文/不坏的土叔 我叫张陵，是天一观的道长。 经常有香客问我，道长，这世上最难降的妖魔是什么？ 我笑而不...
    开封第一讲书人阅读 57,395评论 1赞 288
  • 港岛之恋（遗憾婚礼）
    正文 为了忘掉前任，我火速办了婚礼，结果婚礼上，老公的妹妹穿的比我还像新娘。我一直安慰自己，他们只是感情好，可当我...
    茶点故事阅读 66,487评论 6赞 386
  • 恶毒庶女顶嫁案：这布局不是一般人想出来的
    文/花漫 我一把揭开白布。 她就那样静静地躺着，像睡着了一般。 火红的嫁衣衬着肌肤如雪。 梳的纹丝不乱的头发上，一...
    开封第一讲书人阅读 50,523评论 1赞 293
  • 城市分裂传说
    那天，我揣着相机与录音，去河边找鬼。 笑死，一个胖子当着我的面吹牛，可吹牛的内容都是我干的。 我是一名探鬼主播，决...
    沈念sama阅读 39,525评论 3赞 414
  • 双鸳鸯连环套：你想象不到人心有多黑
    文/苍兰香墨 我猛地睁开眼，长吁一口气：“原来是场噩梦啊……” “哼！你这毒妇竟也来了？” 一声冷哼从身侧响起，我...
    开封第一讲书人阅读 38,300评论 0赞 270
  • 万荣杀人案实录
    序言：老挝万荣一对情侣失踪，失踪者是张志新（化名）和其女友刘颖，没想到半个月后，有当地人在树林里发现了一具尸体，经...
    沈念sama阅读 44,753评论 1赞 307
  • 护林员之死
    正文 独居荒郊野岭守林人离奇死亡，尸身上长有42处带血的脓包…… 初始之章·张勋 以下内容为张勋视角 年9月15日...
    茶点故事阅读 37,048评论 2赞 330
  • 白月光启示录
    正文 我和宋清朗相恋三年，在试婚纱的时候发现自己被绿了。 大学时的朋友给我发了我未婚夫和他白月光在一起吃饭的照片。...
    茶点故事阅读 39,223评论 1赞 343
  • 活死人
    序言：一个原本活蹦乱跳的男人离奇死亡，死状恐怖，灵堂内的尸体忽然破棺而出，到底是诈尸还是另有隐情，我是刑警宁泽，带...
    沈念sama阅读 34,905评论 5赞 338
  • 日本核电站爆炸内幕
    正文 年R本政府宣布，位于F岛的核电站，受9级特大地震影响，放射性物质发生泄漏。R本人自食恶果不足惜，却给世界环境...
    茶点故事阅读 40,541评论 3赞 322
  • 男人毒药：我在死后第九天来索命
    文/蒙蒙 一、第九天 我趴在偏房一处隐蔽的房顶上张望。 院中可真热闹，春花似锦、人声如沸。这庄子的主人今日做“春日...
    开封第一讲书人阅读 31,168评论 0赞 21
  • 一桩弑父案，背后竟有这般阴谋
    文/苍兰香墨 我抬头看了看天上的太阳。三九已至，却和暖如春，着一层夹袄步出监牢的瞬间，已是汗流浃背。 一阵脚步声响...
    开封第一讲书人阅读 32,417评论 1赞 268
  • 情欲美人皮
    我被黑心中介骗来泰国打工， 没想到刚下飞机就差点儿被人妖公主榨干…… 1. 我叫王不留，地道东北人。 一个月前我还...
    沈念sama阅读 47,094评论 2赞 365
  • 代替公主和亲
    正文 我出身青楼，却偏偏与公主长得像，于是被迫代替她去往敌国和亲。 传闻我的和亲对象是个残疾皇子，可洞房花烛夜当晚...
    茶点故事阅读 44,088评论 2赞 352

推荐阅读更多精彩内容

  • 50道SQL练习题及答案与详细分析
    网上流传较广的50道SQL训练，奋斗了不知道多久终于写完了。前18道题的难度依次递增，从19题开始的后半部分算是循...
    Kaidi_G阅读 474,463评论 312赞 452
  • sql 经典50题
    50道SQL练习题及答案与详细分析 https://www.jianshu.com/p/cc23c19ef639[...
    将代码写成诗阅读 2,248评论 0赞 3
  • MySQL：测试题
    一，表关系的练习测试 请创建如下表关系，并建立相关约束 一，创建表结构数据： 创建的话肯定先创建没有关联的表，老师...
    Kevin_Luo阅读 2,792评论 0赞 0
  • 互联网经典SQL面试题及答案解析
    练习前的准备 SQL环境搭建 https://www.jianshu.com/p/edca5142391c 建表语...
    TOMOCAT阅读 23,516评论 3赞 30
  • Mysql_45道练习题
    我使用的Mysql版本是5.7.19。答案可能会因版本会有少许出入。如果答案有问题欢迎在评论区指出并给出你的答案。...
    大石兄阅读 16,477评论 8赞 21