HQL必会50题

平常加班不多，为了提高自己的SQL能力，找了网上流传的比较广泛的SQL50题，先把题目拿过来，计划抽空作为练习用HiveSQL完成。

开始日期：2020-03-18

完成日期：2020-03-21

建表语句

create table student(s_id string,s_name string,s_birth string,s_sex string) row format delimited fields terminated by '\t';

create table course(c_id string,c_name string,t_id string) row format delimited fields terminated by '\t';

create table teacher(t_id string,t_name string) row format delimited fields terminated by '\t';

create table score(s_id string,c_id string,s_score int) row format delimited fields terminated by '\t';

数据：

student.txt

01	赵雷	1990-01-01	男
02	钱电	1990-12-21	男
03	孙风	1990-05-20	男
04	李云	1990-08-06	男
05	周梅	1991-12-01	女
06	吴兰	1992-03-01	女
07	郑竹	1989-07-01	女
08	王菊	1990-01-20	女

teacher.txt

01	张三
02	李四
03	王五

course.txt

01	语文	02
02	数学	01
03	英语	03

score.txt

01	01	80
01	02	90
01	03	99
02	01	70
02	02	60
02	03	80
03	01	80
03	02	80
03	03	80
04	01	50
04	02	30
04	03	20
05	01	76
05	02	87
06	01	31
06	03	34
07	02	89
07	03	98

加载数据：

load data local inpath '~/student.txt' into table student;

load data local inpath '~/course.txt' into table course;

load data local inpath '~/teacher.txt' into table teacher;

load data local inpath '~/score.txt' into table score;

表之间的关系如图：

• 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 
    T0.*,
    T1.s_score AS c01,
    T2.s_score as c02
FROM
    student AS T0,
    (SELECT s_id,s_score FROM score WHERE c_id='01') AS T1,
    (SELECT s_id,s_score FROM score WHERE c_id='02') AS T2
WHERE T0.s_id=T1.s_id
AND T0.s_id=T2.s_id
AND T1.s_score>T2.s_score;

运行结果：

• 2、查询同时存在” 01 “课程和” 02 “课程的情况:

SELECT
    t3.*,
    t1.s_score as c01,
    t2.s_score as c02
FROM 
    (SELECT s_id,s_score FROM score WHERE c_id='01') as t1, -- 选修01课程的同学
    (SELECT s_id,s_score FROM score WHERE c_id='02') as t2, -- 选修01课程的同学
    student as t3
WHERE t1.s_id=t2.s_id
AND t2.s_id=t3.s_id;

运行结果：

• 3、查询存在” 01 “课程但可能不存在” 02 “课程的情况(不存在时显示为 null )

SELECT
    t2.*,
    t1.s_score AS c01,
    t3.s_score AS c02 
FROM
    (SELECT s_id,s_score FROM score WHERE c_id='01') AS t1, -- 选修01课程的同学
    student AS t2
LEFT JOIN (SELECT s_id,s_score FROM score WHERE c_id='02') AS t3 -- 选修01课程的同学
ON t2.s_id = t3.s_id
WHERE t1.s_id=t2.s_id;

运行结果：

• 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
    T1.s_id,
    T1.s_name,
    T2.avg_sc
FROM student AS T1
JOIN(SELECT
        s_id,
        avg(s_score) AS avg_sc 
    FROM score
    GROUP BY s_id
    having avg_sc<60) AS T2
ON T1.s_id=T2.s_id; 

运行结果：

• 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
      T1.s_id,
      T1.s_name,
      T2.co_cnt,
      T3.sum_score
  FROM student AS T1
  LEFT JOIN (SELECT s_id,count(1) AS co_cnt FROM score GROUP BY s_id) AS T2
  ON T1.s_id=T2.s_id
  LEFT JOIN (SELECT s_id,sum(s_score) AS sum_score FROM score GROUP BY s_id) AS T3
  ON T1.s_id=T3.s_id;

  运行结果：

  

• 6、查询”李”姓老师的数量:

  SELECT COUNT(1) 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 
      T1.*,
      T3.c_name,
      T4.t_name
  FROM
      student AS T1,
      score AS T2,
      course AS T3,
      teacher AS T4
  WHERE T1.s_id=T2.s_id
  AND T2.c_id=T3.c_id
  AND T3.t_id=T4.t_id
  AND T4.t_name="张三";

  运行结果：

  

• 8、查询没学过”张三”老师授课的同学的信息:

  SELECT A1.* FROM student AS A1
  LEFT JOIN(SELECT 
                  T1.s_id
              FROM
                  student AS T1,
                  score AS T2,
                  course AS T3,
                  teacher AS T4
              WHERE T1.s_id=T2.s_id
              AND T2.c_id=T3.c_id
              AND T3.t_id=T4.t_id
              AND T4.t_name="张三") AS A2
  ON A1.s_id=A2.s_id
  WHERE A2.s_id IS NULL;

  运行结果：

  

• 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 AS A
  JOIN (SELECT
              T1.s_id
          FROM (SELECT s_id FROM score WHERE c_id='01') AS T1
          JOIN (SELECT s_id FROM score WHERE c_id='02') AS T2
          ON T1.s_id=T2.s_id) AS B
  ON A.s_id=B.s_id;

  运行结果：

  

• 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 AS A
  JOIN (SELECT
              T1.s_id
          FROM (SELECT s_id FROM score WHERE c_id='01') AS T1
          LEFT JOIN (SELECT s_id FROM score WHERE c_id='02') AS T2
          ON T1.s_id=T2.s_id
          WHERE T2.s_id IS NULL) AS B
  ON A.s_id=B.s_id;

运行结果：

• 11、查询没有学全所有课程的同学的信息:

SELECT
    T1.*,
    T2.CNT_C
FROM student AS T1
LEFT JOIN (SELECT s_id,count(1) AS CNT_C FROM score GROUP BY s_id) AS T2
ON T1.s_id=T2.s_id
WHERE T2.CNT_C<3;

• 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 
    t3.*,
    t4.cnt 
FROM student AS t3 
JOIN(
    SELECT s_id,
    COUNT(1) AS cnt 
    FROM score AS t1 
    JOIN (SELECT c_id 
          FROM score
          WHERE s_id ="01") AS t2 
    ON t1.c_id = t2.c_id 
    GROUP BY s_id 
    HAVING cnt>0) AS t4 
ON t3.s_id=t4.s_id;

• 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 
    A1.*,
    A2.CNT_C
FROM student AS A1
JOIN (SELECT T1.s_id,count(1) AS CNT_C
        FROM score AS T1
        JOIN (SELECT c_id FROM score WHERE s_id='01') AS T2
        ON T1.c_id=T2.c_id
        GROUP BY T1.s_id) AS A2
ON A1.s_id=A2.s_id
AND A1.S_id<>'01'
JOIN (SELECT count(1) AS CNT_C FROM score where s_id='01') AS A3
ON A2.CNT_C=A3.CNT_C;

• 14、查询没学过”张三”老师讲授的任一门课程的学生姓名:

SELECT S1.s_name 
FROM student AS S1
LEFT JOIN (SELECT
                A1.s_id
            FROM score AS A1
            JOIN (SELECT T2.c_id 
                    FROM teacher AS T1 
                    JOIN course AS T2
                    ON T1.t_id=T2.t_id
                    WHERE T1.t_name='张三') AS A2
            ON A1.c_id=A2.c_id
            GROUP BY A1.s_id) AS S2
ON S1.s_id=S2.s_id
WHERE S2.s_id IS NULL;

• 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
    T1.s_id,
    T1.s_name,
    T3.AVG_C
FROM student AS T1
JOIN (SELECT
        s_id,
        count(1) AS CNT_C
    FROM score
    WHERE s_score<60
    GROUP BY s_id
    having CNT_C>=2) AS T2
ON T1.s_id=T2.s_id
JOIN (SELECT
            s_id,
            AVG(s_score) AS AVG_C
        FROM score
        GROUP BY s_id) AS T3
ON T1.s_id=T3.s_id;

• 16、检索”01”课程分数小于60，按分数降序排列的学生信息:

SELECT
    T1.*,
    T2.s_score
FROM student AS T1
JOIN (SELECT
            s_id,
            s_score
        FROM score
        WHERE c_id='01'
        AND s_score<60) AS T2
ON T1.S_id=T2.s_id
ORDER BY T2.s_score DESC;

运行结果：

• 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:

尝试用WITH的写法，相对常规写法逻辑还是比较清晰的。

WITH T1 AS (
-- 平均分
    SELECT
        s_id,
        AVG(s_score) AS AVG_S
    FROM score
    GROUP BY s_id
), T2 AS(
-- 语文分数
    SELECT
        s_id,
        s_score
    FROM score
    WHERE c_id='01'
)
, T3 AS(
-- 数学分数
    SELECT
        s_id,
        s_score
    FROM score
    WHERE c_id='02'
), T4 AS(
-- 英语分数
    SELECT
        s_id,
        s_score
    FROM score
    WHERE c_id='03'
)
SELECT
    T0.s_id,
    T0.s_name,
    (CASE WHEN T1.AVG_S IS NULL THEN '0' ELSE round(T1.AVG_S,2) END) AS AVG_S,
    (CASE WHEN T2.s_score IS NULL THEN '0' ELSE T2.s_score END) AS CHINESE,
    (CASE WHEN T3.s_score IS NULL THEN '0' ELSE T3.s_score END) AS MATH,
    (CASE WHEN T4.s_score IS NULL THEN '0' ELSE T4.s_score END) AS ENGLISH
FROM student AS T0
LEFT JOIN T1
ON T1.s_id=T0.s_id
LEFT JOIN T2
ON T2.s_id=T0.s_id
LEFT JOIN T3
ON T3.s_id=T0.s_id
LEFT JOIN T4
ON T4.s_id=T0.s_id
ORDER BY AVG_S 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， ，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）

WITH A1 AS(
    SELECT
        c_id,
        MAX(s_score) AS max_c,
        MIN(s_score) AS min_c,
        ROUND(AVG(s_score),2)  AS avg_c,
        ROUND(SUM(CASE WHEN s_score>=60 THEN 1 ELSE 0 END)/count(s_score),2) AS L1,
        ROUND(SUM(CASE WHEN s_score BETWEEN 70 AND 79 THEN 1 ELSE 0 END)/count(s_score),2)  AS L2,
        ROUND(SUM(CASE WHEN s_score BETWEEN 80 AND 89 THEN 1 ELSE 0 END)/count(s_score),2)  AS L3,
        ROUND(SUM(CASE WHEN s_score>=90 THEN 1 ELSE 0 END)/count(s_score),2)  AS L4
    FROM score GROUP BY c_id
)
SELECT
    A0.c_id,
    A0.c_name,
    A1.max_c,
    A1.min_c,
    A1.avg_c,
    A1.L1,
    A1.L2,
    A1.L3,
    A1.L4
FROM course AS A0
JOIN A1
ON A0.c_id=A1.c_id;

• 19、按各科成绩进行排序，并显示排名:

SELECT
    T1.s_id,
    T3.s_name,
    T1.c_id,
    T2.c_name,
    T1.s_score,
    RANK() OVER (PARTITION BY T1.c_id ORDER BY T1.s_score DESC)
FROM score AS T1
JOIN course AS T2
ON T1.c_id=T2.c_id
JOIN student AS T3
ON T1.s_id=T3.s_id;

• 20、查询学生的总成绩并进行排名:

  SELECT
      T1.s_id,
      T3.s_name,
      sum(T1.s_score) AS SUM_C,
      RANK() OVER(ORDER BY sum(T1.s_score) DESC)
  FROM score AS T1
  JOIN student AS T3
  ON T1.s_id=T3.s_id
  GROUP BY T1.s_id,T3.s_name;

• 21、查询不同老师所教不同课程平均分从高到低显示:

  SELECT
      T1.c_id,
      T3.t_name,
      ROUND(AVG(T1.s_score),2),
      RANK() OVER(ORDER BY AVG(T1.s_score) DESC)
  FROM score AS T1
  JOIN course AS T2
  ON T1.c_id=T2.c_id
  JOIN teacher AS T3
  ON T2.t_id=T3.t_id
  GROUP BY T1.c_id,T3.t_name;

• 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:

  WITH A1 AS(
      SELECT
          T1.s_id,
          T3.s_name,
          T1.c_id,
          T2.c_name,
          T1.s_score,
          RANK() OVER (PARTITION BY T1.c_id ORDER BY T1.s_score DESC) AS rk
      FROM score AS T1
      JOIN course AS T2
      ON T1.c_id=T2.c_id
      JOIN student AS T3
      ON T1.s_id=T3.s_id
  )
  SELECT A1.*
  FROM A1
  WHERE A1.rk=2 OR A1.rk=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

  WITH A1 AS(
      SELECT
          c_id,
          SUM(CASE WHEN s_score BETWEEN 0 AND 59 THEN 1 ELSE 0 END) AS L0, -- [0-59分的人数]
          ROUND(SUM(CASE WHEN s_score BETWEEN 0 AND 59 THEN 1 ELSE 0 END)/COUNT(1),2) AS H0, -- [0-59分的人数占比]
          SUM(CASE WHEN s_score BETWEEN 60 AND 70 THEN 1 ELSE 0 END) AS L1, -- [60-69分的人数]
          ROUND(SUM(CASE WHEN s_score BETWEEN 60 AND 69 THEN 1 ELSE 0 END)/COUNT(1),2) AS H1, -- [60-69分的人数占比]
          SUM(CASE WHEN s_score BETWEEN 70 AND 85 THEN 1 ELSE 0 END)  AS L2, -- [70-85分的人数]
          ROUND(SUM(CASE WHEN s_score BETWEEN 70 AND 85 THEN 1 ELSE 0 END)/COUNT(1),2) AS H2, -- [70-85分的人数占比]
          SUM(CASE WHEN s_score BETWEEN 86 AND 100 THEN 1 ELSE 0 END)  AS L3, -- [86-100分的人数]
          ROUND(SUM(CASE WHEN s_score BETWEEN 86 AND 100 THEN 1 ELSE 0 END)/COUNT(1),2) AS H3 -- [86-100分的人数占比]
      FROM score GROUP BY c_id
  )
  SELECT
      A0.c_id,
      A0.c_name,
      A1.L3,
      A1.H3,
      A1.L2,
      A1.H2,
      A1.L1,
      A1.H1,
      A1.L0,
      A1.H0
  FROM course AS A0
  JOIN A1
  ON A0.c_id=A1.c_id;

• 24、查询学生平均成绩及其名次:

SELECT
    T1.s_id,
    T3.s_name,
    ROUND(AVG(T1.s_score),2) AS AVG_C,
    RANK() OVER(ORDER BY sum(T1.s_score) DESC)
FROM score AS T1
JOIN student AS T3
ON T1.s_id=T3.s_id
GROUP BY T1.s_id,T3.s_name;

• 25、查询各科成绩前三名的记录

  WITH A1 AS(
      SELECT
          T1.s_id,
          T3.s_name,
          T1.c_id,
          T2.c_name,
          T1.s_score,
          RANK() OVER (PARTITION BY T1.c_id ORDER BY T1.s_score DESC) AS rk
      FROM score AS T1
      JOIN course AS T2
      ON T1.c_id=T2.c_id
      JOIN student AS T3
      ON T1.s_id=T3.s_id
  )
  SELECT A1.*
  FROM A1
  WHERE A1.rk<=3;

• 26、查询每门课程被选修的学生数:

  SELECT
      T1.c_id,
      T2.c_name,
      COUNT(1)
  FROM score AS T1
  JOIN course AS T2
  ON T1.c_id=T2.c_id
  GROUP BY T1.c_id,T2.c_name

• 27、查询出只有两门课程的全部学生的学号和姓名:

  WITH A1 AS(
      SELECT
          T1.s_id,
          COUNT(1) AS CNT
      FROM score AS T1
      GROUP BY T1.s_id
      HAVING CNT=2
  )
  SELECT
      A2.s_id,
      A2.s_name,
      A1.CNT
  FROM student AS A2
  JOIN A1 
  ON A1.s_id=A2.s_id;

• 28、查询男生、女生人数:

  SELECT
      T1.s_sex,
      COUNT(1)
  FROM student AS T1
  GROUP BY s_sex;

• 29、查询名字中含有”风”字的学生信息:

  SELECT
      T1.*
  FROM student AS T1
  WHERE T1.s_name LIKE "%风%"

• 30、查询同名同性学生名单，并统计同名人数:

数据中发现没有同名同性的学生，先插入几条

insert into student values
('09' , '张三' , '2017-12-20' , '女'),
('10' , '李四' , '2017-12-25' , '女'),
('11' , '李四' , '2012-06-06' , '女'),
('12' , '赵六' , '2013-06-13' , '女'),
('13' , '孙七' , '2014-06-01' , '女');

然后找同名同性的学生

SELECT
    s_name,
    s_sex,
    COUNT(1) AS CNT_N
FROM student
GROUP BY s_name,s_sex
HAVING CNT_N>1;

• 31、查询1990年出生的学生名单:

SELECT * FROM student WHERE s_birth BETWEEN '1990-01-01' AND '1990-12-31';

• 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:

WITH T1 AS (
    SELECT
        c_id,
        ROUND(AVG(s_score),2) AS AVG_C
    FROM score
    GROUP BY c_id
)
SELECT
    T1.c_id,
    T0.c_name,
    T1.AVG_C
FROM course AS T0
JOIN T1
ON T0.c_id=T1.c_id
ORDER BY T1.AVG_C DESC,T1.c_id;

• 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:

WITH T1 AS (
    SELECT
        s_id,
        AVG(s_score) AS AVG_S
    FROM score
    GROUP BY s_id
    HAVING AVG_S>=85
)
SELECT
    T0.s_id,
    T0.s_name,
    T1.AVG_S
FROM student as T0
JOIN T1
ON T0.s_id=T1.s_id;

• 34、查询课程名称为”数学”，且分数低于60的学生姓名和分数:

WITH T1 AS (
    SELECT
        s_id,
        AVG(s_score) AS AVG_S
    FROM score
    GROUP BY s_id
    HAVING AVG_S>=85
)
SELECT
    T0.s_id,
    T0.s_name,
    ROUND(T1.AVG_S,2) AS AVG_S
FROM student as T0
JOIN T1
ON T0.s_id=T1.s_id;

• 35、查询所有学生的课程及分数情况:

WITH T1 AS (
    SELECT 
        s_id,
        s_score
    FROM score
    WHERE c_id='01'
), T2 AS (
    SELECT 
        s_id,
        s_score
    FROM score
    WHERE c_id='02'
), T3 AS (
    SELECT 
        s_id,
        s_score
    FROM score
    WHERE c_id='03'
)
SELECT
    T0.s_id,
    T0.s_name,
    (CASE WHEN T1.s_score IS NULL THEN "未选修" ELSE T1.s_score END) AS CHINESE,
    (CASE WHEN T2.s_score IS NULL THEN "未选修" ELSE T2.s_score END) AS MATH,
    (CASE WHEN T3.s_score IS NULL THEN "未选修" ELSE T3.s_score END) AS ENGLISH
FROM student AS T0
LEFT JOIN T1
ON T0.s_id=T1.s_id
LEFT JOIN T2
ON T0.s_id=T2.s_id
LEFT JOIN T3
ON T0.s_id=T3.s_id;

• 36、查询任何一门课程成绩在70分以上的学生姓名、课程名称和分数:

SELECT
    -- T0.s_id,
    T1.s_name,
    -- T0.c_id,
    T2.c_name,
    T0.s_score
FROM score AS T0
LEFT JOIN student AS T1
ON T0.s_id=T1.s_id
LEFT JOIN course AS T2
ON T0.c_id=T2.c_id
WHERE T0.s_score>70;

• 37、查询课程不及格的学生:

SELECT
    -- T0.s_id,
    T1.s_name,
    -- T0.c_id,
    T2.c_name,
    T0.s_score
FROM score AS T0
LEFT JOIN student AS T1
ON T0.s_id=T1.s_id
LEFT JOIN course AS T2
ON T0.c_id=T2.c_id
WHERE T0.s_score<60;

• 38、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
    T1.s_id,
    T2.s_name
FROM score AS T1
JOIN student AS T2
ON T1.s_id=T2.s_id
WHERE T1.c_id='01'
AND T1.s_score>=80;

• 39、求每门课程的学生人数:

WITH T1 AS (
    SELECT
        c_id,
        COUNT(1) AS CNT_S
    FROM score
    GROUP BY c_id
)
SELECT
    T2.c_id,
    T2.c_name,
    T1.CNT_S
FROM course AS T2
JOIN T1
ON T1.c_id=T2.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:

WITH T1 AS(
    SELECT
        A4.*,
        A3.s_score,
        RANK() OVER(ORDER BY A3.s_score DESC) AS RK
    FROM teacher AS A1
    JOIN course AS A2
    ON A1.t_id=A2.t_id
    JOIN score AS A3
    ON A3.c_id=A2.c_id
    JOIN student AS A4
    ON A3.s_id=A4.s_id
    WHERE A1.t_name='张三'
)
SELECT
    T1.*
FROM T1
WHERE RK=1;

• 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 T1.s_id,
    T1.c_id,
    T3.c_name,
    T1.s_score
FROM score AS T1,score AS T2,course AS T3
WHERE T1.c_id <> T2.c_id 
AND T1.s_score=T2.s_score
AND T1.c_id=T3.c_id;

• 42、查询每门课程成绩最好的前三名:

WITH A1 AS(
    SELECT
        T1.s_id,
        T1.c_id,
        T1.s_score,
        RANK() OVER(PARTITION BY T1.c_id ORDER BY T1.s_score DESC) AS RK
    FROM score AS T1
)
SELECT
    A1.s_id,
    A3.s_name,
    A2.c_name,
    A1.s_score,
    A1.RK
FROM A1
JOIN course AS A2
ON A1.c_id=A2.c_id
JOIN student AS A3
ON A1.s_id=A3.s_id
WHERE RK<=3;

• 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(1) AS CNT_C
FROM score
GROUP BY c_id
HAVING CNT_C>=5
ORDER BY CNT_C DESC,c_id;

• 44、检索至少选修两门课程的学生学号:

SELECT
    s_id,
    COUNT(1) AS CNT_S
FROM score
GROUP BY s_id
HAVING CNT_S>=2

• 45、查询选修了全部课程的学生信息:

WITH A1 AS(
    SELECT
        T1.s_id,
        COUNT(1) AS CNT_S
    FROM score AS T1
    GROUP BY T1.s_id
    HAVING CNT_S=3
)
SELECT
    A0.*
FROM student A0
JOIN A1
ON A0.s_id=A1.s_id
WHERE A1.CNT_S=3;

• 46、查询各学生的年龄(周岁):

SELECT
    s_name,
    s_birth,
    (YEAR(current_date())-YEAR(s_birth)-
        (CASE WHEN MONTH(current_date())>MONTH(s_birth) THEN 0 
              WHEN MONTH(current_date())=MONTH(s_birth) THEN 
                    (CASE WHEN DAY(current_date())>=DAY(s_birth) THEN 0 ELSE 1 END)
          ELSE 1 END)
    ) AS AGE
FROM student;

• 47、查询本周过生日的学生:

SELECT
    s_name,
    s_birth
FROM student
WHERE s_birth BETWEEN DATE_ADD(NEXT_DAY(current_date(), 'MON'),-1) AND DATE_ADD(NEXT_DAY(current_date(), 'MON'),-7);

今天是3月21号，暂时没有本周过生日的同学，不过本题的思想是找到本周第一天和本周最后一天，HQL实现如下：

SELECT
        DATE_ADD(NEXT_DAY(current_date(), 'MON'),-1) AS SUNDAY,
        DATE_ADD(NEXT_DAY(current_date(), 'MON'),-7) AS MONDAY

• 48、查询下周过生日的学生:

SELECT
    s_name,
    s_birth
FROM student
WHERE s_birth BETWEEN NEXT_DAY(current_date(), 'MON') AND DATE_ADD(NEXT_DAY(current_date(), 'MON'),6);

同样没有下周过生日的同学，下周一到下周日的HQL如下：

SELECT NEXT_DAY(current_date(), 'MON') , DATE_ADD(NEXT_DAY(current_date(), 'MON'),6)

• 49、查询本月过生日的学生:

SELECT
    s_name,
    s_birth
FROM student
WHERE MONTH(s_birth)=MONTH(current_date());

• 50、查询12月份过生日的学生:

SELECT
    s_name,
    s_birth
FROM student
WHERE MONTH(s_birth)=12;