【MySQL】SQL之CASE WHEN用法详解_mysql case when

原文链接：https://blog.csdn.net/rongtaoup/article/details/82183743

前言

建议先看看官方文档_对 case when 的用法

一、简单CASE WHEN函数：

CASE SCORE WHEN 'A' THEN '优' ELSE '不及格' END

# 使用 IF 函数进行替换
IF(SCORE = 'A', '优', '不及格')
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THEN后边的值与ELSE后边的值类型应一致，否则会报错。
如下：
CASE SCORE WHEN ‘A’ THEN ‘优’ ELSE 0 END’优’和0数据类型不一致则报错：
[Err] ORA-00932: 数据类型不一致: 应为 CHAR, 但却获得 NUMBER

简单CASE WHEN函数只能应对一些简单的业务场景，而CASE WHEN条件表达式的写法则更加灵活。

二、CASE WHEN条件表达式函数

类似JAVA中的IF ELSE语句。

格式：

CASE WHEN condition THEN result
 
[WHEN...THEN...]
 
ELSE result
 
END
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SQL语言演示：

CASE 
	 WHEN SCORE = 'A' THEN '优'
     WHEN SCORE = 'B' THEN '良'
     WHEN SCORE = 'C' THEN '中' 
     ELSE '不及格' END

# 等同于
CASE score
    WHEN 'A' THEN '优'
    WHEN 'B' THEN '良'
    WHEN 'C' THEN '中'
    ELSE '不及格' END
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condition是一个返回布尔类型的表达式，
如果表达式返回true，则整个函数返回相应result的值，
如果表达式皆为false，则返回ElSE后result的值，如果省略了ELSE子句，则返回NULL。

三、常用场景

前言

students表的DDL

-- auto-generated definition
create table students
(
    stu_code  varchar(10) null,
    stu_name  varchar(10) null,
    stu_sex   int         null,
    stu_score int         null
);
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students表的DML

# 其中stu_sex字段，0表示男生，1表示女生。
INSERT INTO students (stu_code, stu_name, stu_sex, stu_score) VALUES ('xm', '小明', 0, 88);
INSERT INTO students (stu_code, stu_name, stu_sex, stu_score) VALUES ('xl', '夏磊', 0, 55);
INSERT INTO students (stu_code, stu_name, stu_sex, stu_score) VALUES ('xf', '晓峰', 0, 45);
INSERT INTO students (stu_code, stu_name, stu_sex, stu_score) VALUES ('xh', '小红', 1, 89);
INSERT INTO students (stu_code, stu_name, stu_sex, stu_score) VALUES ('xn', '小妮', 1, 77);
INSERT INTO students (stu_code, stu_name, stu_sex, stu_score) VALUES ('xy', '小一', 1, 99);
INSERT INTO students (stu_code, stu_name, stu_sex, stu_score) VALUES ('xs', '小时', 1, 45);
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---

energy_test表的DDL

-- auto-generated definition
create table energy_test
(
    e_code  varchar(2)    null,
    e_value decimal(5, 2) null,
    e_type  int           null
);
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energy_test表的DML

# 其中，E_TYPE表示能耗类型，0表示水耗，1表示电耗，2表示热耗
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('北京', 28.50, 0);
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('北京', 23.50, 1);
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('北京', 28.12, 2);
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('北京', 12.30, 0);
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('北京', 15.46, 1);
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('上海', 18.88, 0);
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('上海', 16.66, 1);
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('上海', 19.99, 0);
INSERT INTO energy_test (e_code, e_value, e_type) VALUES ('上海', 10.05, 0);
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---

p_price表的DDL

-- auto-generated definition
create table p_price
(
    p_price decimal(5, 2) null comment '价格',
    p_level int           null comment '等级',
    p_limit int           null comment '阈值'
)
    comment '电能耗单价表';
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p_price表的DML

INSERT INTO test.p_price (p_price, p_level, p_limit) VALUES (1.20, 0, 10);
INSERT INTO test.p_price (p_price, p_level, p_limit) VALUES (1.70, 1, 30);
INSERT INTO test.p_price (p_price, p_level, p_limit) VALUES (2.50, 2, 50);
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---

user_col_comments 表的DDL

-- auto-generated definition
create table user_col_comments
(
    column_name varchar(50)  null comment '列名',
    comment     varchar(100) null comment '列的备注'
);
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user_col_comments 表的DML

INSERT INTO test.user_col_comments (column_name, comment) VALUES ('SHI_SHI_CODE', '设施编号');
INSERT INTO test.user_col_comments (column_name, comment) VALUES ('SHUI_HAO', '水耗');
INSERT INTO test.user_col_comments (column_name, comment) VALUES ('RE_HAO', '热耗');
INSERT INTO test.user_col_comments (column_name, comment) VALUES ('YAN_HAO', '盐耗');
INSERT INTO test.user_col_comments (column_name, comment) VALUES ('OTHER', '其他');
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场景1：不同状态展示为不同的值

有分数score，score<60返回不及格，score>=60返回及格，score>=80返回优秀

# 有分数score，score<60返回不及格，score>=60返回及格，score>=80返回优秀
SELECT
    stu_name,
    (CASE WHEN stu_score < 60 THEN '不及格'
        WHEN stu_score >= 60 AND stu_score < 80 THEN '及格'
        WHEN stu_score >= 80 THEN '优秀'
        ELSE '异常' END) AS REMARK
FROM students;
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注意：如果你想判断score是否null的情况，WHEN score = null THEN ‘缺席考试’，这是一种错误的写法，正确的写法应为：
CASE WHEN score IS NULL THEN '缺席考试' ELSE '正常' END

场景2：统计不同状态下的值

现老师要统计班中，有多少男同学，多少女同学，并统计男同学中有几人及格，女同学中有几人及格，要求用一个SQL输出结果。其中stu_sex字段，0表示男生，1表示女生。

SELECT
	sum(CASE WHEN STU_SEX = 0 THEN 1 ELSE 0 END) AS MALE_COUNT,
	sum(CASE WHEN STU_SEX = 1 THEN 1 ELSE 0 END) AS FEMALE_COUNT,
	sum(CASE WHEN STU_SCORE >= 60 AND STU_SEX = 0 THEN 1 ELSE 0 END) AS MALE_PASS,
	sum(CASE WHEN STU_SCORE >= 60 AND STU_SEX = 1 THEN 1 ELSE 0 END) AS FEMALE_PASS
FROM
	students;
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输出结果如下：

注意点：

• 用的是 ：sum 而不是count
• THEN 1 ELSE 0的位置不能改变：否则会有以下效果：

  sum(CASE WHEN stu_sex = 0 THEN '1' ELSE '0' END) AS '男性',
  
  改变了 
  sum(CASE WHEN stu_sex = 0 THEN '0' ELSE '1' END) AS '女性'：
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• 字符 ‘0’ 和 数值 0，使用 都是一样的

场景3：配合聚合函数做统计

现要求统计各个城市，总共使用了多少水耗、电耗、热耗，使用一条SQL语句输出结果
有能耗表如下：其中，E_TYPE表示能耗类型，0表示水耗，1表示电耗，2表示热耗

select e_code,
       sum(case when e_type = 0 then e_value else 0 end) as '水耗',
       sum(case when e_type = 1 then e_value else 0 end) as '电耗',
       sum(case when e_type = 2 then e_value else 0 end) as '热耗'
from energy_test
group by e_code;
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输出结果如下：

场景4：CASE WHEN中使用子查询

根据城市用电量多少，计算用电成本。假设电能耗单价分为三档，根据不同的能耗值，使用相应价格计算成本。
当能耗值小于10时，使用P_LEVEL=0时的P_PRICE的值，能耗值大于10小于30使用P_LEVEL=1时的P_PRICE的值…

energy_test 我修改了e_type 为1的值的两条数据的e_value。

select e_code, e_value,
     (CASE WHEN e_value <= (SELECT p_limit FROM p_price WHERE p_level = 0)
        THEN (SELECT p_price FROM p_price WHERE p_level = 0)
     WHEN e_value > (SELECT p_limit FROM p_price WHERE p_level = 0) AND e_value <= (SELECT p_limit FROM p_price WHERE p_level = 1)
        THEN (SELECT P_PRICE FROM p_price WHERE P_LEVEL = 1)
     WHEN e_value > (SELECT p_limit FROM p_price WHERE p_level = 1) AND e_value <= (SELECT p_limit FROM p_price WHERE p_level = 2)
        THEN (SELECT p_price FROM p_price WHERE P_LEVEL = 2) end ) as price
from energy_test
where e_type = 1;
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输出结果如下：

场景5：经典行转列，结合max聚合函数

行转列中 SUM作用：无用，但是select后得跟聚合函数，不能去掉sum。直接写max或者min也行。

select
    max(case when column_name = 'SHI_SHI_CODE' then comment else ''end) as SHI_SHI_CODE_COMMENT,
    max(case when column_name = 'SHUI_HAO' then comment else ''end) as SHUI_HAO_COMMENT,
    max(case when column_name = 'RE_HAO' then comment else ''end) as RE_HAO_COMMENT,
    max(case when column_name = 'YAN_HAO' then comment else ''end) as YAN_HAO_COMMENT,
    max(case when column_name = 'OTHER' then comment else '' end) as OTHER_COMMENT
from user_col_comments;
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输出结果如下：

四、练习题

LeetCode练习题

• LeetCode_1873. 计算特殊奖金
• LeetCode_627. 变更性别
• LeetCode_608. 树节点
• LeetCode_1393. 股票的资本损益

练习题一

题目描述

create table test
(
    year  int           null,
    mouth int           null,
    amout decimal(2, 1) null
);

INSERT INTO test.test (year, mouth, amout) VALUES (1991, 1, 1.1);
INSERT INTO test.test (year, mouth, amout) VALUES (1991, 2, 1.2);
INSERT INTO test.test (year, mouth, amout) VALUES (1991, 3, 1.3);
INSERT INTO test.test (year, mouth, amout) VALUES (1991, 4, 1.4);
INSERT INTO test.test (year, mouth, amout) VALUES (1992, 1, 2.1);
INSERT INTO test.test (year, mouth, amout) VALUES (1992, 2, 2.2);
INSERT INTO test.test (year, mouth, amout) VALUES (1992, 3, 2.3);
INSERT INTO test.test (year, mouth, amout) VALUES (1992, 4, 2.4);
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全部数据如下：

要求，最后用SQL语句转化为如下结果：

SQL 实现

select  t.year,
       sum(case when  t.mouth = 1 then amout else 0 end ) as m1,
       sum(case when  t.mouth = 2 then amout else 0 end ) as m2,
       sum(case when  t.mouth = 3 then amout else 0 end ) as m3,
       sum(case when  t.mouth = 4 then amout else 0 end ) as m4
from test t
group by t.year
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思路

行转列第一个就是想到了case when 语法。

那么，看题目，以年份year作为筛选条件，去找每个月份mouth 对应的金额amout。根据题目的意思，如果月份mouth 是1 ，那么找出来的金额amout 就是 m1。

此时就有了初步的轮廓：

select  t.year,
      (case when  t.mouth = 1 then amout else 0 end ) as m1,
      (case when  t.mouth = 2 then amout else 0 end ) as m2,
      (case when  t.mouth = 3 then amout else 0 end ) as m3,
      (case when  t.mouth = 4 then amout else 0 end ) as m4
from test t
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此时，得到的结果是：

得到这一步就很好理解了，我们就需要一个统计函数，就可以得到想要的结果了。

select  t.year,
       sum(case when  t.mouth = 1 then amout else 0 end ) as m1,
       sum(case when  t.mouth = 2 then amout else 0 end ) as m2,
       sum(case when  t.mouth = 3 then amout else 0 end ) as m3,
       sum(case when  t.mouth = 4 then amout else 0 end ) as m4
from test t
group by t.year
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完美解决：

练习题二

题目描述

-- auto-generated definition
create table plant_plan_rule_info
(
    rc_code              varchar(50) null,
    start_date           datetime    null,
    plant_plan_rule_type int         null,
    id                   int         null
);


INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14018, 'florescence', '2024-01-10 00:00:00', '0');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14020, 'SEED', '2024-01-10 00:00:00', '0');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14023, 'BOOT_STAGE', '2024-03-02 00:00:00', '0');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14025, 'HEADING_AND_FILLING_PERIOD', '2024-05-05 00:00:00', '0');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14027, 'SPRAY_STAGE', '2024-04-04 00:00:00', '0');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14029, 'DRIVE_POWDER', '2024-05-05 00:00:00', '0');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14031, 'MATERNAL_HARVESTING', '2024-06-07 00:00:00', '0');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14033, 'florescence', '2024-01-10 00:00:00', '1');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14035, 'BOOT_STAGE', '2024-03-02 00:00:00', '1');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14037, 'HEADING_AND_FILLING_PERIOD', '2024-05-05 00:00:00', '1');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14039, 'SPRAY_STAGE', '2024-04-04 00:00:00', '1');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14041, 'DRIVE_POWDER', '2024-05-05 00:00:00', '1');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14043, 'PATERNAL_CIRCUMCISION', '2024-06-05 00:00:00', '1');
INSERT INTO plant_plan_rule_info (id, rc_code, start_date, plant_plan_rule_type) VALUES (14159, 'SEED', '2024-02-02 00:00:00', '1');
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不同 plant_plan_rule_type（只会为 0 1 ）下，比较相同 rcCode 他们 在start_date 的差值。

要求：最后输出的结果如下：

SQL 实现

select rc_code, abs(DATEDIFF(maleStartDate, femaleStartDate)) as diffDay
from (select rc_code,
             max(if(plant_plan_rule_type = 1, start_date, null)) as maleStartDate,
             max(if(plant_plan_rule_type = 0, start_date, null)) as femaleStartDate
      from plant_plan_rule_info i
      group by rc_code) a;
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思路

步骤一：想把它转化为一行，那么，先把每一行都要有个字段：maleStartDate 、femaleStartDate；

select rc_code,
       if(plant_plan_rule_type = 1, start_date, null) as maleStartDate,
       if(plant_plan_rule_type = 0, start_date, null) as femaleStartDate
from plant_plan_rule_info i
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步骤二：再通过聚合函数 + 分组，就可以实现变为一行。

选择哪一种聚合函数：

• min() : 那么， maleStartDate 、femaleStartDate 只会展示为0的，不符合预期
• sum() ：统计这一组值的和，不符合
• count() ：统计这一组值的个数，不符合
• max() ：符合，统计这一组中最大的值，那么我们就可以去除掉0了

select rc_code,
       max(if(plant_plan_rule_type = 1, start_date, null)) as maleStartDate,
       max(if(plant_plan_rule_type = 0, start_date, null)) as femaleStartDate
from plant_plan_rule_info i
group by rc_code;
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步骤三：到这一步了，就没啥好说的了，搞个虚表，使用sql函数统计日期差值，为了避免出现负数，取绝对值就行

select rc_code, abs(DATEDIFF(maleStartDate, femaleStartDate)) as diffDay
from (select rc_code,
             max(if(plant_plan_rule_type = 1, start_date, null)) as maleStartDate,
             max(if(plant_plan_rule_type = 0, start_date, null)) as femaleStartDate
      from plant_plan_rule_info i
      group by rc_code) a;
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小小延伸

要求：我现在要展示的列如下：

• rc_code : plant_plan_rule_info 的值
• malePlantPlanParentsRuleId : 当rc_code =1的时候，id的值:
• femalePlantPlanParentsRuleId：当rc_code =0的时候，id的值
• maleStartDate: 当rc_code =1的时候，start_date的值
• femaleStartDate: 当rc_code =0的时候，start_date的值
• diffDay：不同 plant_plan_rule_type（只会为 0 1 ）下，比较相同 rcCode 他们 在start_date 的差值

最终展示的结果如下：

答案：

select rc_code,
       malePlantPlanParentsRuleId,
       femalePlantPlanParentsRuleId,
       maleStartDate,
       femaleStartDate,
       abs(DATEDIFF(maleStartDate, femaleStartDate)) as diffDay
from (select rc_code,
             max(if(plant_plan_rule_type = 1, id, 0)) as malePlantPlanParentsRuleId,
             max(if(plant_plan_rule_type = 0, id, 0)) as femalePlantPlanParentsRuleId,
             max(if(plant_plan_rule_type = 1, start_date, null)) as maleStartDate,
             max(if(plant_plan_rule_type = 0, start_date, null)) as femaleStartDate
      from plant_plan_rule_info i
      group by rc_code) a;
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为什么maleStartDate 和femaleStartDate 为null，而 malePlantPlanParentsRuleId 和 femalePlantPlanParentsRuleId 为0呢？

---

因为现在与Mybatis字段是实体类映射的时候，maleStartDate和 femaleStartDate Date无法映射为0，所以做null处理；
而malePlantPlanParentsRuleId 和 femalePlantPlanParentsRuleId 为Long，可以映射为0，也可以进行null处理。