上个月和本月使用 MySQL CASE 求和答案

【问题标题】：Using MySQL CASE to SUM last month and this month totals上个月和本月使用 MySQL CASE 求和
【发布时间】：2018-07-27 05:20:43
【问题描述】：

我有一个查询要从 2 个表中生成课程名称和总数的列表。它工作正常，但现在我想在同一个查询中获得本月和上个月的总数，所以我添加了几个案例。问题是这些案例返回的是空值。

这是我的查询：

$Sql = "SELECT t1.coursename, t1.id, SUM(t2.amount) AS total,

CASE 
WHEN YEAR(t2.date) = YEAR(CURRENT_DATE - INTERVAL 1 MONTH) AND MONTH(t2.date) = MONTH(CURRENT_DATE - INTERVAL 1 MONTH) THEN SUM(t2.amount)
ELSE 0 END 
AS lastMonthTotal,      

CASE
WHEN YEAR(t2.date) = YEAR(CURRENT_DATE) AND MONTH(t2.date) = MONTH(CURRENT_DATE) THEN SUM(t2.amount) 
ELSE 0 END 
AS thisMonthTotal

FROM t1
LEFT JOIN t2 ON t2.courseid = t1.courseid
WHERE $pdoCoursesString GROUP BY t1.courseid";

【问题讨论】：

将所有这些案例条件放在 sum(amount) 下，并检查 group by 子句中的列

标签： mysql sql sum case

【解决方案1】：

我认为您应该将 CASE 语句放在 SUM 函数中。

SQL 小提琴：http://sqlfiddle.com/#!9/4e10e2/5

SQL：

测试数据：

create table t1(courseid int, coursename varchar(10));
create table t2(courseid int, date date, amount int);
insert into t1 values(1,'courseA');
insert into t1 values(2,'courseB');
insert into t2 values(1,'20180211',12);
insert into t2 values(1,'20180111',16);
insert into t2 values(2,'20180101',1);
insert into t2 values(2,'20180201',1);
insert into t2 values(2,'20180101',1);

查询：

SELECT t1.coursename, t1.courseid, SUM(t2.amount) AS total,
SUM(
  CASE 
  WHEN YEAR(t2.date) = YEAR(CURRENT_DATE - INTERVAL 1 MONTH) 
  AND MONTH(t2.date) = MONTH(CURRENT_DATE - INTERVAL 1 MONTH) 
  THEN 
    t2.amount
  ELSE 
    0 
  END
) AS lastmonthtotal,
SUM(
  CASE 
  WHEN YEAR(t2.date) = YEAR(CURRENT_DATE) 
  AND MONTH(t2.date) = MONTH(CURRENT_DATE) 
  THEN 
    t2.amount
  ELSE 
    0 
  END
) AS thismonthtotal
FROM t1
LEFT JOIN t2 ON t2.courseid = t1.courseid
GROUP BY t1.courseid
;

结果：

| coursename | courseid | total | lastmonthtotal | thismonthtotal |
|------------|----------|-------|----------------|----------------|
|    courseA |        1 |    28 |             16 |             12 |
|    courseB |        2 |     3 |              2 |              1 |

【讨论】：

案例表达式，不是语句。
感谢您的回答和 sql fiddle。这真的很有帮助

【解决方案2】：

sum(case when {condition} then {field} else 0 end)

【讨论】：

请提供一些关于答案如何帮助 OP 的解释。

【解决方案3】：

将大小写放在 sum 中。

总和（当...的情况）

【讨论】：

【解决方案4】：

将CASE 放在SUM 中，而不是相反。

$Sql = "SELECT t1.coursename, t1.id, SUM(t2.amount) AS total,

SUM(CASE WHEN YEAR(t2.date) = YEAR(CURRENT_DATE - INTERVAL 1 MONTH) AND MONTH(t2.date) = MONTH(CURRENT_DATE - INTERVAL 1 MONTH) 
         THEN t2.amount
         ELSE 0 END) AS lastMonthTotal,      

SUM(CASE
    WHEN YEAR(t2.date) = YEAR(CURRENT_DATE) AND MONTH(t2.date) = MONTH(CURRENT_DATE) 
    THEN t2.amount 
    ELSE 0 END) AS thisMonthTotal

FROM t1
LEFT JOIN t2 ON t2.courseid = t1.courseid
WHERE $pdoCoursesString GROUP BY t1.courseid";

【讨论】：

【解决方案5】：

您应该在您的选择子句中对具有未聚合函数的相同列进行分组，例如 t1.coursename、t1.id 并使用 sum ( case... ) 例如：

$Sql = "SELECT t1.coursename, t1.id, SUM(t2.amount) AS total,

   sum(  CASE 
    WHEN YEAR(t2.date) = YEAR(CURRENT_DATE - INTERVAL 1 MONTH) AND MONTH(t2.date) = MONTH(CURRENT_DATE - INTERVAL 1 MONTH) THEN t2.amount
    ELSE 0 END )
    AS lastMonthTotal,      

    sum( CASE
    WHEN YEAR(t2.date) = YEAR(CURRENT_DATE) AND MONTH(t2.date) = MONTH(CURRENT_DATE) THEN t2.amount
    ELSE 0 END )
    AS thisMonthTotal
FROM t1
LEFT JOIN t2 ON t2.courseid = t1.courseid
WHERE $pdoCoursesString 
GROUP BY t1.coursename, t1.id";

【讨论】：