Question

我正在尝试在查询中使用@ variable1 + @ variable2，但实际上得到的结果为0。

MariaDB 服务器版本：10.2.21

set @start_at = '2019-01-01';
set @end_at = '2019-01-16';
set @receivable = 0;
set @invoiced = 0;
SELECT DISTINCT Customer.custnr 'Customer Number',
                Address.name    'Name',
                @receivable := sum(case
                      WHEN [condition1 <= @start_at]
                        AND Transactions.`key` not in [subquery]
                        THEN Transactions.amount
                      ELSE 0 END) 'Account Receivable',
                @invoiced := sum(case
                      WHEN [condition1 between @start_at and @end_at]
                        AND [condition2]
                        AND [condition3]
                        AND Transactions.`key` not in [subquery]
                        THEN Transactions.amount
                      ELSE 0 END) 'Invoiced',
                @receivable + @invoiced 'Total'
FROM LocalCust
       INNER JOIN Customer
                  on Customer.`key` = LocalCust.customerkey
       INNER JOIN Address
                  on Address.`key` = Customer.addresskey
       INNER JOIN Location
                  on Location.`key` = LocalCust.localkey
       INNER JOIN Transactions
                  on Transactions.localcustkey = LocalCust.`Key`
GROUP BY Transactions.localcustkey;

结果：

Answer 1

使用子查询，根本不使用变量：

SELECT x.*, (Account_Receivable + Invoice) as Total
FROM (SELECT c.custnr as Customer_Number, a.name, 
             sum(case when condition1 <= @start_at and
                           t.`key` not in [subquery]
                      then t.amount
                      else 0 
                 end) as Account_Receivable,
             sum(case when condition1 between @start_at and @end_at and
                           [condition2] and
                           [condition3] and
                           t.`key` not in [subquery]
                      then t.amount
                      else 0 
                 end) as Invoiced
      FROM LocalCust lc JOIN
           Customer c
           on c.`key` = lc.customerkey JOIN
           Address a
           on a.`key` = c.addresskey JOIN
           Location l
           on l.`key` = lc.localkey join
           Transactions t
           on t.localcustkey = lc.`Key`
      GROUP BY c.custnr, a.name
     ) x;

注意：

表别名使查询更易于编写和阅读。
SELECT DISTINCT几乎不需要GROUP BY。
GROUP BY键应与SELECT中未聚合的列匹配。
选择不需要转义的列别名。也就是说，没有空格。

Answer 2

您不能只将@ receivable，@ invoiced和@receivable + @invoiced放在同一选择语句中。（它们不会按顺序存储值。它们将同时执行。）

首先，您需要将值存储在@ receivable，@ invoiced中，然后使用子查询来计算总数：

set @start_at = '2019-01-01';
set @end_at = '2019-01-16';
set @receivable = 0;
set @invoiced = 0;

SELECT *, A.[Account Receivable] + A.[Invoiced]  AS TOTAL FROM (
SELECT DISTINCT Customer.custnr 'Customer Number',
                Address.name    'Name',
                @receivable := sum(case
                      WHEN [condition1 <= @start_at]
                        AND Transactions.`key` not in [subquery]
                        THEN Transactions.amount
                      ELSE 0 END) 'Account Receivable',
                @invoiced := sum(case
                      WHEN [condition1 between @start_at and @end_at]
                        AND [condition2]
                        AND [condition3]
                        AND Transactions.`key` not in [subquery]
                        THEN Transactions.amount
                      ELSE 0 END) 'Invoiced'
FROM LocalCust
       INNER JOIN Customer
                  on Customer.`key` = LocalCust.customerkey
       INNER JOIN Address
                  on Address.`key` = Customer.addresskey
       INNER JOIN Location
                  on Location.`key` = LocalCust.localkey
       INNER JOIN Transactions
                  on Transactions.localcustkey = LocalCust.`Key`
GROUP BY Transactions.localcustkey) A;

局部变量求和前几列

2 个答案: