SQL Server组由谁来解决？

需要查询帮助来整理此“简单”任务；）

tbl1

您的查询->

结果tbl2：

Name (Credit–Debit)/2014 (Credit–Debit)/2015 (Credit–Debit)/2016
Diaz          0              700                   300
Maya        200              500                     0
Jane        700                0                   200

非常感谢！！

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D.

您可以使用以下

解释

从日期中提取年份

查找贷方-借方

对201420152016年的金额应用透视

数据设置

我使用贷记和借记的数据类型作为货币，如果你想在你的案例中使用int

declare @data table(
    Name nvarchar(50) not null,
    TransDate date not null,
    Credit money not null,
    Debit money not null
)

insert into @data(Name,TransDate,Credit,Debit)
values ( 'Diaz','2015-01-01',500,200),
       ( 'Diaz','2015-05-01',500,100),
       ( 'Diaz','2016-01-01',500,200),
       ( 'Maya','2014-01-01',500,300),
       ( 'Maya','2015-04-01',500,300),
       ( 'Maya','2015-11-01',500,200),
       ( 'Jane','2014-02-01',500,200),
       ( 'Jane','2014-03-01',500,100),
       ( 'Jane','2016-08-01',500,200)

质疑

结果

Name    Credit-Debit/2014   Credit-Debit/2015   Credit-Debit/2016
Diaz    0.00    700.00  300.00
Jane    700.00  0.00    300.00
Maya    200.00  500.00  0.00

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希望这能帮助您

假设您的已知年数有限，一种方法是使用

案例

和

聚合

SELECT
  name,
  COALESCE(SUM(CASE
    WHEN YEAR(TransDate) = '2014' THEN credit - debit
  END), 0) AS credit_debit_2014,
  COALESCE(SUM(CASE
    WHEN YEAR(TransDate) = '2015' THEN credit - debit
  END), 0) AS credit_debit_2015,
  COALESCE(SUM(CASE
    WHEN YEAR(TransDate) = '2016' THEN credit - debit
  END), 0) AS credit_debit_2016
FROM table1
GROUP BY name

结果：

+------+-------------------+-------------------+-------------------+
| name | credit_debit_2014 | credit_debit_2015 | credit_debit_2016 |
+------+-------------------+-------------------+-------------------+
| Diaz |                 0 |               700 |               300 |
| Jane |               700 |                 0 |               200 |
| Maya |               200 |               500 |                 0 |
+------+-------------------+-------------------+-------------------+

---

我喜欢使用条件聚合来解决这些问题。即使给定年份的

贷方

或

借方

为

空

，以下各项也应有效：

SELECT name,
       SUM(CASE WHEN YEAR(TransDate) = 2014
                THEN COALESCE(credit, 0) - COALESCE(debit, 0)
                ELSE 0
           END) as credit_debit_2014,
       SUM(CASE WHEN YEAR(TransDate) = 2015
                THEN COALESCE(credit, 0) - COALESCE(debit, 0)
                ELSE 0
           END) as credit_debit_2015,
       SUM(CASE WHEN YEAR(TransDate) = 2016
                THEN COALESCE(credit, 0) - COALESCE(debit, 0)
                ELSE 0
           END) as credit_debit_2016
FROM table1
GROUP BY name;

---

你有多少年了？只有三个，或者还有更多？如果这是一个简单的任务，为什么你不试一下呢self@qotqn还有更多。@Imran Ali Khan您显然错过了“简单”中的引语，我喜欢您的解决方案。只是一个问题：在这种情况下，Coalesce和isnull是可互换的？@Diaz。在这种情况下，它们是非常可互换的。两者有细微的区别，但在这里并不相关。

+------+-------------------+-------------------+-------------------+
| name | credit_debit_2014 | credit_debit_2015 | credit_debit_2016 |
+------+-------------------+-------------------+-------------------+
| Diaz |                 0 |               700 |               300 |
| Jane |               700 |                 0 |               200 |
| Maya |               200 |               500 |                 0 |
+------+-------------------+-------------------+-------------------+

SELECT name,
       SUM(CASE WHEN YEAR(TransDate) = 2014
                THEN COALESCE(credit, 0) - COALESCE(debit, 0)
                ELSE 0
           END) as credit_debit_2014,
       SUM(CASE WHEN YEAR(TransDate) = 2015
                THEN COALESCE(credit, 0) - COALESCE(debit, 0)
                ELSE 0
           END) as credit_debit_2015,
       SUM(CASE WHEN YEAR(TransDate) = 2016
                THEN COALESCE(credit, 0) - COALESCE(debit, 0)
                ELSE 0
           END) as credit_debit_2016
FROM table1
GROUP BY name;