sql group by 语句用法

合计函数 (比如 SUM) 常常需要添加 GROUP BY 语句。
GROUP BY 语句
GROUP BY 语句用于结合合计函数，根据一个或多个列对结果集进行分组。

SQL GROUP BY 语法
SELECT column_name, aggregate_function(column_name)
FROM table_name
WHERE column_name operator value
GROUP BY column_name
SQL GROUP BY 实例
我们拥有下面这个 “Orders” 表：

O_Id OrderDate OrderPrice Customer
1 2008/12/29 1000 Bush
2 2008/11/23 1600 Carter
3 2008/10/05 700 Bush
4 2008/09/28 300 Bush
5 2008/08/06 2000 Adams
6 2008/07/21 100 Carter

现在，我们希望查找每个客户的总金额（总订单）。

我们想要使用 GROUP BY 语句对客户进行组合。

我们使用下列 SQL 语句：

SELECT Customer,SUM(OrderPrice) FROM Orders
GROUP BY Customer
结果集类似这样：

Customer SUM(OrderPrice)
Bush 2000
Carter 1700
Adams 2000

很棒吧，对不对？

让我们看一下如果省略 GROUP BY 会出现什么情况：

SELECT Customer,SUM(OrderPrice) FROM Orders结果集类似这样：

Customer SUM(OrderPrice)
Bush 5700
Carter 5700
Bush 5700
Bush 5700
Adams 5700
Carter 5700

上面的结果集不是我们需要的。

那么为什么不能使用上面这条 SELECT 语句呢？解释如下：上面的 SELECT 语句指定了两列（Customer 和 SUM(OrderPrice)）。”SUM(OrderPrice)” 返回一个单独的值（”OrderPrice” 列的总计），而 “Customer” 返回 6 个值（每个值对应 “Orders” 表中的每一行）。因此，我们得不到正确的结果。不过，您已经看到了，GROUP BY 语句解决了这个问题。
GROUP BY 一个以上的列
我们也可以对一个以上的列应用 GROUP BY 语句，就像这样：

SELECT Customer,OrderDate,SUM(OrderPrice) FROM Orders
GROUP BY Customer,OrderDate

综合实例

> create table employee(
2> ID int,
3> name nvarchar (10),
4> salary int,
5> start_date datetime,
6> city nvarchar (10),
7> region char (1))
8> GO
1>
2> insert into employee (ID, name, salary, start_date, city, region)
3> values (1, ‘Jason’, 40420, ’02/01/94′, ‘New York’, ‘W’)
4> GO

(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (2, ‘Robert’,14420, ’01/02/95′, ‘Vancouver’,’N’)
3> GO

(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (3, ‘Celia’, 24020, ’12/03/96′, ‘Toronto’, ‘W’)
3> GO

(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (4, ‘Linda’, 40620, ’11/04/97′, ‘New York’, ‘N’)
3> GO

(1 rows affected)
1> insert into来2源gaodaima#com搞(代@码&网 employee (ID, name, salary, start_date, city, region)
2> values (5, ‘David’, 80026, ’10/05/98′, ‘Vancouver’,’W’)
3> GO

(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (6, ‘James’, 70060, ’09/06/99′, ‘Toronto’, ‘N’)
3> GO

(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (7, ‘Alison’,90620, ’08/07/00′, ‘New York’, ‘W’)
3> GO

(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (8, ‘Chris’, 26020, ’07/08/01′, ‘Vancouver’,’N’)
3> GO

(1 rows affected)
1> insert into employee (ID, name, salary, start_date, city, region)
2> values (9, ‘Mary’, 60020, ’06/09/02′, ‘Toronto’, ‘W’)
3> GO

(1 rows affected)
1>
2> * from employee
3> GO
ID name salary start_date city region
———– ———- ———– ———————– ———- ——
1 Jason 40420 1994-02-01 00:00:00.000 New York W
2 Robert 14420 1995-01-02 00:00:00.000 Vancouver N
3 Celia 24020 1996-12-03 00:00:00.000 Toronto W
4 Linda 40620 1997-11-04 00:00:00.000 New York N
5 David 80026 1998-10-05 00:00:00.000 Vancouver W
6 James 70060 1999-09-06 00:00:00.000 Toronto N
7 Alison 90620 2000-08-07 00:00:00.000 New York W
8 Chris 26020 2001-07-08 00:00:00.000 Vancouver N
9 Mary 60020 2002-06-09 00:00:00.000 Toronto W

(9 rows affected)
1>
2> –GROUP BY clause with an aggregator ‘SUM()’.
3>
4> SELECT region, SUM(Salary)
5> FROM Employee
6> WHERE ID BETWEEN 1 AND 10
7> GROUP BY Region
8> GO
region
—— ———–
N 151120
W 295106

(2 rows affected)
1>
2>
3> drop table employee
4> GO
1>

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