显示每个经理下的所有员工角色
在SQL Server 2008中，我有一个包含5列的表tblStock：
Manager，Employee，Value1，Value2，Value3。
样本数据将是，
我编写了以下代码来获取上面的示例输出：
Highcharts.chart('containerChart', {
chart: {
type: 'bar',
marginLeft: 0,
marginTop: 100
},
title: {
text: 'Reject Category',
align: 'left',
x: 30,
y: 35,
style: {
color: '#666666',
fontWeight: 'normal',
fontSize: 12,
fontFamily: 'Lucida Grande,Lucida Sans Unicode, Arial, Helvetica, sans-serif',
}
},
tooltip: {
enabled: false,
valueSuffix: ' %',
}
},
exporting: {
enabled: false,
},
credits: {
enabled: false,
},
plotOptions: {
bar: {
dataLabels: {
enabled: false
},
},
series: {
pointWidth: 25,
}, states: {
hover: {
enabled: false
}
}
},
},
xAxis: [{
categories: arrcategory,
width: 100,
left: 140,
height: 34 * ctglength,
lineWidth: 0,
minorGridLineWidth: 0,
minorTickLength: 0,
// tickLength: 0,
title: {
text: null,
},
labels: {
overflow: 'justify',
enabled: true,
style: {
fontWeight: 'bold',
}
},
gridLineWidth: 0
}, {
categories: true,
linkedTo: 0,
width: 100,
height: 34 * ctglength,
left: 358,
visible: false
}],
yAxis: [{
min: 0,
max: 100,
width: 400,
left: 140,
//gridLineWidth: 0,
labels: {
overflow: 'justify',
enabled: false
},
title: {
text: '# of Transactions',
"textAlign": 'right',
"rotation": 0,
x: 15,
y: -310
},
}, {
min: 0,
max: 100,
left: 700,
width: 400,
// gridLineWidth: 0,
labels: {
overflow: 'justify',
enabled: false
},
title: {
text: '$ in Premium',
"textAlign": 'right',
"rotation": 0,
x: 10,
y: -330
},
}],
series: [{
data: arrdata1,
color: '#E95E4F',
showInLegend: false,
//enableMouseTracking: false,
dataLabels: {
overflow: false,
enabled: true,
crop: false,
}, {
data: arrdata2,
color: '#E95E4F',
showInLegend: false,
//enableMouseTracking: false,
dataLabels: {
overflow: false,
enabled: true,
crop: false,
},
yAxis: 1,
xAxis: 1
}]
});
我想从上表中获得以下输出，
在这个输出中，我提到的绿色行是属于特定管理者的员工的总和。
任何人都知道这方面的解决方案。
5 个答案:
答案 0 :(得分：2)
这是一种方法：
SELECT Name, Role, Value1, Value2, Value3
FROM
(
SELECT Manager As Name,
'Manager' As Role,
SUM(Value1) As Value1,
SUM(Value2) AS Value2,
SUM(Value3) AS Value3,
Manager
FROM #tblDepartment
GROUP BY Manager
UNION ALL
SELECT Employee As Name,
'Employee' As Role,
Value1,
Value2,
Value3,
Manager
FROM #tblDepartment
) t
ORDER BY Manager,Role DESC, Name
这与您的订单不完全相同，但员工都在正确的经理之下。
答案 1 :(得分：2)
获得结果的另一种方法
WITH mCTE AS (
SELECT a.Manager, a.Manager AS Employee, 'Manager' AS Role, SUM(a.Value1) AS Value1, SUM(a.Value2) AS Value2, SUM(a.Value3) AS Value3, 1 AS r
FROM #tblDepartment AS a
GROUP BY a.Manager
UNION ALL
SELECT a.Manager, a.Employee, 'Employee' AS Role, a.Value1, a.Value2, a.Value3, 2 AS r
FROM #tblDepartment AS a
)
SELECT Employee, Role, Value1, Value2, Value3
FROM mCTE
ORDER BY Manager, r
答案 2 :(得分：1)
考虑到您的订单，这是最接近您的结果：
select Name, Role, Value1, Value2, Value3
from (
select t.MgrId, t.manager as Name, 'Manager' as Role, sum(t.Value1) Value1, sum(t.Value2) Value2, sum(t.Value3) Value3
from (select *, dense_rank () OVER(ORDER BY Manager desc) AS MgrId from #tblDepartment ) t
group by t.MgrId, t.manager
union all
select t.MgrId, t.employee as Name, 'Employee' as Role, sum(t.Value1) Value1, sum(t.Value2) Value2, sum(t.Value3) Value3
from (select *, dense_rank () OVER(ORDER BY Manager desc) AS MgrId from #tblDepartment ) t
group by t.MgrId, t.employee
) t
order by t.MgrId, t.Role desc
您可以查看工作版本here。 排名函数从SQL Server 2008开始提供，因此它也适用于您。
答案 3 :(得分：1)
试试这个：
declare @tb table(manager varchar(50),employee varchar(50), value1 int,value2 int,value3 int)
insert into @tb
SELECT 'Theva' Manager, 'Lawlrence' Employee, 10 Value1, 20 Value2, 60 Value3
UNION ALL
SELECT 'Theva', 'David', 20, 35, 42
UNION ALL
SELECT 'Theva', 'Ragav', 45, 35, 86
UNION ALL
SELECT 'Prem', 'Vino', 69, 99, 45
UNION ALL
SELECT 'Prem', 'Lara', 27, 99, 45
UNION ALL
SELECT 'Anzal', 'Ranjani', 65, 55, 12
UNION ALL
SELECT 'Anzal', 'Priya', 55, 47, 89
UNION ALL
SELECT 'Anzal', 'Vinoth', 98, 53, 56
UNION ALL
SELECT 'Rafeek', 'Ashok', 48, 75, 45
select case when len(employee)>0 then employee else a.manager end as Name,case when len(employee)>0 then 'Employee' else 'Manager' end as [Role],b.value1,b.value2,b.value3 from
(select distinct manager,'Manager' as Role from @tb
union all
select employee,'employee' as Role from @tb) as a
left join
(select distinct a.manager,employee,value1,value2,value3 from
(select manager,'' as employee,sum(value1)value1,sum(value2)value2,sum(value3)value3 from @tb group by manager
union all
select a.manager,b.employee,b.value1,b.value2,b.value3 from
(select manager,sum(value1)value1,sum(value2)value2,sum(value3)value3 from @tb group by manager) as a
left join
(select * from @tb) as b
on a.manager = b.manager) as a) as b
on a.manager = b.manager where value1 is not null
结果：
Anzal Manager 218 155 157
Priya Employee 55 47 89
Ranjani Employee 65 55 12
Vinoth Employee 98 53 56
Prem Manager 96 198 90
Lara Employee 27 99 45
Vino Employee 69 99 45
Rafeek Manager 48 75 45
Ashok Employee 48 75 45
Theva Manager 75 90 188
David Employee 20 35 42
Lawlrence Employee 10 20 60
Ragav Employee 45 35 86
答案 4 :(得分：1)
这可能不是OP的理想答案，我尝试使用SQL 2012重新创建它，这就是我想出来的。
IF OBJECT_ID('tempdb..#tblDepartment') IS NOT NULL DROP TABLE #tblDepartment
SELECT * INTO #tblDepartment
FROM
(
SELECT 'Theva' Manager, 'Lawlrence' Employee, 10 Value1, 20 Value2, 60 Value3
UNION ALL
SELECT 'Theva', 'David', 20, 35, 42
UNION ALL
SELECT 'Theva', 'Ragav', 45, 35, 86
UNION ALL
SELECT 'Prem', 'Vino', 69, 99, 45
UNION ALL
SELECT 'Prem', 'Lara', 27, 99, 45
UNION ALL
SELECT 'Anzal', 'Ranjani', 65, 55, 12
UNION ALL
SELECT 'Anzal', 'Priya', 55, 47, 89
UNION ALL
SELECT 'Anzal', 'Vinoth', 98, 53, 56
UNION ALL
SELECT 'Rafeek', 'Ashok', 48, 75, 45
)TAB
DECLARE @T AS TABLE
(
ID INT IDENTITY(1, 1)
PRIMARY KEY ,
NAME VARCHAR(50) ,
Role VARCHAR(50) ,
Value1 INT ,
Value2 INT ,
Value3 INT
);
WITH CTE
AS ( SELECT Manager ,
Employee ,
SUM(Value1) AS Value1 ,
SUM(Value2) AS Value2 ,
SUM(Value3) AS Value3
FROM #tblDepartment
GROUP BY Manager ,
Employee
WITH ROLLUP
)
INSERT INTO @T
( NAME, Role, Value1, Value2, Value3 )
SELECT Name ,
Role ,
Value1 ,
Value2 ,
Value3
FROM ( SELECT ISNULL(Employee, Manager) NAME ,
IIF(Employee IS NULL, 'Manager', 'Employee') ROLE ,
Value1 ,
Value2 ,
Value3
FROM CTE
) T
WHERE t.NAME IS NOT NULL
SELECT Name ,
Role ,
Value1 ,
Value2 ,
Value3 FROM @T
ORDER BY ID DESC
结果：
Name Role Value1 Value2 Value3
--------------------- ------------------ ----------- ----------- -----------
Theva Manager 75 90 188
Ragav Employee 45 35 86
Lawlrence Employee 10 20 60
David Employee 20 35 42
Rafeek Manager 48 75 45
Ashok Employee 48 75 45
Prem Manager 96 198 90
Vino Employee 69 99 45
Lara Employee 27 99 45
Anzal Manager 218 155 157
Vinoth Employee 98 53 56
Ranjani Employee 65 55 12
Priya Employee 55 47 89
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