Sql server TSQL：基于另一列的值递增列值（SQL Server）
我有一个旧的存储过程，我正试图修改它 这是正在修改的表中的数据转储 每次PayrollRateID第2列（此处显示的示例值为25、27、28等）递增时，“Incrementer”列需要递增 ID列是使用行号生成的 编写存储过程的人正在使用一列向下遍历表中的每一行，并在每次PayrollRateID值出现时设置递增列的值。不幸的是，这需要几分钟来完成20000多行。这只是几天的数据，如果我用一个月的时间运行，那么，你可以想象我遇到的性能问题 存储过程的逻辑是以递增值0开始，并以示例-25中的第一个PayrollRateID为起点，在PayrollRateID更改时，给出所有这些0，然后递增递增列中的计数，并在示例-27中给出所有下一个值a 1，在我的例子28中，当它变为下一个值时，给递增列a 2，以此类推Sql server TSQL：基于另一列的值递增列值（SQL Server）,sql-server,sql-server-2008,tsql,Sql Server,Sql Server 2008,Tsql,我有一个旧的存储过程，我正试图修改它 这是正在修改的表中的数据转储 每次PayrollRateID第2列（此处显示的示例值为25、27、28等）递增时，“Incrementer”列需要递增 ID列是使用行号生成的 编写存储过程的人正在使用一列向下遍历表中的每一行，并在每次PayrollRateID值出现时设置递增列的值。不幸的是，这需要几分钟来完成20000多行。这只是几天的数据，如果我用一个月的时间运行，那么，你可以想象我遇到的性能问题 存储过程的逻辑是以递增值0开始，并以示例-25中的第一个
Date
PayrollRateID
DayNum
Variance
VarianceID
Incrementer
ID
2011-07-25 00:00:00.000 25 1 1.00 0 0 1
2011-07-26 00:00:00.000 25 2 1.00 0 0 2
2011-07-27 00:00:00.000 25 3 1.00 0 0 3
2011-07-28 00:00:00.000 25 4 1.00 0 0 4
2011-07-29 00:00:00.000 25 5 1.00 0 0 5
2011-07-30 00:00:00.000 25 6 1.00 0 0 6
2011-07-31 00:00:00.000 25 7 1.00 0 0 7
2011-08-01 00:00:00.000 25 1 1.00 0 0 8
2011-08-02 00:00:00.000 25 2 1.00 0 0 9
2011-08-03 00:00:00.000 25 3 1.00 0 0 10
2011-08-04 00:00:00.000 25 4 1.00 0 0 11
2011-08-05 00:00:00.000 25 5 1.00 0 0 12
2011-08-06 00:00:00.000 25 6 1.00 0 0 13
2011-08-07 00:00:00.000 25 7 1.00 0 0 14
2011-08-08 00:00:00.000 25 1 1.00 0 0 15
2011-08-09 00:00:00.000 25 2 1.00 0 0 16
2011-08-10 00:00:00.000 25 3 1.00 0 0 17
2011-08-11 00:00:00.000 25 4 1.00 0 0 18
2011-08-12 00:00:00.000 25 5 1.00 0 0 19
2011-08-13 00:00:00.000 25 6 1.00 0 0 20
2011-08-14 00:00:00.000 25 7 1.00 0 0 21
2011-07-25 00:00:00.000 27 1 1.00 0 1 22
2011-07-26 00:00:00.000 27 2 1.00 0 1 23
2011-07-27 00:00:00.000 27 3 1.00 0 1 24
2011-07-28 00:00:00.000 27 4 1.00 0 1 25
2011-07-29 00:00:00.000 27 5 1.00 0 1 26
2011-07-30 00:00:00.000 27 6 1.00 0 1 27
2011-07-31 00:00:00.000 27 7 1.00 0 1 28
2011-08-01 00:00:00.000 27 1 1.00 0 1 29
2011-08-02 00:00:00.000 27 2 1.00 0 1 30
2011-08-03 00:00:00.000 27 3 1.00 0 1 31
2011-08-04 00:00:00.000 27 4 1.00 0 1 32
2011-08-05 00:00:00.000 27 5 1.00 0 1 33
2011-08-06 00:00:00.000 27 6 1.00 0 1 34
2011-08-07 00:00:00.000 27 7 1.00 0 1 35
2011-08-08 00:00:00.000 27 1 1.00 0 1 36
2011-08-09 00:00:00.000 27 2 1.00 0 1 37
2011-08-10 00:00:00.000 27 3 1.00 0 1 38
2011-08-11 00:00:00.000 27 4 1.00 0 1 39
2011-08-12 00:00:00.000 27 5 1.00 0 1 40
2011-08-13 00:00:00.000 27 6 1.00 0 1 41
2011-08-14 00:00:00.000 27 7 1.00 0 1 42
2011-07-25 00:00:00.000 28 1 1.00 0 2 43
2011-07-26 00:00:00.000 28 2 1.00 0 2 44
2011-07-27 00:00:00.000 28 3 1.00 0 2 45
2011-07-28 00:00:00.000 28 4 1.00 0 2 46
2011-07-29 00:00:00.000 28 5 1.00 0 2 47
2011-07-30 00:00:00.000 28 6 1.00 0 2 48
2011-07-31 00:00:00.000 28 7 1.00 0 2 49
2011-08-01 00:00:00.000 28 1 1.00 0 2 50
2011-08-02 00:00:00.000 28 2 1.00 0 2 51
2011-08-03 00:00:00.000 28 3 1.00 0 2 52
2011-08-04 00:00:00.000 28 4 1.00 0 2 53
2011-08-05 00:00:00.000 28 5 1.00 0 2 54
2011-08-06 00:00:00.000 28 6 1.00 0 2 55
2011-08-07 00:00:00.000 28 7 1.00 0 2 56
2011-08-08 00:00:00.000 28 1 1.00 0 2 57
2011-08-09 00:00:00.000 28 2 1.00 0 2 58
2011-08-10 00:00:00.000 28 3 1.00 0 2 59
2011-08-11 00:00:00.000 28 4 1.00 0 2 60
2011-08-12 00:00:00.000 28 5 1.00 0 2 61
2011-08-13 00:00:00.000 28 6 1.00 0 2 62
2011-08-14 00:00:00.000 28 7 1.00 0 2 63
2011-07-25 00:00:00.000 34 1 1.00 0 3 64
2011-07-26 00:00:00.000 34 2 1.00 0 3 65
2011-07-27 00:00:00.000 34 3 1.00 0 3 66
2011-07-28 00:00:00.000 34 4 1.00 0 3 67
2011-07-29 00:00:00.000 34 5 1.00 0 3 68
2011-07-30 00:00:00.000 34 6 1.00 0 3 69
2011-07-31 00:00:00.000 34 7 1.00 0 3 70
2011-08-01 00:00:00.000 34 1 1.00 0 3 71
2011-08-02 00:00:00.000 34 2 1.00 0 3 72
2011-08-03 00:00:00.000 34 3 1.00 0 3 73
2011-08-04 00:00:00.000 34 4 1.00 0 3 74
2011-08-05 00:00:00.000 34 5 1.00 0 3 75
2011-08-06 00:00:00.000 34 6 1.00 0 3 76
2011-08-07 00:00:00.000 34 7 1.00 0 3 77
2011-08-08 00:00:00.000 34 1 1.00 0 3 78
2011-08-09 00:00:00.000 34 2 1.00 0 3 79
2011-08-10 00:00:00.000 34 3 1.00 0 3 80
2011-08-11 00:00:00.000 34 4 1.00 0 3 81
2011-08-12 00:00:00.000 34 5 1.00 0 3 82
2011-08-13 00:00:00.000 34 6 1.00 0 3 83
2011-08-14 00:00:00.000 34 7 1.00 0 3 84
2011-07-25 00:00:00.000 38 1 1.00 0 4 85
2011-07-26 00:00:00.000 38 2 1.00 0 4 86
2011-07-27 00:00:00.000 38 3 1.00 0 4 87
2011-07-28 00:00:00.000 38 4 1.00 0 4 88
2011-07-29 00:00:00.000 38 5 1.00 0 4 89
2011-07-30 00:00:00.000 38 6 1.00 0 4 90
2011-07-31 00:00:00.000 38 7 1.00 0 4 91
2011-08-01 00:00:00.000 38 1 1.00 0 4 92
2011-08-02 00:00:00.000 38 2 1.00 0 4 93
2011-08-03 00:00:00.000 38 3 1.00 0 4 94
2011-08-04 00:00:00.000 38 4 1.00 0 4 95
2011-08-05 00:00:00.000 38 5 1.00 0 4 96
2011-08-06 00:00:00.000 38 6 1.00 0 4 97
2011-08-07 00:00:00.000 38 7 1.00 0 4 98
2011-08-08 00:00:00.000 38 1 1.00 0 4 99
2011-08-09 00:00:00.000 38 2 1.00 0 4 100
2011-08-10 00:00:00.000 38 3 1.00 0 4 101
2011-08-11 00:00:00.000 38 4 1.00 0 4 102
2011-08-12 00:00:00.000 38 5 1.00 0 4 103
2011-08-13 00:00:00.000 38 6 1.00 0 4 104
2011-08-14 00:00:00.000 38 7 1.00 0 4 105
2011-07-25 00:00:00.000 40 1 1.00 0 5 106
2011-07-26 00:00:00.000 40 2 1.00 0 5 107
2011-07-27 00:00:00.000 40 3 1.00 0 5 108
2011-07-28 00:00:00.000 40 4 1.00 0 5 109
2011-07-29 00:00:00.000 40 5 1.00 0 5 110
2011-07-30 00:00:00.000 40 6 1.00 0 5 111
2011-07-31 00:00:00.000 40 7 1.00 0 5 112
2011-08-01 00:00:00.000 40 1 1.00 0 5 113
2011-08-02 00:00:00.000 40 2 1.00 0 5 114
2011-08-03 00:00:00.000 40 3 1.00 0 5 115
2011-08-04 00:00:00.000 40 4 1.00 0 5 116
2011-08-05 00:00:00.000 40 5 1.00 0 5 117
2011-08-06 00:00:00.000 40 6 1.00 0 5 118
2011-08-07 00:00:00.000 40 7 1.00 0 5 119
2011-08-08 00:00:00.000 40 1 1.00 0 5 120
2011-08-09 00:00:00.000 40 2 1.00 0 5 121
2011-08-10 00:00:00.000 40 3 1.00 0 5 122
2011-08-11 00:00:00.000 40 4 1.00 0 5 123
2011-08-12 00:00:00.000 40 5 1.00 0 5 124
2011-08-13 00:00:00.000 40 6 1.00 0 5 125
2011-08-14 00:00:00.000 40 7 1.00 0 5 126
2011-07-25 00:00:00.000 41 1 1.00 0 6 127
2011-07-26 00:00:00.000 41 2 1.00 0 6 128
2011-07-27 00:00:00.000 41 3 1.00 0 6 129
2011-07-28 00:00:00.000 41 4 1.00 0 6 130
2011-07-29 00:00:00.000 41 5 1.00 0 6 131
2011-07-30 00:00:00.000 41 6 1.00 0 6 132
2011-07-31 00:00:00.000 41 7 1.00 0 6 133
2011-08-01 00:00:00.000 41 1 1.00 0 6 134
2011-08-02 00:00:00.000 41 2 1.00 0 6 135
2011-08-03 00:00:00.000 41 3 1.00 0 6 136
2011-08-04 00:00:00.000 41 4 1.00 0 6 137
2011-08-05 00:00:00.000 41 5 1.00 0 6 138
2011-08-06 00:00:00.000 41 6 1.00 0 6 139
2011-08-07 00:00:00.000 41 7 1.00 0 6 140
2011-08-08 00:00:00.000 41 1 1.00 0 6 141
2011-08-09 00:00:00.000 41 2 1.00 0 6 142
2011-08-10 00:00:00.000 41 3 1.00 0 6 143
2011-08-11 00:00:00.000 41 4 1.00 0 6 144
2011-08-12 00:00:00.000 41 5 1.00 0 6 145
2011-08-13 00:00:00.000 41 6 1.00 0 6 146
2011-08-14 00:00:00.000 41 7 1.00 0 6 147
2011-07-25 00:00:00.000 42 1 1.00 0 7 148
2011-07-26 00:00:00.000 42 2 1.00 0 7 149
2011-07-27 00:00:00.000 42 3 1.00 0 7 150
2011-07-28 00:00:00.000 42 4 1.00 0 7 151
2011-07-29 00:00:00.000 42 5 1.00 0 7 152
2011-07-30 00:00:00.000 42 6 1.00 0 7 153
2011-07-31 00:00:00.000 42 7 1.00 0 7 154
2011-08-01 00:00:00.000 42 1 1.00 0 7 155
2011-08-02 00:00:00.000 42 2 1.00 0 7 156
2011-08-03 00:00:00.000 42 3 1.00 0 7 157
2011-08-04 00:00:00.000 42 4 1.00 0 7 158
2011-08-05 00:00:00.000 42 5 1.00 0 7 159
2011-08-06 00:00:00.000 42 6 1.00 0 7 160
2011-08-07 00:00:00.000 42 7 1.00 0 7 161
2011-08-08 00:00:00.000 42 1 1.00 0 7 162
2011-08-09 00:00:00.000 42 2 1.00 0 7 163
2011-08-10 00:00:00.000 42 3 1.00 0 7 164
2011-08-11 00:00:00.000 42 4 1.00 0 7 165
2011-08-12 00:00:00.000 42 5 1.00 0 7 166
2011-08-13 00:00:00.000 42 6 1.00 0 7 167
2011-08-14 00:00:00.000 42 7 1.00 0 7 168
2011-07-25 00:00:00.000 44 1 1.00 0 8 169
2011-07-26 00:00:00.000 44 2 1.00 0 8 170
2011-07-27 00:00:00.000 44 3 1.00 0 8 171
2011-07-28 00:00:00.000 44 4 1.00 0 8 172
2011-07-29 00:00:00.000 44 5 1.00 0 8 173
SQL Server 2008。不确定我是否理解存储过程中的逻辑，但是，您是否可以创建另一个表，例如[PayrollIncrementer]，该表具有ID标识种子值，并在需要增加ID时插入表中
CREATE TABLE [dbo].[PayrollIncrementer]
(
[Id] INT NOT NULL IDENTITY(1),
[DateAdded] DATETIME NOT NULL
)
BEGIN
-- PROC BEGIN
DECLARE @id INT
INSERT INTO [dbo].[PayrollIncrementer] ([DateAdded]) VALUES (GETDATE())
SET @id = SCOPE_IDENTITY()
-- remainder of proc & use the identity
-- PROC END
END
Update a
Set Incrementer = b.Incrementer
From dbo.YourTable a
Join
(
Select PayRollId,
Row_Number() Over(Order By PayRollId) as [Incrementer]
From dbo.YourTable
Group By PayRollId
)b on a.PayRollId = b.PayRollId
用密集秩代替行数怎么样？这将一次完成所有操作。我无法完整地运行它，但内部选择提供了与我之前得到的相同的效果，即递增器将为每个PayrollRateID递增，然后为下一个ID重置。我所追求的效果是，所有PayrollRateID都获得相同的值，递增器为每个PayrollRateID提供不同的值PayrollRateID@cometbill当前位置更新了我的答案，我想这会帮你解决问题。在第二行，a.递增器关闭，工作正常。否则我会得到一个“无法绑定多部分标识符a.Incrementer”。错误。回答得好。你也可以使用稠密等级-我认为这会更容易。从表中选择[Date]、PayrollRateID、DayNum、Variance、DENSE\u按PayrollRateID排序作为VarianceID、IncrementerID