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Generate random int value from 3 to 6
...S(CHECKSUM(NEWID()) % 6) + 1
3 through 6
SELECT ABS(CHECKSUM(NEWID()) % 4) + 3
Dynamic (Based on Eilert Hjelmeseths Comment)
SELECT ABS(CHECKSUM(NEWID()) % (@max - @min + 1)) + @min
Updated based on comments:
NEWID generates random string (for each row in return)
CHECKSUM takes value of ...
Mod in Java produces negative numbers [duplicate]
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answered Mar 21 '11 at 23:42
andrewmuandrewmu
12.8k44 gold badges3535 silver badges3636 bronze badges
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Express.js req.body undefined
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340
UPDATE July 2020
express.bodyParser() is no longer bundled as part of express. You need to inst...
Swift: Pass array by reference?
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74
Structs in Swift are passed by value, but you can use the inout modifier to modify your array (s...
How do I remove a substring from the end of a string in Python?
...h and slicing:
url = 'abcdc.com'
if url.endswith('.com'):
url = url[:-4]
Or using regular expressions:
import re
url = 'abcdc.com'
url = re.sub('\.com$', '', url)
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UPDATE and REPLACE part of a string
...work:
UPDATE dbo.xxx
SET Value = REPLACE(Value, '123\', '')
WHERE ID <=4
(I also added the \ in the replace as I assume you don't need that either)
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Array.push() if does not exist?
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114
You could extend the Array prototype with a custom method:
// check if an element exists in arr...
Select SQL Server database size
...og_size_mb = CAST(SUM(CASE WHEN type_desc = 'LOG' THEN size END) * 8. / 1024 AS DECIMAL(8,2))
, row_size_mb = CAST(SUM(CASE WHEN type_desc = 'ROWS' THEN size END) * 8. / 1024 AS DECIMAL(8,2))
, total_size_mb = CAST(SUM(size) * 8. / 1024 AS DECIMAL(8,2))
FROM sys.master_files WITH(NOWAIT)
WHE...
REDHAT 6.4 X64下ORACLE 11GR2静默安装 - 数据库(内核) - 清泛网 - 专注C/C++及内核技术
REDHAT 6.4 X64下ORACLE 11GR2静默安装前言在某些情况下，我们不具备桌面条件，只能在命令行窗口下安装oracle第一REDHAT6.4安装1.2虚拟机安装REDHAT名称随便自己起我没有外部存储...前言
在某些情况下，我们不具备桌面条件，只能在命...
Algorithm to generate all possible permutations of a list?
... at which point there is only one possible order.
So with the list [1,2,3,4] all the permutations that start with 1 are generated, then all the permutations that start with 2, then 3 then 4.
This effectively reduces the problem from one of finding permutations of a list of four items to a list of ...