Here's another option with an additional validation for strings without digits.

SELECT col1

,LEFT( PartialString, CHARINDEX( ' ', PartialString + ' '))

FROM ...

RVarn (7/20/2016)

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The basic syntax is this -

ALTER TABLE table_name

ADD CONSTRAINT constraint_name

CHECK (column_name condition);

Example -

ALTER TABLE employees

ADD CONSTRAINT check_value

CHECK (fieldname IN ('Y', 'N', NULL));

Example-

ALTER TABLE...

Ever thought about using a bit column?

From BOL

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Any scenarios not specified in the list above are not supported, including but not limited to the following:

Installing SQL Server 2005 and SQL Server 2016 on same computer (side...

Sean Lange (7/19/2016)

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Ray K (7/19/2016)

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My wife is dying to...

Is it possible? Maybe

Is it supported? No.

https://msdn.microsoft.com/en-us/library/ms143393.aspx

Have you checked fragmentation?

Apparently, the quotes and minus signs were modified by a word processor.

Here's a corrected version after performing 4 Replace All.

USE master

GO

PRINT '*******************************************************************************************'

PRINT '*******************************************************************************************'

PRINT '<SQL Server Instance Name>'

SELECT @@SERVERNAME

GO

PRINT '*******************************************************************************************'

PRINT '*******************************************************************************************'

PRINT...

Not a fan of Star Trek (and I guess my wife is grateful for that), so we're going to watch Ice Age instead.

Just be sure to understand it completely and ask any questions that you might have. 😉

Congratulations Brandie! Now ask for vacations 😀

Or the input.

Or cast the output as string, but don't do this, unless you don't plan on doing anything else with this data.

SELECT DATEADD(HOUR,-5, CAST('12:00:01' AS time)) AS cast_input,

...

Use DATEADD()

shirolkar.anand (7/19/2016)

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Hi,

you can try with below approach-

declare @t table(Parent_ID char(1),Child_ID char(1))

insert @t values ('A','B'),('C','D'),('B','C'),('B','E'),('E','F')

select t1.Parent_ID,t1.Child_ID,isnull(t2.count+t3.t3_cnt,1) Level from @t t1 left join

(select Parent_ID,count(Child_ID)count from @t group by Parent_ID) t2

on t1.Parent_ID=t2.Parent_ID

left join...

Great, just be sure to understand it perfectly. 😉