Hi,
I'm not a big friend of MSSQL, but I have to do one query I've done for
mySQL.
But I don't know how...
I have to select 'user' from 'db' where first letter is E or N, second is B
or 0 and after that there are 6 or 7 digits I know.
How can I do that?
In mySQL it would be something like:
SELECT * FROM `table` WHERE `account` regexp '^[EN][B0]123456$' ORDER BY
`Id`;
Thanks in advance,
Martin
10  73763 
where ( account like 'EB%
or account like 'NB%' )
and ( account like 'E0%
or account like 'N0%' )
and len( account ) between 6 and 7
--
Tony Rogerson
SQL Server MVP http://sqlserverfaq.com - free video tutorials
"Martin Lukasik" <ma****@milea.pl.i.hate.this.spam> wrote in message
news:68**************************@ZOO.CO.UK... Hi,
I'm not a big friend of MSSQL, but I have to do one query I've done for
mySQL.
But I don't know how...
I have to select 'user' from 'db' where first letter is E or N, second is
B or 0 and after that there are 6 or 7 digits I know.
How can I do that?
In mySQL it would be something like:
SELECT * FROM `table` WHERE `account` regexp '^[EN][B0]123456$' ORDER BY
`Id`;
Thanks in advance,
Martin
"Tony Rogerson" <to**********@sqlserverfaq.com> wrote in message
news:e2*******************@news.demon.co.uk... where ( account like 'EB%
or account like 'NB%' )
and ( account like 'E0%
or account like 'N0%' )
and len( account ) between 6 and 7
Thank you,
Is there any easier way?
What if first characters are A, D, C, 4, X, Z, Y? It's not really nice to
type every permutation.
There must be some regular expression...
Martin
where account like '[EN][BO][0-9][0-9][0-9][0-9][0-9][0-9]'
or account like '[EN][BO][0-9][0-9][0-9][0-9][0-9][0-9][0-9]'
On Mon, 24 Apr 2006 16:37:44 +0100, Martin Lukasik wrote: Hi,
I'm not a big friend of MSSQL, but I have to do one query I've done for
mySQL.
But I don't know how...
(snip)In mySQL it would be something like:
SELECT * FROM `table` WHERE `account` regexp '^[EN][B0]123456$' ORDER BY
`Id`;
Hi Martin,
Try
SELECT * -- Use column list instead!!
FROM table
WHERE account LIKE '[EN][B0]123456%'
ORDER BY Id
--
Hugo Kornelis, SQL Server MVP
Hang on Martin, you didn't ask for that.
Whats your actual requirement?
You can use true regular expressions by using CLR function written in say C#
/ VB.NET.
LEFT( account, 1 ) IN ( .... )
The LIKE is more performant because it can better use an index.
--
Tony Rogerson
SQL Server MVP http://sqlserverfaq.com - free video tutorials
"Martin Lukasik" <ma****@milea.pl.i.hate.this.spam> wrote in message
news:26***************************@ZOO.CO.UK...
"Tony Rogerson" <to**********@sqlserverfaq.com> wrote in message
news:e2*******************@news.demon.co.uk... where ( account like 'EB%
or account like 'NB%' )
and ( account like 'E0%
or account like 'N0%' )
and len( account ) between 6 and 7
Thank you,
Is there any easier way?
What if first characters are A, D, C, 4, X, Z, Y? It's not really nice to
type every permutation.
There must be some regular expression...
Martin
Check out the following link from the code project. http://www.codeproject.com/managedcpp/xpregex.asp
It provides C++ external functions for SQL Server 2000 that I assume is
usable from 2005 as well. Unlike some onelse who said that LIKE is
faster it is not, this flies!
Andy Blum
> Hi Martin,
Try
SELECT * -- Use column list instead!!
FROM table
WHERE account LIKE '[EN][B0]123456%'
ORDER BY Id
That's exactly what I was looking for.
Thank you!
m. http://www.sqlservercentral.com/colu...olkitpart2.asp
"Martin Lukasik" <ma****@milea.pl.i.hate.this.spam> wrote in message
news:68**************************@ZOO.CO.UK... Hi,
I'm not a big friend of MSSQL, but I have to do one query I've done for
mySQL.
But I don't know how...
I have to select 'user' from 'db' where first letter is E or N, second is
B or 0 and after that there are 6 or 7 digits I know.
How can I do that?
In mySQL it would be something like:
SELECT * FROM `table` WHERE `account` regexp '^[EN][B0]123456$' ORDER BY
`Id`;
Thanks in advance,
Martin
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