How to Remove Carriage Returns from database field?

I have saved user input from an html text area into a MS SQL 2000
database in the process replacing the \n character with a <br> tag.
Unfortunately I did not replace the \r at the same time so now have
many fields that have the following
Line 1
Line 2
in SQL table it looks like Line 1x<br>Line 2x<br> ... etc where the x
appears as a little square box.
I then present this data in an html table which works fine. If the user
copy pastes into Excel then each of the \r characters cause Excel to
create a new cell. This is not fine.
Can someone tell me how I can access the \r characters in the table so
that I can replace them.
ie Select * From tab where col like '%\r'
does not contain any records
thanks for the helpHi Peter !
See if the values are related to the value of which is a carriage
return CHAR(13)
http://www.asciitable.com, then REPLACE it with the REPLACE function to
whatever you want to display.
HTH, Jens Suessmeyer.|||Try:
update MyTable
set
MyCol = replace (MyCol, char (13), '')
where
MyCol like '%' + char (13)
--
Tom
----
Thomas A. Moreau, BSc, PhD, MCSE, MCDBA
SQL Server MVP
Columnist, SQL Server Professional
Toronto, ON Canada
www.pinpub.com
.
<peter.rietmann@.swisscom.com> wrote in message
news:1129721023.545769.275420@.o13g2000cwo.googlegroups.com...
I have saved user input from an html text area into a MS SQL 2000
database in the process replacing the \n character with a <br> tag.
Unfortunately I did not replace the \r at the same time so now have
many fields that have the following
Line 1
Line 2
in SQL table it looks like Line 1x<br>Line 2x<br> ... etc where the x
appears as a little square box.
I then present this data in an html table which works fine. If the user
copy pastes into Excel then each of the \r characters cause Excel to
create a new cell. This is not fine.
Can someone tell me how I can access the \r characters in the table so
that I can replace them.
ie Select * From tab where col like '%\r'
does not contain any records
thanks for the help|||Just to be a nut picker ;-).
You don=B4t need that:
where MyCol like '%' + char (13)
Jens Suessmeyer.|||that's great, it has solved my problem. thanks