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 Invalid IP Address

Post time: 2020-2-5 22:30:01
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Is there such a function in sql server, which is to remove the carriage return from the string with carriage return.
as me
      is having a
      Questions to ask "
Becomes "I have a question to ask"
Does anyone know?
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 Germany

Post time: 2020-3-24 13:30:02
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select replace ("I
      is having a
      Question to ask ", carriage return symbol, '')
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 China

Post time: 2020-3-24 14:15:02
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select replace ('"I
      is having a
      Question to ask "', carriage return symbol,")
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 China

Post time: 2020-3-27 22:15:01
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In sql,
char (10)-newline character
char (13)-carriage return

for example:
declare @ t1 varchar (50)
set @ t1 = 'aaaaa' + char (13) + 'bbbbbb'
print @ t1
result--------------------
aaaaa
bbbbbb


Just replace this carriage return:
declare @ t1 varchar (50)
set @ t1 = 'aaaaa' + char (13) + 'bbbbbb'
set @ t1 = replace (@ t1, char (13), '')
print @ t1

result------------------
aaaaabbbbbb
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 Invalid IP Address

Post time: 2020-3-28 22:30:01
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declare @str varchar (100)
set @ str = 'me
is having a
Questions to ask '

select replace (replace (@str, char (13), ''), char (10), '')
--result
I have a question to ask

(1 row (s) affected)
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 China

Post time: 2020-3-29 08:15:01
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Upstairs solution!
Ha ha
Add a little bit ~

declare @ t1 varchar (50)
set @ t1 = 'aaaaa
bbbbbb '
set @ t1 = replace (replace (@ t1, char (13), ''), char (10), '')
print @ t1
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 China

Post time: 2020-4-1 23:30:01
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set ANSI_NULLS ON
set QUOTED_IDENTIFIER ON
GO
create function [dbo]. [fn_removeChr13Chr10]
(
@str varchar (max) = 'helloworld',
@ IsRemoveChr13 bit = 1,
@ IsRemoveChr10 bit = 1
)
RETURNS varchar (max)
as
begin

--- remove the key char (13) [return] in the string
if @ IsRemoveChr13 = 1
begin
select @ str = replace (@ str, char (13), '')
end

--- remove the key char (10) [line feed] in the string
if @ IsRemoveChr10 = 1
begin
select @ str = replace (@ str, char (10), '')
end

return @str

end
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