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CRLF removal
Posted: Fri Feb 05, 2021 2:27 am
by danwinslow
I have a string in the structure Am called Text. I want to remove any occurrences of the $0d$0a character pattern (cr/lf).
This does not seem to work. The string text is unchanged. What am I doing wrong?
Code: Select all
Am\Text=ReplaceString(Am\text, Str($0d)+Str($0a), "")
*edit*
Ach, never mind. I need Chr() I think.
*re-edit*
Er nope. Is this a UTF vs. Ascii type issue? I notice the strings are two byte characters.
*re-re-edit
Yeah, Am\Text=ReplaceString(Am\text, Str($0d)+chr(0)+Str($0a)+str(0), "") sort of works, except it replaces them with spaces (Chr(20)+Chr(0)) rather than trimming.
Re: CRLF removal
Posted: Fri Feb 05, 2021 3:31 am
by Mijikai
Use the constant:
Code: Select all
EnableExplicit
Global str.s
str = "hello world!" + #CRLF$ + "123"
MessageRequester("",str)
str = ReplaceString(str,#CRLF$,#Null$)
MessageRequester("",str)
End
Re: CRLF removal
Posted: Fri Feb 05, 2021 6:26 am
by TI-994A
danwinslow wrote:...to remove any occurrences of the $0d$0a character pattern (cr/lf)...
The PureBasic
#CRLF$ string constant is just a convenient compound constant representing the
carriage return character
(CR = ASCII-13 = HEX-0D), and the
line feed character
(LF = ASCII-10 = HEX-0A). Their values can be called with the
Chr() function, as follows:
Code: Select all
;line feeds
Chr(10) ;ASCII
Chr($0A) ;HEX
;carriage returns
Chr(13) ;ASCII
Chr($0D) ;HEX
;interchangeable values of compound constant #CRLF$
If #CRLF$ = Chr(13) + Chr(10) : Debug "same value" : EndIf
If #CRLF$ = Chr($0D) + Chr($0A) : Debug "same value" : EndIf
If #CRLF$ = Chr(13) + Chr($0A) : Debug "same value" : EndIf
If #CRLF$ = Chr($0D) + Chr(10) : Debug "same value" : EndIf
As per your intended search/replace usage, they all work quite interchangeably:
Code: Select all
Structure AM_struct
text1.s
text2.s
EndStructure
Define AM.AM_struct
AM\text1 = "A" + Chr($0D) + Chr($0A) +
"string" + Chr($0D) + Chr($0A) +
"with" + Chr($0D) + Chr($0A) +
"many" + Chr($0D) + Chr($0A) +
"linefeeds."
AM\text2 = "Another" + #CRLF$ +
"string" + #CRLF$ +
"with" + #CRLF$ +
"many" + #CRLF$ +
"linefeeds."
;finding Chr($0D)+Chr($0A) with #CRLF$
AM\text1 = ReplaceString(AM\text1, #CRLF$, " ")
;finding #CRLF$ with Chr($0D)+Chr($0A)
AM\text2 = ReplaceString(AM\text2, Chr($0D) + Chr($0A), " ")
Debug AM\text1
Debug AM\text2
Re: CRLF removal
Posted: Fri Feb 05, 2021 7:08 am
by Marc56us
I want to remove any occurrences of the $0d$0a character pattern (cr/lf).
"
Think simple"
RemoveString() 
Code: Select all
A$ = "Hello " + #CRLF$ + "World!"
Debug A$
A$ = RemoveString(A$, #CRLF$)
Debug ""
Debug A$
Re: CRLF removal
Posted: Fri Feb 05, 2021 1:45 pm
by danwinslow
Thanks all.
As it mentions above, I did try replacestring, which didn't work:
Code: Select all
Am\Text=ReplaceString(Am\text, Str($0d)+Str($0a), "")
I don't know whether it was the empty string as 3rd param or not using the #CRLF$ constant, but it did not work.
Re: CRLF removal
Posted: Fri Feb 05, 2021 1:59 pm
by Kiffi
danwinslow wrote:Code: Select all
Am\Text=ReplaceString(Am\text, Str($0d)+Str($0a), "")
read carefully:
Code: Select all
Am\Text=ReplaceString(Am\text, Chr($0d)+Chr($0a), "")
or:
Code: Select all
Am\Text=RemoveString(Am\text, Chr($0d)+Chr($0a))
Re: CRLF removal
Posted: Fri Feb 05, 2021 2:04 pm
by infratec
Your version will never work
We showed you above how it works.
What does Str()
It converts a number to a string
So you do Str(13) + Str(10) which results in "1310"
For testing:
Code: Select all
Debug Str($0d) + Str($0a)
Debug #CR$ + #LF$
Debug "----"
Re: CRLF removal
Posted: Fri Feb 05, 2021 4:29 pm
by mk-soft
You don't need the Chr function. For the control characters 1 to 31, these are available as constants.
Code: Select all
Debug Asc(#TAB$)
Debug Asc(#CR$)
Debug Asc(#LF$)
Debug Asc(#STX$)
Debug Asc(#ETX$)
; etc
Re: CRLF removal
Posted: Fri Feb 05, 2021 10:30 pm
by danwinslow
Yep, was using ReplaceString instead of RemoveString. RemoveString worked much better.
To the question above, Str() converts a number to its string representation.
Thanks for the help.
Dan