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Hex editing

 
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# 1  
Old 12-07-2007
Question Hex editing
Hi,
I am new to UNIX. I have a text file where each line ends on the hexadecimal character "0A". In the file there are some records that contain the Hex characters "0D0A" which I need to replace by Hex "20".

Is there a simple way to do this?

Regards,
Swanie
# 2  
Old 12-07-2007
From the top
0A = newline character which is expected in unix text files
0D0A = DOS or windows textfile cr/lf

Changing the cr/lf to 20 is the same as joining the line beneath the cr/lf line to the cr/lf line. Is that what you really want: appending lines?

dos2ux (or dos2unix) is a utility that will get rid of DOS carriage return/linefeed characters and make them standard newlines.
# 3  
Old 12-08-2007
dos2ux
Hi Jim,
Thank you for responding. That is exactly what I want to do, join the line with the CR/LF to the next line. I am using NCR UNIX which does not appear to have the utility you mention:

# man dos2ux
ERROR: Manual entry does not exist for page dos2ux
# man dos2unix
ERROR: Manual entry does not exist for page dos2unix
#

Is there any other way?

Regards,
Swanie
Last edited by Swanie; 12-08-2007 at 12:16 AM.. Reason: Misspelling
# 4  
Old 12-08-2007
If you really want to convert CRLFs to a space and your file does not contain the string "@@", the following should work

Code:
 tr "\r\n" "@@" < infile | sed 's/@@/ /g'  > outfile

If "@@" is present in your file, you can substitute another unique string to achieve the same result
 
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