Date: Fri, 13 Jun 2003 10:14:46 -0400
From: Jason Tishler <jason(_at_)tishler(_dot_)net>
What about od piped through fgrep?
$ od -c crlf | fgrep '\r'
0000000 h e l l o \r \n b y e \r \n
$ od -c lf | fgrep '\r'
$
Or, just fgrep if you are willing to deal with explaining how to enter a
"^M" to the shell:
This is accomplished in bash and the version of pdksh I use by typing
^V^M
both characters being the control characters entered by typing them
into the keyboard. ^V escapes the shell's input and allows a literal
control character to be entered. I would put them into single quotes as
you have below. Without quotes, you're venturing into POSIX "undefined
behavior" territory. I can't speak for any of the other shells.
$ fgrep '^M' crlf
hello
bye
$ fgrep '^M' lf
$
Jason
I'd assume rather than simply detecting these retarded carriage returns,
one would like to remove them.
Assume "crlf" is your file containing the carriage returns (DOS-style) and
"lf" is the output file to contain no carriage returns (UNIX-style).
$ sed 's/^M$//' < crlf > lf
This uses the ^V^M trick I mentioned above. It's very important that the
two files are different filenames, else you'll overwrite your recipe stuff.
At the risk of turning this into a sed-users list, this is a commonly asked
question there, and googling will turn up a multitude of results for more
information.
Date: Fri, 13 Jun 2003 18:01:32 +0200
From: Dallman Ross <dman(_at_)nomotek(_dot_)com>
I suggest that people use the Unix 'file' command to see if a
file contains a CRLF. Unfortunately not all versions of 'file'
support this. Does anyone have suggestions for easy ways for a
newbie to find out if a file contains a CRLF?
I would think "file" would be fine, but how about:
grep -sq ^M$ spample && echo y
That's a real ^M in there. I actually created it by typing
^C^J in my shell terminal.
--
dman
Hmm, I'm not familiar with the use of that key sequence, but that doesn't
mean it doesn't work. (It just means I'm not experienced enough. ;-)
Chris
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