I have a log/text file which is full of highlighted ^B lines when opened with less. Usually I see ^A and I have the following sed command to replace ^A with |. I found it a while ago and do not know how it works.
sed 's/\x1/\|/g'
This seems to be the first time I have seen ^B and the sed command does not work for it.
How can I modify my sed command to handle ^A, ^B or any other possible combination?
^B is \x2, so you can just change your \x1 to \x2 to fix it. More generally, you can use "man ascii" to see the entire ASCII table and find out that ^B (hex 02) is hex 40 less than B (hex 42). Similarly, ^[ (ESC, hex 1B) is hex 40 less than [ (hex 5B).
To handle a range of characters, use brackets. For example, sed 's/[\x1-\x1f]/|/g' will replace all low-numbered control characters with vertical bars.
As others have already mentioned, \x1 is the ASCII escape sequence for the character Control-A as represented by ^A in most displays. Rather than having to deal with [ranges of] specific ASCII escape sequences, consider using a POSIX character class:
sed 's/[[:cntrl:]]/|/g'
to just change all control-characters in each input line to |s so you don't have to worry about which control characters are present or what their associated ASCII escape values are.
For example:
$ printf '\x01foo\x02bar\x03\n' | cat -A
^Afoo^Bbar^C$
$ printf '\x01foo\x02bar\x03\n' | sed 's/[[:cntrl:]]/|/g' | cat -A
|foo|bar|$
See the POSIX regexp spec for more info on bracket expressions, [...], and the character class [:cntrl:].
You don't need the \ before | btw, there's nothing special about | in sed replacement text, nor even in a Basic Regular Expression as sed uses by default.
\x1matches it. ^B is printed for ASCII code 2 so you'd need\x02.sed -n 'l0' fileand the such sequences are represented by\nnne.gprintf '\x01foo\x02bar\x03baz' | sed -n l0will output\001foo\002bar\003baz$. Thus to replace the first control sequence usingprintf '\x01foo\x02bar\x03baz' | sed 's/\o001/X/'will outputXfoobarbaz.