[SOLVED] sed separator (delimiter) replacement not working with p command

Issue

I try to replace / with # separators in sed print command but don't succeed. Please look at the third line beneath:

u@debian:~$ echo A | sed -n '/A/p'
A
u@debian:~$ echo A | sed -n '#A#p'
u@debian:~$ echo A | sed -n s#A#B#p
B
u@debian:~$ 

How to replace it?

Solution

Preamble

Reading the title of your question, I think you have to read quietly the address chapter* in info sed!

Understanding the difference between addressing and commands! s, like p are commands!

So your request is about addressing for executing a command.

Address range and address for commands

Little sample: comment all lines that contain badWord:

sed -e '/badWord/s/^/# /' -i file

More complete sample:

info sed |
  sed -ne '
    /^4.3/,/^5/{
        /^\(\o47\|\o342\o200\o230\){
           :a;
            N;
            /\n / ! ba;
            N;
            p
         }
    }'

• From the line that begin by 4.3 to the line that begin by 5,
  • On lines that begin by a quote ' (or a UTF8 open quote: ‘),
    • Place a label a.
    • Append next line
    • If current buffer does not contain a newline followed by one space, the branch to label a.
    • Append one more line
    • print current buffer.

'/REGEXP/'
     This will select any line which matches the regular expression
     REGEXP.  If REGEXP itself includes any '/' characters, each must be
'\%REGEXP%'
     (The '%' may be replaced by any other single character.)

'/REGEXP/I'
'\%REGEXP%I'
     The 'I' modifier to regular-expression matching is a GNU extension
     which causes the REGEXP to be matched in a case-insensitive manner.
'/REGEXP/M'
'\%REGEXP%M'
     The 'M' modifier to regular-expression matching is a GNU 'sed'
     extension which directs GNU 'sed' to match the regular expression
'/[0-9]/p' matches lines with digits and prints them.  Because the
second line is changed before the '/[0-9]/' regex, it will not match and
will not be printed:

     $ seq 3 | sed -n 's/2/X/ ; /[0-9]/p'
     1
'0,/REGEXP/'
     A line number of '0' can be used in an address specification like
     '0,/REGEXP/' so that 'sed' will try to match REGEXP in the first
'ADDR1,+N'
     Matches ADDR1 and the N lines following ADDR1.

'ADDR1,~N'
     Matches ADDR1 and the lines following ADDR1 until the next line
     whose input line number is a multiple of N.  The following command

Please, RTFM:

Have a look at info sed, search for * sed addresses, then * Regexp Addresses:

‘/REGEXP/’
     This will select any line which matches the regular expression
     REGEXP.  If REGEXP itself includes any ‘/’ characters, each must be
     escaped by a backslash (‘\’).

...

‘\%REGEXP%’
     (The ‘%’ may be replaced by any other single character.)

     This also matches the regular expression REGEXP, but allows one to
     use a different delimiter than ‘/’.  This is particularly useful if
     the REGEXP itself contains a lot of slashes, since it avoids the
     tedious escaping of every ‘/’.  If REGEXP itself includes any
     delimiter characters, each must be escaped by a backslash (‘\’).

In fine, regarding your question:

So you have to precede your 1st delimiter by a backslash \:

$ echo A | sed -ne '\#A#p'
A

Answered By - F. Hauri - Give Up GitHub

Answer Checked By - Robin (WPSolving Admin)