Extract text between two characters (multiline input)

Question

If I have the text:

aaaaaaaaa
#some info
other rows
end row#
another info

How could I extract only the text between the characters # obtaining only:

some info
other rows
end row

I was trying with sed in this way:

echo -e "aaaaaaaaa\n#some info\nother rows\nend row#\nanother info" |
sed -n -e '/\#/,/\#/p'

but it gives me also the character #. Is there a way to remove # using sed?

The only thing missing in those duplicates would be removing the delimiters (# here) — Jeff Schaller, Oct 17 '18 at 12:43

pLumo · Accepted Answer · 2018-10-17T10:47:33.707

5

You can use perl:

echo -e "aaaaaaaaa\n#some info\nother rows\nend row#\nanother info" |\
perl -0777 -ne '/#([^#]*)#/ && print $1,"\n"; '

Explanation:

-0777 slurp the whole file as one line (enables multiline matching)
/#([^#]*)#/ match non-# characters [^#] between too # and with the brackets add it as first matching group.
&& print $1,"\n" if found, print first matching group and a final newline.

edited Oct 17 '18 at 10:47

answered Oct 17 '18 at 10:36

pLumo

You could also use a non-greedy quantifier /#(.*?)#/ and you should probably use && instead of ; to separate the match and the print -- that will prevent a single newline from being printed if the match is not found. – glenn jackman Oct 17 '18 at 10:46
I tried with the non-greedy quantifier, but doesn't work. Good idea though with the &&. – pLumo Oct 17 '18 at 10:48
That's a very common source of bugs, assuming the RE matches and $1 has s value – glenn jackman Oct 17 '18 at 10:52
what do you mean with RE? could you make an example? Thanks to all of you! – tamarindoz Oct 17 '18 at 12:37
RE = regular expression – pLumo Oct 17 '18 at 12:40
"assuming the RE matches and $1 has s value" for example if its call inside an script where $1 is the argument of the script? do you mean this or ? thanksss! – tamarindoz Oct 17 '18 at 12:56
No, that is prevented by the single quotes. He means I should not do do_something; do_something_dependent; but rather do_something && do_something_dependent;. In this case print $1 is dependent on the pattern match (/#([^#]*)#/). If you don't know what I am talking about, read this. – pLumo Oct 17 '18 at 13:01

score 5 · Answer 2 · edited Oct 17 '18 at 11:15

5

Slight adaption to your sed one liner:

echo -e "aaaaaaaaa\n#some info\nother rows\nend row#\nanother info" |
sed -n '/^#/,/#$/ {s/#//;p;}'

Output:

some info
other rows
end row

edited Oct 17 '18 at 11:15

agc

answered Oct 17 '18 at 11:09

RudiC

thanks, but if the '#' is not the first or last character? In this case sed doesn't work !? – tamarindoz Oct 17 '18 at 12:51

2 Answers2