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Below is the text in the file:
Pseudo name=Apple
Code=42B
state=fault
Pseudo name=Prance
Code=43B
state=good
I need to grep for "42B" and get the output from the above text like:
Pseudo name=Apple
Code=42B
state=fault
Does anyone have idea on how to achieve this using grep/awk/sed?
With awk
awk -v RS='' '/42B/' file
RS= changes the input record separator from a newline to blank lines. If any field in an record contains /42B/ print the record.
'' (the null string) is a magic value used to represent blank lines according to POSIX:
If RS is null, then records are separated by sequences consisting of a
<newline>plus one or more blank lines, leading or trailing blank lines shall not result in empty records at the beginning or end of the input, and a<newline>shall always be a field separator, no matter what the value of FS is.
The output paragraphs will not be separated since the output separator remains a single newline. To ensure that there is a blank line between output paragraphs, set the output record separator to two newlines:
awk -v RS='' -v ORS='\n\n' '/42B/' file
FILENAME), it's not a bad idea to use redirection as that avoids problems for filename containing = or starting with - (or being -), makes for consistent error messages, and avoids running awk or performing other redirections if the input file can't be opened.
– Stéphane Chazelas
Nov 17 '15 at 10:22
Assuming the data is structured so that it's always the line before and after that you want you can make use of grep's -A (after) and -B (before) switches to tell it to include the 1 line before the match and 1 line after it:
$ grep -A 1 -B 1 "42B" sample.txt
Pseudo name=Apple
Code=42B
state=fault
If you want the same number lines before and after the search term you can use the -C (context) switch:
$ grep -C 1 "42B" sample.txt
Pseudo name=Apple
Code=42B
state=fault
If you'd like to be more stringent when matching the multiple lines you can use the tool pcregrep, to match a pattern over multiple lines:
$ pcregrep -M 'Pseudo.*\n.*42B.*\nstate.*' sample.txt
Pseudo name=Apple
Code=42B
state=fault
The above pattern matches as follows:
-M - multiple lines'Pseudo.*\n.*42B.*\nstate.*' - matches a group of strings where the first string starts with the word "Pseudo" followed by any characters up until a end of line \n, followed by any characters up until the string "42B" followed by any characters up until another end of line (\n), followed by the string "state"followed by any characters.
-C (context) can be used as a shortcut, if -A and -B are the same.
– David Baggerman
Jul 14 '13 at 0:37
There is probably a similarly easy way to do it with awk, but in perl:
cat file | perl -ne 'BEGIN { $/="\n\n" }; print if $_ =~ /42B/;'
That basically says to split the file into chunks delimited by blank lines, then to only print those chunks that match your regular expression.
cat; perl -00 -ne 'print if /42B/' file
– tripleee
Jul 14 '13 at 7:02
The grep of some flavours of Unix have the -p flag for "paragraph". I know AIX does.
grep -p 42B <myfile>
would do exactly what you're asking for there. YMMV and GNU grep doesn't have this flag.
An other perl solution, without a trailing empty line:
perl -00ne 'if ($_ =~ /42B/) {chomp($_); printf "%s\n",$_}' foo
Example
% perl -00ne 'if ($_ =~ /42B/) {chomp($_); printf "%s\n",$_}' foo
Pseudo name=Apple
Code=42B
state=fault
% cat foo
Pseudo name=Apple
Code=42B
state=fault
Pseudo name=Prance
Code=43B
state=good
perl -00 -ne 'print if /42B/' file.
– mivk
Jan 4 '16 at 15:49
