how to extract word from grep result in shell? - shell

Question

Using shell i want to search and print only sub-string with next word to that sub-string.
e.g. logfile has line "today is monday and this is:1234 so I am in."
if grep -q "this is :" ./logfile; then
#here i want to print only sub-string with next word i.e. "this is:1234"
#echo ???
fi

Answer 1

You can use sed with \1 to display the matched string in \(..\):
sed 's/.*\(this is:[0-9a-zA-Z]*\).*/\1/' logfile
EDIT: The above command is only fine for 1 line input.
When you have a file with more lines, you only want to print the lines that match:
sed -n 's/.*\(this is:[0-9a-zA-Z]*\).*/\1/p' logfile
When you have a large file and only want to see the first match, you can combine this command with head -1, but you would like to stop scanning/parsing after the first match. You can use q to quit, but you only want to quit after a match.
sed -n '/.*\(this is:[0-9a-zA-Z]*\).*/{s//\1/p;q}'

Answer 2

You can use a regular expression with a look-behind, if you want only the next word:
$ grep --perl-regexp -o '(?<=(this is:))(\S+)' ./logfile
1234
If you want both, then just:
$ grep --perl-regexp -o 'this is:\S+' ./logfile
this is:1234
The -o option instructs grep to return only the matching part.
In the commands above, we assumed that a "word" is a sequence of non-space characters. You can adjust that according to your needs.

Answer 3

If you have a system with GNU extensions (but aren't certain it was compiled with optional PCRE support), consider:
if result=$(grep -E -m 1 -o 'this is:[^[:space:]]+' logfile); then
echo "value is: ${result#*:}"
fi
${varname#value} expands to the contents of varname, but with value stripped from the beginning if present. Thus, ${result#*:} strips everything up to the first colon in result.
However, this may not work on systems without the non-POSIX options -o or -m.
If you want to support non-GNU systems, awk is a tool worth considering: Unlike answers requiring nonportable extensions (like grep -P), this should work on any modern platform (tested with GNU awk, recent BSD awk, and mawk; also, no warnings with with gawk --posix --lint):
# note that the constant 8 is the length of "this is:"
# GNU awk has cleaner syntax, but trying to be portable here.
if value=$(awk '
BEGIN { matched=0; } # by default, this will trigger END to exit as failure
/this is:/ {
match($0, /this\ is:([^[:space:]]+)/);
print substr($0, RSTART+8, RLENGTH-8);
matched=1; # tell END block to use zero exit status
exit(0); # stop processing remaining file contents, jump to END
}
END { if(matched == 0) { exit(1); } }
'); then
echo "Found value of $value"
else
echo "Could not find $value in file"
fi

Answer 4

You can look for everything up to, but not including the next space like this:
grep -Eo "this is:[^[:space:]]+" logfile
The [] introduces the set of characters you are looking for and the ^ at the start complements the set, so the set of characters you are looking for is a blank space, but complemented, i.e. not a blank space. The + says there must be at least one or more such characters.
The -E tells grep to use extended regular expressions and the -o means to only print the matched part.