03-24-2010
Finding a string in a text file and posting part of the line
What would be the most succinct way of doing this (preferably in 1 line, maybe 2):
searching the first 10 characters of every line in a text file for a specific string, and if it was found, print out characters 11-20 of the line on which the string was found.
In this case, it's known that there are no duplicates in the file, so it either finds the string on one line or none of them.
I'm guessing one of grep, sed or awk can be used but I can't figure out the best way.
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GREP(1) General Commands Manual GREP(1)
NAME
grep - search a file for lines containing a given pattern
SYNOPSIS
grep [-elnsv] pattern [file] ...
OPTIONS
-e -e pattern is the same as pattern
-c Print a count of lines matched
-i Ignore case
-l Print file names, no lines
-n Print line numbers
-s Status only, no printed output
-v Select lines that do not match
EXAMPLES
grep mouse file # Find lines in file containing mouse
grep [0-9] file # Print lines containing a digit
DESCRIPTION
Grep searches one or more files (by default, stdin) and selects out all the lines that match the pattern. All the regular expressions
accepted by ed and mined are allowed. In addition, + can be used instead of * to mean 1 or more occurrences, ? can be used to mean 0 or 1
occurrences, and | can be used between two regular expressions to mean either one of them. Parentheses can be used for grouping. If a
match is found, exit status 0 is returned. If no match is found, exit status 1 is returned. If an error is detected, exit status 2 is
returned.
SEE ALSO
cgrep(1), fgrep(1), sed(1), awk(9).
GREP(1)