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Is there a way to make grep output "words" from files that match the search expression?
If I want to find all the instances of, say, "th" in a number of files, I can do:
grep "th" * but the output will be something like (bold is by me);
some-text-file : the cat sat on the mat some-other-text-file : the quick brown fox yet-another-text-file : i hope this explains it thoroughly
What I want it to output, using the same search, is:
the the the this thoroughly Is this possible using grep? Or using another combination of tools?
299
The easiest of the two commands is to use grep's -w option. This will find only lines that contain your target word as a complete word. Run the command "grep -w hub" against your target file and you will only see lines that contain the word "hub" as a complete word.
Grep: Print only the words of the line that matched the regular expression, one per line. We used the following parameters on our command: -h, –no-filename : Suppress the prefixing of file names on output. This is the default when there is only one file (or only standard input) to search.
If there's no match, that should generally be considered a failure, so a return of 0 would not be appropriate. Indeed, grep returns 0 if it matches, and non-zero if it does not.
-v, --invert-match Invert the sense of matching, to select non-matching lines. -w, --word-regexp Select only those lines containing matches that form whole words.
Try grep -o:
grep -oh "\w*th\w*" * Edit: matching from Phil's comment.
From the docs:
-h, --no-filename Suppress the prefixing of file names on output. This is the default when there is only one file (or only standard input) to search. -o, --only-matching Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line. If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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