How do I grep only numbers in a line?
grep -o will print only the matching part of the line. Otherwise grep will print any lines with the pattern. the accepted answer ignores there could be numbers in the previous lines of the file, it does work for the example data, but what if the file was remote? In this example output.
How do I grep only numbers in Linux?
3 Answers. grep -r -L “[^0-9 ]” . [^0-9 ] will match anything that doesn’t contain digits or spaces (thought that would be fitting or else all files that contain spaces and number would be discarded). The -L flag is –files-without-match , so this basically gives you files that DO contain digits or spaces.
How to make grep select only numeric values?
Output lines use the same delimiters as input, and delimiters are null bytes if -z (–null-data) is also used (see Other Options ). grep will print any lines matching the pattern you provide. If you only want to print the part of the line that matches the pattern, you can pass the -o option:
How to print only part of a pattern in grep?
-P Interpret PATTERNS as Perl-compatible regular expressions (PCREs). -o Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line. grep -o will print only the matching part of the line. Otherwise grep will print any lines with the pattern.
When to use-X or-e in grep?
The -x option to grep forces the pattern to match across a complete line. I’m using -E to enable extended regular expression to be able to use +. The pattern [ [:digit:]]+ would match at least one digit. The type of digit that it matches may depend on your locale.
How does the-O flag work in grep?
Here’s the details on the -o (or –only-matching flag) works from the grep manual page. Print only the matched (non-empty) parts of matching lines, with each such part on a separate output line. Output lines use the same delimiters as input, and delimiters are null bytes if -z (–null-data) is also used (see Other Options ).