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I use the df command in a bash script:
df . -B MB | tail -1 | awk {'print $4'} | grep .[0-9]* This script returns:
99% But I need only numbers (to make the next comparison). If I use the grep regex without the dot:
df . -B MB | tail -1 | awk {'print $4'} | grep .[0-9]* I receive nothing. How to fix?
995
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 -n ( or --line-number ) option tells grep to show the line number of the lines containing a string that matches a pattern. When this option is used, grep prints the matches to standard output prefixed with the line number.
By default, grep is case-sensitive. This means that the uppercase and lowercase characters are treated as distinct. To ignore the case when searching, invoke grep with the -i option. If the search string includes spaces, you need to enclose it in single or double quotation marks.
If you try:
echo "99%" |grep -o '[0-9]*' It returns:
99 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).
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