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?
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).
\d like echo "99%" |grep -o '\d*'
-E)\d effective, you need -P. And if there is a portable version available, use the portable one.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:
-o, --only-matching Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Like this:
echo 'Here is a line mentioning 99% somewhere' | grep -o '[0-9]+'
How about:
df . -B MB | tail -1 | awk {'print $4'} | cut -d'%' -f1
No need to used grep here, Try this:
df . -B MB | tail -1 | awk {'print substr($5, 1, length($5)-1)'}
function getPercentUsed() {
$sys = system("df -h /dev/sda6 --output=pcent | grep -o '[0-9]*'", $val);
return $val[0];
}
Don't use more commands than necessary, leave away tail, grep and cut. You can do this with only (a simple) awk
PS: giving a block-size en print only de persentage is a bit silly ;-) So leave also away the "-B MB"
df . |awk -F'[multiple field seperators]' '$NF=="Last field must be exactly --> mounted patition" {print $(NF-number from last field)}'
in your case, use:
df . |awk -F'[ %]' '$NF=="/" {print $(NF-2)}'
output: 81
If you want to show the percent symbol, you can leave the -F'[ %]' away and your print field will move 1 field further back
df . |awk '$NF=="/" {print $(NF-1)}'
output: 81%
You can use Perl style regular expressions as well. A digit is just \d then.
grep -Po "\\d+" filename
-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.
