[linux] How to make GREP select only numeric values?

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?

No need to used grep here, Try this:

df . -B MB | tail -1 | awk {'print substr($5, 1, length($5)-1)'}

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.

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%

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