awk output to variable and change directory - perl

Question

In the below script. am not able to change the directory.i need the output like above 70% disk inside that directory which one is consuming more space.
#!/usr/bin/perl
use strict;
use warnings;
my $test=qx("df -h |awk \+\$5>=70 {print \$6} ");
chdir($test) or die "$!";
print $test;
system("du -sh * | grep 'G'");

Answer 1

No need to call awk in your case because Perl is quite good at splitting and printing certain lines itself. Your code has some issues:
The code qx("df -h |awk \+\$5>=70 {print \$6} ") tries to execute the string "df -h | awk ..." as a command which fails because there is no such command called "df -h | awk". When I run that code I get sh: 1: df -h |awk +>=70 {print } : not found. You can fix that by dropping the quotes " because qx() already is quoting. The variable $test is empty afterwards, so the chdir changes to your $HOME directory.
Then you'll see the next error: awk: line 1: syntax error at or near end of line, because it calls awk +\$5>=70 {print \$6}. Correct would be awk '+\$5>=70 {print \$6}', i.e. with ticks ' around the awk scriptlet.
As stated in a comment, df -h splits long lines into two lines. Example:
Filesystem 1K-blocks Used Available Use% Mounted on
/long/and/possibly/remote/file/system
10735331328 10597534720 137796608 99% /local/directory
Use df -hP to get guaranteed column order and one line output.
The last system call shows the directory usage (space) for all lines containing the letter G. I reckon that's not exactly what you want.
I suggest the following Perl script:
#!/usr/bin/env perl
use strict;
use warnings;
foreach my $line ( qx(df -hP) ) {
my ($fs, $size, $used, $avail, $use, $target) = split(/\s+/, $line);
next unless ($use =~ /^\d+\s*\%$/); # skip header line
# now $use is e.g. '90%' and we drop the '%' sign:
$use =~ s/\%$//;
if ($use > 70) {
print "almost full: $target; top 5 directories:\n";
# no need to chdir here. Simply use $target/* as search pattern,
# reverse-sort by "human readable" numbers, and show the top 5:
system("du -hs $target/* 2>/dev/null | sort -hr | head -5");
print "\n\n";
}
}

Answer 2

#!/usr/bin/perl
use strict;
use warnings;
my #bigd = map { my #f = split " "; $f[5] }
grep { my #f = split " "; $f[4] =~ /^(\d+)/ && $1 >= 70}
split "\n", `df -hP`;
print "big directories: $_\n" for #bigd;
for my $bigd (#bigd) {
chdir($bigd);
my #bigsubd = grep { my #f = split " "; $f[0] =~ /G/ }
split "\n", `du -sh *`;
print "big subdirectories in $bigd:\n";
print "$_\n" for #bigsubd;
}
I belive you wanted to do something like this.