Question

这就是我要做的事情：

$variable = `grep -i "Text: Added to directory" '$FOO/result.txt' | awk '{print $6}' | tr -d "'"`
print $variable;

输出：

Text: Added to directory /path/to/directory/
Use of uninitialized value $6 in concatenation (.) or string

如何才能获取“/ path / to / directory”而不是“Text：添加到目录/ path / to / directory /”？

Answer 1

当然Perl可以做grep，awk和tr所能做的事情。

open my $fh, "<", "$FOO/result.txt" or die "can't open file: $!\n";
while (<$fh>) {
    next unless /pattern/i;
    (my $six = (split)[5]) =~ tr/'//d;
    print $six, "\n";
}
close $fh;

Answer 2

您可能需要调整引用转义但这应该是：

awk IGNORECASE=1 '/yourpattern/{ gsub(/\'/, \'\'); print $6 }' $FOO/result.txt

AWK非常多才多艺。

在Perl脚本中使用AWK？

2 个答案: