如何使用变量执行Perl qx函数 [英] How to execute Perl qx function with variable

问题描述

如何使Perl的qx函数与我的$opt变量一起执行?

How do I get Perl's qx function to execute with my $opt variable?

(工作之前):

my @df_output = qx (df -k /tmp);

我想使用-k，-g或-H:

my @df_output = qx (df -$opt /tmp);

推荐答案

您应该使用的工具，但是:永远不要使用qx.这是古老而危险的；不管您输入的内容是经过外壳的，所以很容易受到外壳注入的影响，或者如果/bin/sh并非您所期望的那样，您会感到惊讶.

What you have should work, but: don't ever use qx. It's ancient and dangerous; whatever you feed to it goes through the shell, so it's very easy to be vulnerable to shell injection or run into surprises if /bin/sh isn't quite what you expected.

使用open()的多参数形式，完全绕过外壳.

Use the multi-arg form of open(), which bypasses the shell entirely.

open my $fh, '-|', 'df', "-$opt", '/tmp' or die "Can't open pipe: $!";
my @lines = <$fh>;  # or read in a loop, which is more likely what you want
close $fh or die "Can't close pipe: $!";

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