Question

#!/usr/bin/perl

use strict;
use warnings;

my $group_name = <STDIN>;
my $group_id = <STDIN>;
my $choice = <STDIN>;

  print "* My menu *\n";
  print "* *\n";
  print "* 1. Create a Unix Group  *\n";
  print "* 2. Delete a Unix Group  *\n";
  print "* 3. Create a Unix User *\n";
  print "* 4. Delete a Unix USer *\n";
  print "* 5. Quit *\n";
  print "* *\n";
  print "******************************************************\n";
  print "* Enter Your Choice >\n";

  $choice = <STDIN> ;

  chomp $choice;
  if ($choice == 1) {
  print "**********************************\n";
  print " Create A Unix Group\n";
  print "**********************************\n";
  print " Enter The Group Name to Create >\n";
  chomp ($group_name = <STDIN>);
  print " Enter the Group ID to Create >\n";
  chomp ($group_id = <STDIN> );

  if ( ! 'grep -i $group_name /etc/group' ) {
      system ("groupadd -g $group_id $group_name");
  print "Group Created Successfully!\n"
      } else {
  print "Group Already Exists !\n";
      }
      }

当我运行脚本并输入组ID时，它返回组已经存在。我输入的内容并不重要，说实话它不断回归集团已经存在，我知道这个集团并不存在。任何Suggestiongs？

Answer 1

问题在于：

if ( ! 'grep -i $group_name /etc/group' )

在这里您使用引用'...'，因此perl会将其视为文字字符串。布尔上下文中非空的字符串将被标记为true，因此if条件总是失败，执行else分支。

要从perl运行系统命令，您必须使用backstick `...`或使用system()内置函数：

if ( ! `grep -i $group_name /etc/group` )

小组已经存在

1 个答案: