如果在命令行中没有输入 arguments，则显示错误

Question

我需要编写一个程序来读取 Linux 中的 passwd 文件。 用户输入程序的名称和所需的用户名，然后程序打印出相应的 userID 和 homeDir。 如果未提供用户名或输入了多个用户，程序将显示错误消息。

Good ex)
User types:
lookupUser.pl jdoe123
Result:
UID: 123456
HomeDir: /home/jdoe123

Bad ex)
User types:
lookupUser.pl
Result:
Enter one, and only one username.

Bad ex)
User types:
lookupUser.pl jdoes123 ssanta456
Result:
Enter one, and only one username.

在某些时候，我已经完成了所有工作，但我开始清理代码，现在当我没有输入用户名时，我收到一个错误“ Use of uninitialized value $user in scalar chomp at./lookupUser.pl line 6. ”。

任何帮助是极大的赞赏。

#!/usr/bin/perl
use strict;
use warnings FATAL => 'all';

my $user = $ARGV[0];
chomp ($user);

if ($#ARGV == -1 || $#ARGV >= 1){
    print "Enter one, and only one username.\n";
}

elsif (getpwnam("$user")) {
    (my $name, my $passwd, my $uid, my $gid, my $quota, my $comment,
        my $gcos, my $dir, my $shell) = getpwnam $user;
    #print "Name = $name\n";
    #print "Password = $passwd\n";
    print "UID = $uid\n";
    #print "GID = $gid\n";
    #print "Quota = $quota\n";
    #print "Comment = $comment\n";
    #print "Gcos = $gcos\n";
    print "HOME DIR = $dir\n";
    #print "Shell = $shell\n";
}
else {
    print "Username $user does not exist.\n";
}

Answer 1

仅当恰好有 1 个参数传递给程序时，才检查用户名。

use strict;
use warnings FATAL => 'all';

if (@ARGV == 1 ) {
    my $user = $ARGV[0];
    chomp ($user);
    if (getpwnam("$user")) {
        (my $name, my $passwd, my $uid, my $gid, my $quota, my $comment,
            my $gcos, my $dir, my $shell) = getpwnam $user;
        print "UID = $uid\n";
        print "HOME DIR = $dir\n";
    }
    else {
        print "Username $user does not exist.\n";
    }
}
else {
    print "Enter one, and only one username.\n";
}