如何使用 Perl 输入密码并将字符替换为“*"? [英] How can I enter a password using Perl and replace the characters with '*'?
问题描述
我有一个要求用户输入密码的 Perl 脚本.我怎样才能在用户输入字符时只回显 '*' 来代替他们输入的字符?
I have a Perl script that requires the user to enter a password. How can I echo only '*' in place of the character that the user types, as they type it?
我使用的是 Windows XP/Vista.
I'm using Windows XP/Vista.
推荐答案
您可以使用 Term::ReadKey.这是一个非常简单的示例，其中包含对退格键和删除键的一些检测.我已经在 Mac OS X 10.5 上对其进行了测试，但根据 ReadKey 手册应该在 Windows 下工作.手册 指出在 Windows 下使用非阻塞读取(ReadKey(-1)) 将失败.这就是为什么我使用基本上是 getc 的 ReadKey(0) (更多关于 getc 在 libc 手册).
You can play with Term::ReadKey. Here is a very simple example, with some detection for backspace and delete key. I've tested it on Mac OS X 10.5 but according to the ReadKey manual it should work under Windows. The manual indicates that under Windows using non-blocking reads (ReadKey(-1)) will fail. That's why I'm using ReadKey(0) who's basically getc (more on getc in the libc manual).
#!/usr/bin/perl
use strict;
use warnings;
use Term::ReadKey;
my $key = 0;
my $password = "";
print "
Please input your password: ";
# Start reading the keys
ReadMode(4); #Disable the control keys
while(ord($key = ReadKey(0)) != 10)
# This will continue until the Enter key is pressed (decimal value of 10)
{
# For all value of ord($key) see http://www.asciitable.com/
if(ord($key) == 127 || ord($key) == 8) {
# DEL/Backspace was pressed
#1. Remove the last char from the password
chop($password);
#2 move the cursor back by one, print a blank character, move the cursor back by one
print " ";
} elsif(ord($key) < 32) {
# Do nothing with these control characters
} else {
$password = $password.$key;
print "*(".ord($key).")";
}
}
ReadMode(0); #Reset the terminal once we are done
print "
Your super secret password is: $password
";
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