PHP和Javascript密码匹配验证

Question

尝试获取页面以动态显示密码是否匹配(回声“匹配”或“不匹配”)。(类似于之前向here提出的问题。)

查看password1和password2是否与输入“密码”或“确认密码”输入框相同。

编辑： ($password2 == $password1)似乎没有按照我的意愿检查比赛。password1和password2的发布工作正常(在输入框中输入“不匹配”时，#反馈更改为“不匹配”，但“匹配”从未显示)。任何有关如何修复此代码的帮助都将不胜感激。

check.php

<?php
include __DIR__ . '/mysqli_connect.php';

$password2 = mysqli_real_escape_string($dbc, $_POST['password2']);
$password1 = mysqli_real_escape_string($dbc, $_POST['password1']);

if ($password2==NULL && $password1==NULL) {
    echo '';
} elseif ($password2 == $password1) {
    echo 'Match';
} else {
    echo 'Do not match';
}

register.php

<script type="text/javascript" src="/jquery-3.2.1.min.js"></script>

<script>
$(document).ready(function() {
    $('#feedback').load('/includes/check.php').show();

    $('#password1').keyup(function() {
        $.post('/includes/check.php', { password1: form.password1.value }, 
        function(result) {
            $('#feedback').html(result).show;
        });
    });
});

$(document).ready(function() {
    $('#feedback').load('/includes/check.php').show();

    $('#password2').keyup(function() {
        $.post('/includes/check.php', { password2: form.password2.value }, 
        function(result) {
            $('#feedback').html(result).show;
        });
    });
});
</script>

<form action="register.php" method="post" name="form">

<input id="password1" class="signup_input_box" type="password" name="password1" maxlength="20" value="<?php if (isset($trimmed['password1'])) echo $trimmed['password1']; ?>">

<input id="password2" class="signup_input_box" type="password" name="password2" maxlength="20" value="<?php if (isset($trimmed['password2'])) echo $trimmed['password2']; ?>">
<div id="feedback"></div>

<p><center><input type="submit" name="submit" value="Register"></center>
</form>

Answer 1

使用MySql投递跳过，只使用JS检查密码输入的值，并在键盘上显示文本。Reference here.