Question

尝试让页面动态显示密码是否匹配（echo'Match'或'Do not match'）。（与上一个问题类似here。）

查看密码1和密码2是否与您在“密码”或“确认密码”输入框中输入的内容相同。

编辑：似乎elseif（$ password2 == $ password1）没有像我预期的那样检查匹配。发布password1和password2可以正常工作（当您在任一输入框中键入时，#feedback会更改为“不匹配”，但“匹配”从不显示）。有关如何修复此代码的任何帮助表示赞赏。

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帖子跳过并仅使用JS来检查密码输入的值并在keyup上显示文本。 Reference here.

PHP和Javascript验证密码匹配

1 个答案: