php到AJAX表单提交

Question

在AJAX新手这里有这个代码，请帮助，我的代码不断去sec_reg.php页面，即使密码不匹配，甚至当表单是有效的，我希望用户即使他提交表格，也要留在当前页面上。这里是我的代码

这里是我的形式

<h4>ADD ANOTHER ADMIN</h4> 
    <form action="sec_reg.php" method="post" name="registration_form"> 
    <br> 
    <p> 

    <strong>Email</strong> 
    <br> 
    <br> 
    <input class="acc_input" type="text" id="email" name="email"placeholder="Email"> 

    </p> 
    <br> 
    <p> 
    <strong>  Password</strong> 
    <br> 
    <br> 
    <input class="acc_input" type="password" name="password" id="password" placeholder="Password"> 
    <br /><br /> 
    <strong>  Confirm Password</strong> 
    <br> 
    <br> 
    <input class="acc_input" type="password" name="cpassword" id="cpassword" placeholder="Confirm Password"> 

    <input type="hidden" name="p" id="p" value=""> 
    <br> 

    </p> 
    <button type="submit" class="btnsubmit" onclick="formhash(this.form, 
this.form.password, this.form.p);" ><strong>Register</strong></button> 

    </form> 

这里是forhash脚本（密码需要发送的安全性之前，必须哈希）

<script src="sha512.js"></script> 
<script> 
function formhash (form, password) 
{ 
var pass1 = document.getElementById("password").value; 
var pass2 = document.getElementById("cpassword").value; 
var ok = true; 
if (password != cpassword) { 
    //alert("Passwords Do not match"); 
    document.getElementById("password").style.borderColor = "#E34234"; 
    document.getElementById("cpassword").style.borderColor = "#E34234"; 
    ok = false; 

} 
else { 


var p = document.createElement("input");  
form.appendChild(p); 
p.name="p"; 
p.type="hidden"; 
p.value=hex_sha512(password.value); 
password.value=""; 
form.submit(); 

} 
    } 
</script> 

这里是我的sec_reg.php

<?php 
// Include database connection and functions here. 
include '../Connections/mabini150_Conn.php'; 
if (isset($_POST['p'])) 
{ 
include 'login_Function.php'; 
// The hashed password from the form 
$password = $_POST['p']; 
// Create a random salt 
$random_salt = hash('sha512', uniqid(mt_rand(1, mt_getrandmax()), true)); 
// Create salted password (Careful with the chilli) 
$password = hash('sha512', $password.$random_salt); 
$username='nousername'; 
$email = $_POST['email']; 
if ($insert_stmt = $mysqli->prepare("INSERT INTO members (username, email, password, salt) VALUES (?, ?, ?, ?)")) 
{ 
$insert_stmt->bind_param('ssss', $username, $email, $password, $random_salt); 
// Execute the prepared query. 
$insert_stmt->execute(); 

Answer 1

您需要调用后返回false，formhash（）;在你的onclick属性中。

<button type="submit" class="btnsubmit" 
    onclick="formhash(this.form, this.form.password, this.form.p); return false;" ><strong>Register</strong></button> 

否则按钮会提交不管什么。