在PHP中访问Ajax发送的POST值
在PHP中访问Ajax发送的POST值
提问于 2016-10-24 09:34:43
我正在使用Ajax从表单中获取POST值。但是，当我尝试在提交时在数据库中插入表单值时，它没有被插入。我仍然不知道为什么它不能工作。
这是我的HTML
<form method="post" action="" id="home-sign-up-form">
<input type="text" name="suFirstName" placeholder="First Name" class="text-input-minor" id="sign-up-first-name-text-input">
<input type="text" name="suLastName" placeholder="Last Name" class="text-input-minor" id="sign-up-last-name-text-input">
<input type="text" name="suEmail" placeholder="Email" class="text-input-minor" id="sign-up-email-text-input">
<input type="password" name="suPassword" placeholder="Password" class="text-input-minor" id="sign-up-password-text-input">
<input type="password" name="suConfirmPassword" placeholder="Confirm Password" class="text-input-minor" id="sign-up-confirm-password-text-input">
<input type="text" name="suDisplayName" placeholder="Display Name (you can change this later)" class="text-input-minor" id="sign-up-display-name-text-input">
<br><font class="text-error" id="sign-up-error-text"></font><br>
<label><input type="checkbox" name="suRememberMe" value="yes" id="sign-up-remember-me-checkbox"><font id="sign-up-remember-me-text">Remember me</font></label>
<input name="signUp" type="submit" value="Sign Up" id="sign-up-submit">
</form>我的JS (第一个console.log确实通过并工作)：
if (validForm)
{
console.log("valid form");
console.log(JSON.stringify($('#home-sign-up-form')[0].seriaize()));
$.ajax(
{
type:'POST',
url:'form-submit.php',
data:$('#home-sign-up-form')[0].serialize(),
success:function(response)
{
$suForm.hide();
$tosppText.hide();
$mailSentIcon.show();
$emailSentText.show();
$emailSentTextEmail.text($suEmail);
$suBox.css("padding-left", "10px");
$suBox.css("padding-right", "10px");
}
});
}和我的PHP/MySQL：
<?php require 'dbconnect.php'; ?>
if (isset($_POST['suEmail']))
{
echo "<script type='text/javascript'>alert('got');</script>";
$suFirstName = mysqli_real_escape_string($_POST['suFirstName']);
$suLastName = mysqli_real_escape_string($_POST['suLastName']);
$suEmail = mysqli_real_escape_string($_POST['suEmail']);
$suPassword = mysqli_real_escape_string($_POST['suPassword']);
$suDisplayName = mysqli_real_escape_string($_POST['suDisplayName']);
$code = substr(md5(mt_rand()),0,15);
$query = $connection->query("INSERT INTO users (firstName,lastName,email,password,displayName,confirmCode,verified)Values('{$suFirstName}','{$suLastName}','{$suEmail}','{$suPassword}','{$suDisplayName}','{$confirmCode},'{$verified}')");
}在PHP代码中的警报，所以我假设它没有得到'signUp‘POST变量。非常感谢！感谢任何帮助！:D
回答 1
Stack Overflow用户
发布于 2016-10-24 12:30:45
$("#home-sign-up-form").submit(function(event) {
alert("Handler for .submit() called.");
event.preventDefault();
$.ajax({
type: 'POST',
url: 'form-submit.php',
data: $('#home-sign-up-form').serialize(),
success: function(response) {
console.log(response);
var data = JSON.parse(response);
$suForm.hide();
$tosppText.hide();
$mailSentIcon.show();
$emailSentText.show();
//$emailSentTextEmail.text($suEmail);
$emailSentTextEmail.text(data.suEmail);
$suBox.css("padding-left", "10px");
$suBox.css("padding-right", "10px");
}
});
});<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<form method="post" action="" id="home-sign-up-form">
<input type="text" name="suFirstName" placeholder="First Name" class="text-input-minor" id="sign-up-first-name-text-input">
<input type="text" name="suLastName" placeholder="Last Name" class="text-input-minor" id="sign-up-last-name-text-input">
<input type="text" name="suEmail" placeholder="Email" class="text-input-minor" id="sign-up-email-text-input">
<input type="password" name="suPassword" placeholder="Password" class="text-input-minor" id="sign-up-password-text-input">
<input type="password" name="suConfirmPassword" placeholder="Confirm Password" class="text-input-minor" id="sign-up-confirm-password-text-input">
<input type="text" name="suDisplayName" placeholder="Display Name (you can change this later)" class="text-input-minor" id="sign-up-display-name-text-input">
<br><font class="text-error" id="sign-up-error-text"></font>
<br>
<label>
<input type="checkbox" name="suRememberMe" value="yes" id="sign-up-remember-me-checkbox"><font id="sign-up-remember-me-text">Remember me</font>
</label>
<input name="signUp" type="submit" value="Sign Up" id="sign-up-submit">
</form>
<?php
if (isset($_POST['suEmail']))
{
$con=mysqli_connect("localhost","root","cakpep","backoffice");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$suFirstName = mysqli_real_escape_string($con,$_POST['suFirstName']);
$suLastName = mysqli_real_escape_string($con,$_POST['suLastName']);
$suEmail = mysqli_real_escape_string($con,$_POST['suEmail']);
$suPassword = mysqli_real_escape_string($con,$_POST['suPassword']);
$suDisplayName = mysqli_real_escape_string($con,$_POST['suDisplayName']);
$code = substr(md5(mt_rand()),0,15);
$query = $connection->query("INSERT INTO users (firstName,lastName,email,password,displayName,confirmCode,verified)Values('{$suFirstName}','{$suLastName}','{$suEmail}','{$suPassword}','{$suDisplayName}','{$confirmCode},'{$verified}')");
echo json_encode($_POST);
}
?>
在提交表单时添加此触发器，然后在php上返回带有json字符串的响应回显json_encode($_POST);
- and，然后在响应中ajax将json文本解析为对象，如下所示$("#home-sign-up-form").submit(function(event) {
- and $(“#home-sign-up-form”).submit(function(event){
- and=JSON.parse(
- )；$emailSentTextEmail.text(data.suEmail);
- i希望这就是您想要的..
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原文链接：
https://stackoverflow.com/questions/40209590
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