很抱歉再次询问，但我检查了几乎所有有关该主题的问题并且仍然无法解决。这是我的form.php和creditform.php
Edit1。将.get更改为.post
Edit2。我收到了未定义变量的错误＆＃34; name＆＃34; ＆＃34;电子邮件＆＃34; ＆＃34;地址＆＃34; ＆＃34;收入＆＃34; creditform.php 等等（简称所有这些） 我想要做的只是将所有输入插入到db。中的表中。
HTML
<!-- End crumbs-->
<div class="container wrap wow fadeInUp">
<div class="row">
<div class="col-sm-5 col-md-6 left-app">
<form id="form" action="php/creditform.php" method="post">
<input type="text" placeholder="Name" name="name" required>
<input type="email" placeholder="Email" name="email" required>
<input type="text" placeholder="Address" name="address" required>
<input type="number" placeholder="Monthly income before taxes" name="income" required>
<input type="number" placeholder="Amount Needed" name="amount_needed" required>
<input type="number" placeholder="Phone" name="phone" required>
<div class="row">
<div class="container">
<input type="submit" name="submit" value="Submit" class="button"></div></div>
<div id="result"></div>
</form>
</div>
<script>
$(document).ready(function($) {
'use strict';
$('#form').submit(function(event) {
event.preventDefault();
var url = $(this).attr('action');
var datos = $(this).serialize();
$.post(url, datos, function(resultado) {
$('#result').html(resultado);
});
});
</script>
form.php的
<?php
include('db.config.php');
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
$name = $email = $address = $income = $amount_needed = $phone = '';
if(isset($_POST['submit'])){
$name = addslashes($_POST['name']);
$email = addslashes($_POST['email']);
$address = addslashes($_POST['address']);
$income = addslashes($_POST['income']);
$amount_needed = addslashes($_POST['amount_needed']);
$phone = addslashes($_POST['phone']);
// check form fields
if(empty($name)){
$error .= 'Enter name <br />';
}
if(empty($email)){
$error .= 'Enter email <br />';
}
// check if errors exist
if(!empty($error)){
echo $error;
} else {
// process form as normal
$sql = "INSERT INTO `" . DBN . "`.`creditapp` (`name`, `email`, `address`, `income`, `amount_needed`, `phone`) VALUES ('$name', '$email', '$address', '$income', '$amount_needed', '$phone')";
$db->Query($sql);
}
}
}
print_r($_POST);
?>
CREDITFORM.PHP
<?php
if(isset($_POST['submit'])){
$name = $_POST['name'];
$email = $_POST['email'];
$address = $_POST['address'];
$income = $_POST['income'];
$amount_needed = $_POST['amount_needed'];
$phone = $_POST['phone'];
}
print_r($_POST);
?>
答案 0 :(得分：3)
您需要向POST数组发送$_POST个请求才能包含任何内容。
$.get(url, datos, function(resultado) {
$('#result').html(resultado);
});
这会发送GET请求（请检查$_GET）。您想在此处使用$.post。
$.post(url, datos, function(resultado) {
$('#result').html(resultado);
});
答案 1 :(得分：2)
以下是一个事件对Php 文件的简单 Ajax调用：点击按钮。
在您的示例中，您必须使用POST方法，因为您使用：
$_POST['something'];
Javascript客户端：
$("body").on("click", "#mybutton", function() {
var mydata = $("#form").serialize();
$.ajax({
type: "POST",
url: "/api/api.php",
data: {data : mydata},
timeout: 6e3,
error: function(a, b) {
if ("timeout" == b) $("#err-timedout").slideDown("slow"); else {
$("#err-state").slideDown("slow");
$("#err-state").html("An error occurred: " + b);
}
},
success: function(a) {
var e = $.parseJSON(a);
if (true == e["success"]) {
$("#result").html(e['message']);
// here is what you want, callback Php response content in Html DOM
}
}
});
return false;
});
接下来在您的Php代码中，只需在任何成功函数之后执行：
if ($result) {
echo json_encode(array(
'success' => true,
'msg' => "Nice CallBack by Php sent to client Side by Ajax Call"
));
}
答案 2 :(得分：1)
好的，我在ajax-json-php上练习了更多，最后这是我的解决方案和我的最新代码。
感谢所有的答案和建议。希望这对某人也有帮助。
仍然愿意改进任何建议。
<?php
include('db.config.php');
if(isset($_POST['submit'])){
$name = addslashes($_POST['name']);
$email = addslashes($_POST['email']);
$address = addslashes($_POST['address']);
$income = addslashes($_POST['income']);
$amount_needed = addslashes($_POST['amount_needed']);
$phone = addslashes($_POST['phone']);
// process form as normal
$sql = "INSERT INTO `" . DBN . "`.`creditapp` (`name`, `email`, `address`, `income`, `amount_needed`, `phone`) VALUES ('$name', '$email', '$address', '$income', '$amount_needed', '$phone')";
$db->Query($sql);
die();
<div class="container wrap wow fadeInUp">
<div class="row">
<div class="col-sm-5 col-md-6 left-app">
<form id="form" action="creditapp.php" method="post">
<input type="text" placeholder="Name" name="name" id="name" required>
<input type="email" placeholder="Email" name="email" id="email" required>
<input type="text" placeholder="Address" name="address" id="address" required>
<input type="text" placeholder="Monthly income before taxes" id="income" name="income" required>
<input type="text" placeholder="Amount Needed" name="amount_needed" id="amount_needed" required>
<input type="text" placeholder="Phone" name="phone" id="phone" required>
<div class="row">
<div class="container">
<input type="submit" name="submit" value="Submit" class="button sub"></div>
</div>
</form>
</div>
<script>
$(document).ready(function() {
$('.sub').click(function(e){
e.preventDefault();
var name = $('#name').val(),
email = $('#email').val(),
address = $('#address').val(),
income = $('#income').val(),
amount_needed = $('#amount_needed').val(),
phone = $('#phone').val();
$.ajax({
url: "creditapp.php",
type: "POST",
data: {
name: name,
email: email,
address: address,
income: income,
amount_needed: amount_needed,
phone: phone,
submit: "submit"
}
}).done(function(msg){
$('.right-info').html('<pre>' + msg + '</pre>');
})
})
});
</script>
答案 3 :(得分：0)
你遇到的问题是你的javascript。您等待提交，而不是执行ajax GET 请求，您需要 POST 。
<script>
$(document).ready(function($) {
'use strict';
$('#form').submit(function(event) {
event.preventDefault();
var url = $(this).attr('action');
var datos = $(this).serialize();
$.post(url, datos, function(resultado) {
$('#result').html(resultado);
});
});
</script>