我试图将数据从表单发送到数据库。这是我使用的形式:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input type="submit" value="Send" />
</form>
典型的方法是提交表单，但这会导致浏览器重定向。使用jQuery和Ajax，是否有可能捕获所有表单的数据并将其提交给PHP脚本(例如，form. PHP)?
当前回答
HTML:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" class="inputs" name="bar" type="text" value="" />
<input type="submit" value="Send" onclick="submitform(); return false;" />
</form>
JavaScript:
function submitform()
{
var inputs = document.getElementsByClassName("inputs");
var formdata = new FormData();
for(var i=0; i<inputs.length; i++)
{
formdata.append(inputs[i].name, inputs[i].value);
}
var xmlhttp;
if(window.XMLHttpRequest)
{
xmlhttp = new XMLHttpRequest;
}
else
{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
}
}
xmlhttp.open("POST", "insert.php");
xmlhttp.send(formdata);
}
其他回答
自从引入了Fetch API，就没有理由再用jQuery Ajax或xmlhttprequest来实现了。要POST表单数据到php脚本在香草JavaScript你可以做以下:
异步函数postData() { 尝试{ Const res = await fetch('../php/contact.php'， { 方法:“文章”, new FormData(document.getElementById('form')) }) if (!res.ok)抛出新的错误('网络响应不正常'); } catch (err) { console.log (err) ｝ ｝ <form id="form" action="javascript:postData()" > <input id="name" name="name" placeholder=" name" type="text" required> <input type="submit" value=" submit" > > < /形式
下面是一个非常基本的php脚本示例，它获取数据并发送电子邮件:
<?php
header('Content-type: text/html; charset=utf-8');
if (isset($_POST['name'])) {
$name = $_POST['name'];
}
$to = "test@example.com";
$subject = "New name submitted";
$body = "You received the following name: $name";
mail($to, $subject, $body);
我使用如下所示的方法。它像文件一样提交所有东西。
$(document).on("submit", "form", function(event)
{
event.preventDefault();
var url = $(this).attr("action");
$.ajax({
url: url,
type: 'POST',
dataType: "JSON",
data: new FormData(this),
processData: false,
contentType: false,
success: function (data, status)
{
},
error: function (xhr, desc, err)
{
console.log("error");
}
});
});
我想分享一个详细的方法，如何张贴与PHP + Ajax以及错误抛出返回失败。
首先，创建两个文件，例如form.php和process.php。
我们将首先创建一个表单，然后使用jQuery .ajax()方法提交。其余的将在评论中解释。
form.php
<form method="post" name="postForm">
<ul>
<li>
<label>Name</label>
<input type="text" name="name" id="name" placeholder="Bruce Wayne">
<span class="throw_error"></span>
<span id="success"></span>
</li>
</ul>
<input type="submit" value="Send" />
</form>
使用jQuery客户端验证表单并将数据传递给process.php。
$(document).ready(function() {
$('form').submit(function(event) { //Trigger on form submit
$('#name + .throw_error').empty(); //Clear the messages first
$('#success').empty();
//Validate fields if required using jQuery
var postForm = { //Fetch form data
'name' : $('input[name=name]').val() //Store name fields value
};
$.ajax({ //Process the form using $.ajax()
type : 'POST', //Method type
url : 'process.php', //Your form processing file URL
data : postForm, //Forms name
dataType : 'json',
success : function(data) {
if (!data.success) { //If fails
if (data.errors.name) { //Returned if any error from process.php
$('.throw_error').fadeIn(1000).html(data.errors.name); //Throw relevant error
}
}
else {
$('#success').fadeIn(1000).append('<p>' + data.posted + '</p>'); //If successful, than throw a success message
}
}
});
event.preventDefault(); //Prevent the default submit
});
});
现在我们来看看process.php
$errors = array(); //To store errors
$form_data = array(); //Pass back the data to `form.php`
/* Validate the form on the server side */
if (empty($_POST['name'])) { //Name cannot be empty
$errors['name'] = 'Name cannot be blank';
}
if (!empty($errors)) { //If errors in validation
$form_data['success'] = false;
$form_data['errors'] = $errors;
}
else { //If not, process the form, and return true on success
$form_data['success'] = true;
$form_data['posted'] = 'Data Was Posted Successfully';
}
//Return the data back to form.php
echo json_encode($form_data);
项目文件可以从http://projects.decodingweb.com/simple_ajax_form.zip下载。
HTML:
<form name="foo" action="form.php" method="POST" id="foo">
<label for="bar">A bar</label>
<input id="bar" class="inputs" name="bar" type="text" value="" />
<input type="submit" value="Send" onclick="submitform(); return false;" />
</form>
JavaScript:
function submitform()
{
var inputs = document.getElementsByClassName("inputs");
var formdata = new FormData();
for(var i=0; i<inputs.length; i++)
{
formdata.append(inputs[i].name, inputs[i].value);
}
var xmlhttp;
if(window.XMLHttpRequest)
{
xmlhttp = new XMLHttpRequest;
}
else
{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function()
{
if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
{
}
}
xmlhttp.open("POST", "insert.php");
xmlhttp.send(formdata);
}
这是一篇非常好的文章，包含了关于jQuery表单提交你需要知道的一切。
文章摘要:
简单HTML表单提交
HTML:
<form action="path/to/server/script" method="post" id="my_form">
<label>Name</label>
<input type="text" name="name" />
<label>Email</label>
<input type="email" name="email" />
<label>Website</label>
<input type="url" name="website" />
<input type="submit" name="submit" value="Submit Form" />
<div id="server-results"><!-- For server results --></div>
</form>
JavaScript:
$("#my_form").submit(function(event){
event.preventDefault(); // Prevent default action
var post_url = $(this).attr("action"); // Get the form action URL
var request_method = $(this).attr("method"); // Get form GET/POST method
var form_data = $(this).serialize(); // Encode form elements for submission
$.ajax({
url : post_url,
type: request_method,
data : form_data
}).done(function(response){ //
$("#server-results").html(response);
});
});
HTML Multipart/ Form -data表单提交
要将文件上传到服务器，我们可以使用XMLHttpRequest2提供的FormData接口，该接口构造一个FormData对象，可以使用jQuery Ajax轻松地将其发送到服务器。
HTML:
<form action="path/to/server/script" method="post" id="my_form">
<label>Name</label>
<input type="text" name="name" />
<label>Email</label>
<input type="email" name="email" />
<label>Website</label>
<input type="url" name="website" />
<input type="file" name="my_file[]" /> <!-- File Field Added -->
<input type="submit" name="submit" value="Submit Form" />
<div id="server-results"><!-- For server results --></div>
</form>
JavaScript:
$("#my_form").submit(function(event){
event.preventDefault(); // Prevent default action
var post_url = $(this).attr("action"); // Get form action URL
var request_method = $(this).attr("method"); // Get form GET/POST method
var form_data = new FormData(this); // Creates new FormData object
$.ajax({
url : post_url,
type: request_method,
data : form_data,
contentType: false,
cache: false,
processData: false
}).done(function(response){ //
$("#server-results").html(response);
});
});
我希望这能有所帮助。
