我的代码中有一个错误，我的页面中还包含一个js文件，它阻止了$(document(.ready(function((｛…
我正在尝试汇总这个登录表单：
<form class="form" id="AjaxForm">
<input type="text" name="username" placeholder="Username">
<input type="password" name="password" placeholder="Password">
<button type="submit" id="login-button">Login</button>
</form>
通过ajax使用以下代码：
var request;
$("#AjaxForm").submit(function(event){
// Abort any pending request
if (request) {
request.abort();
}
// setup some local variables
var $form = $(this);
// Let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");
// Serialize the data in the form
var serializedData = $form.serialize();
// Let's disable the inputs for the duration of the Ajax request.
// Note: we disable elements AFTER the form data has been serialized.
// Disabled form elements will not be serialized.
$inputs.prop("disabled", true);
// Fire off the request to /form.php
request = $.ajax({
url: "login.php",
type: "post",
data: serializedData
});
// Callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
// Log a message to the console
console.log("Hooray, it worked!");
});
// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
// Log the error to the console
console.error(
"The following error occurred: "+
textStatus, errorThrown
);
});
// Callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
// Reenable the inputs
$inputs.prop("disabled", false);
});
// Prevent default posting of form
event.preventDefault();
});
我在这里找到了：带有PHP 的jQuery Ajax POST示例
我正试图将它发布到login.php，它会检查它是否是有效的用户名和密码。但当我按下登录按钮时，它只是把用户名和密码放在url中，什么也不做。当我添加action="login.php"method="POST">时，它会提交表单，但不会通过ajax提交，因为当我注释ajax代码时，它仍然会提交。我正在努力防止这种情况发生。对我的问题有什么见解吗？
编辑：暂时住在这里：http://5f6738d9.ngrok.io/test/public/index.html用户名和密码正在测试
检查事件是否绑定在$(document).on('ready' ...中，否则事件不会激发，表单将正常提交或不通过AJAX提交。
你的代码应该看起来像：
$(document).on('ready', function () {
var request;
$("#AjaxForm").submit(function(event){
// Abort any pending request
if (request) {
request.abort();
}
// setup some local variables
var $form = $(this);
// Let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");
// Serialize the data in the form
var serializedData = $form.serialize();
// Let's disable the inputs for the duration of the Ajax request.
// Note: we disable elements AFTER the form data has been serialized.
// Disabled form elements will not be serialized.
$inputs.prop("disabled", true);
// Fire off the request to /form.php
request = $.ajax({
url: "login.php",
type: "post",
data: serializedData
});
// Callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
// Log a message to the console
console.log("Hooray, it worked!");
});
// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
// Log the error to the console
console.error(
"The following error occurred: "+
textStatus, errorThrown
);
});
// Callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
// Reenable the inputs
$inputs.prop("disabled", false);
});
// Prevent default posting of form
event.preventDefault();
});
});
请注意，从jQuery 1.8开始，这些回调事件实际上已被弃用。
您还需要确保在任何情况下都在表单上设置了POST和action属性。
您的提交按钮是一个标准的提交类型按钮，这意味着您的表单将正常提交。根据您的HTML代码，它只会将表单提交到同一个URL。JS代码将没有时间执行。你所需要做的就是通过添加来取消默认的HTML表单提交
event.preventDefault();
您需要在提交侦听器中添加此第一项内容。所以你的JS代码会像这个一样开始
$("#AjaxForm").submit(function(event){
event.preventDefault();
// Abort any pending request
if (request) {
request.abort();
}
//....
尝试使用以下代码：
$(document).ready(function(){
$('#AjaxForm').on('submit', function(event){
event.preventDefault();
if(request){
request.abort();
request = false;
}
var $form = $(this);
var serializedData = $form.serialize();
var $inputs = $form.find("input, select, button, textarea");
$inputs.prop("disabled", true);
var request = $.ajax({
url: 'login.php',
type: 'POST',
data: serializedData,
success: function (data, textStatus, jqXHR) {
// login was successful so maybe refresh the page
window.location.reload();
},
error: function (jqXHR, textStatus, errorThrown) {
// display form errors received from server
},
complete: function (jqXHR, textStatus) {
request = false;
}
});
});
});
就我个人而言，我会使用这个：
$(document).ready(function(){
$("#AjaxForm").on("submit",function(e){
e.preventDefault();
var $form = $(this);
var $cacheData = $form.find("input, submit");
var serializedData = $form.serialize();
$.ajax({
url: $form.attr("action"),
type: $form.attr("method"),
data: serializedData,
xhrFields: {
onprogress: function(e){
$cacheData.prop("disabled", true);
console.log(e.loaded / e.total*100 + "%");
}
},
done: function(text){
if(text == "Succes!"){
alert(text);
} else {
alert(text);
}
},
fail: function(xhr, textStatus, errorThrown){
alert(textStatus + " | " + errorThrown);
},
always: function(){
$cacheData.prop("disabled", false);
}
});
});
});
这允许你做一些有用的事情：
- 您可以在控制台中监控进度
- Ajax成功并不意味着不能返回错误。例如：密码错误。所以现在您可以简单地回显错误，这个脚本将显示它们。登录正常时回显"成功！">
请记住，此脚本要求您在表单中设置HTML属性action和method。