我是AJAX的新手,并使用此SO答案中的代码jQuery Ajax POST example with PHP与WordPress网站上的表单集成。它工作正常,但我无法将其与jquery验证集成
我尝试将上面页面中的javascript放入<{1}}函数下面的
submitHandler
我的表单在第一次点击时验证。然后,如果我键入输入并提交没有任何反应,我必须再次单击表单以使用AJAX正确提交。下面是一个jsfiddle。感谢任何帮助。
我的代码jsfiddle认为它会将错误记录到控制台,因为form.php没有链接
答案 0 :(得分:11)
submitHandler的工作是提交表单,而不是注册表单提交事件处理程序。
在触发formm submit事件时调用submitHandler,在您的情况下,而不是提交您正在注册提交处理程序的表单,以便在第一次触发表单提交事件时不提交表单。当它第二次被触发时,首先由验证器处理提交事件,然后触发你注册的处理程序,触发ajax请求。
在submitHandler中你只需要发送ajax请求就不需要注册事件处理程序
$("#add-form").validate({
submitHandler: function (form) {
// setup some local variables
var $form = $(form);
// 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
$inputs.prop("disabled", true);
// fire off the request to /form.php
request = $.ajax({
url: "forms.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!");
alert("success awesome");
$('#add--response').html('<div class="alert alert-success"><button type="button" class="close" data-dismiss="alert">×</button><strong>Well done!</strong> You successfully read this important alert message.</div>');
});
// 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 occured: " + textStatus, errorThrown);
});
// callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
// reenable the inputs
$inputs.prop("disabled", false);
});
}
});
答案 1 :(得分:3)
致电$("#add-form").submit(function(){...})
不提交表单。它绑定了一个处理程序,该处理程序说明当用户提交表单时该做什么。这就是你必须提交两次的原因:第一次调用validate插件的提交处理程序,它验证数据并运行你的函数,第二次调用你第一次添加的提交处理程序。
不要将代码包装在.submit()
中,只需直接在submitHandler:
函数中执行即可。变化:
var $form = $(this);
为:
var $form = $(form);
您不需要event.PreventDefault()
,验证插件也会为您执行此操作。
答案 2 :(得分:2)
$("#add-form").validate({
submitHandler: function (form) {
var request;
// bind to the submit event of our form
// 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
$inputs.prop("disabled", true);
// fire off the request to /form.php
request = $.ajax({
url: "forms.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!");
alert("success awesome");
$('#add--response').html('<div class="alert alert-success"><button type="button" class="close" data-dismiss="alert">×</button><strong>Well done!</strong> You successfully read this important alert message.</div>');
});
// 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 occured: " + 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);
});
}
});