【发布时间】:2020-05-25 10:35:29
【问题描述】:
如果与 html 放在同一个文件上,我的 jQuery 将无法正常工作,输出将只有一个值 password,并且不会显示 email,但如果当我点击 Login 按钮时放置在不同的文件中,值 console.log(datatopost) 将在控制台上消失(如刷新)。
HTML 代码
<!-- Login Modal -->
<form id="loginForm" method="post">
<div class="modal fade" id="loginModal" tabindex="-1" role="dialog" aria-labelledby="exampleModalLabel" aria-hidden="true">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<h5 class="modal-title" id="exampleModalLabel">Login to your Account</h5>
<!-- SIGN IN ERROR MESSAGE-->
<div id="loginerrorMessage"></div>
<button type="button" class="close" data-dismiss="modal" aria-label="Close">
<span aria-hidden="true">×</span>
</button>
</div>
<div class="modal-body">
<div class="form-group">
<label for="loginemail" class="sr-only">Email address</label>
<input type="email" class="form-control" id="loginemail" aria-describedby="emailHelp" placeholder="Enter email" autocomplete="on">
</div>
<div class="form-group">
<label for="loginpassword" class="sr-only">Confirm Password</label>
<input type="password" class="form-control" name="password" id="loginpassword" placeholder="Enter Password" autocomplete="on">
</div>
<h6>Login as:</h6>
<div class="form-check form-check-inline">
<input type="radio" name="loginRadio" class="form-check-input" id="checkbank" value="bank">
<label class="form-check-label" for="checkbank">Bank</label>
</div>
<div class="form-check form-check-inline">
<input class="form-check-input" type="radio" name="loginRadio" id="checkcustomer" value="customer">
<label class="form-check-label" for="checkcustomer">Customer</label>
</div>
<div class="form-check form-check-inline">
<input class="form-check-input" type="radio" name="loginRadio" id="checkadmin" value="admin">
<label class="form-check-label" for="checkadmin">Admin</label>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-primary mr-auto" data-target="#signupModal" data-toggle="modal" data-dismiss="modal">Register</button>
<input type="submit" value="Login" class="btn btn-info" name="login">
<!-- <button type="submit" class="btn btn-info">Login</button> -->
<button type="button" class="btn btn-danger" data-dismiss="modal">Close</button>
</div>
</div>
</div>
</div>
</form>
jquery 代码
$(document).ready(function(){
$('#loginForm').on('submit', function(e){
//Stop the form from submitting itself to the server.
e.preventDefault();
let datatopost = $(this).serializeArray();
console.log(datatopost);
});
});
拜托我只是一个初学者...
【问题讨论】:
标签: javascript php html jquery