我有一个Jquery函数可以根据当前表单激活和停用表单。当我单击“提交”按钮时,如果将其设置为ID,则可以在页面加载中获取activeform的值,但第二个表单中没有任何值。当我将其设置为class时,它似乎根本不会加载任何东西,并且当我单击提交时我也没有收到任何警报。我的jQuery是这样的:
<script>
$(document).ready(function() {
var activeform = "register";
$("#loginlink").click(function(e) {
var activeform = "login";
$("#login").show("blind", 1000);
$("#register").hide("blind", 1000);
})
$("#registerlink").click(function(e) {
var activeform = "register";
$("#register").show("blind", 1000);
$("#login").hide("blind", 1000);
})
$(".submit").click(function(e) {
e.preventDefault();
alert(activeform);
});
});
</script>
以及我表单的代码:
<p>
<form id="register">
<div class='row'>
<div class="col-xs-4"></div>
<div class="col-xs-4">
<label for username><i class="fa fa-user"></i> Username</label>
<input class="form-control" type="text" id="username"></div>
<div class="col-xs-4"></div>
</div>
<div class="row">
<div class="col-xs-4"></div>
<div class="col-xs-4">
<label for password><i class="fa fa-key"></i> Password</label>
<input class="form-control" type="password" id="password"></div>
<div class="col-xs-4"></div>
</div>
<div class="row">
<div class="col-xs-4"></div>
<div class="col-xs-4">
<label for email><i class="fa fa-envelope"></i> Email</label>
<input class="form-control" type="text" id="email"></div>
<div class="col-xs-4"><i class="fa fa-info-circle" title="You may be notified of delayed or denied payments." data-toggle="tooltip"></i></div>
</div>
<div class="row">
<div class="col-xs-4"></div>
<div class="col-xs-4">
<label for submit><i class="fa fa-info" data-toggle="tooltip" title="By Registering you Agree to be Bound by our Terms of Service"></i></label>
<button type="button" class="btn btn-success form-control" class="submit">Register</button></div>
<div class="col-xs-4"></div>
</div>
</form>
<form id="login" style="display:none;">
<div class='row'>
<div class="col-xs-4"></div>
<div class="col-xs-4">
<label for username><i class="fa fa-user"></i> Username</label>
<input class="form-control" type="text" id="username"></div>
<div class="col-xs-4"></div>
</div>
<div class="row">
<div class="col-xs-4"></div>
<div class="col-xs-4">
<label for password><i class="fa fa-key"></i> Password</label>
<input class="form-control" type="password" id="password"></div>
<div class="col-xs-4"></div>
</div>
<div class="row">
<div class="col-xs-4"></div>
<div class="col-xs-4">
<br />
<button type="button" class="btn btn-success form-control" class="submit">Login</button></div>
<div class="col-xs-4"></div>
</div>
</form>
</p>
登录| 注册最后但并非最不重要的。这是一个小提琴。:) https://jsfiddle.net/rm6j5da9/2/
您activeform
每次在click事件中都在重新创建变量。删除var
关键字。您已经将其声明为文档就绪范围内的全局变量。
$(function(){
var activeform = "register";
$("#loginlink").click(function(e) {
activeform = "login";
$("#login").show("blind", 1000);
$("#register").hide("blind", 1000);
});
$("#registerlink").click(function(e) {
activeform = "register";
$("#register").show("blind", 1000);
$("#login").hide("blind", 1000);
});
$(".submit").click(function(e) {
e.preventDefault();
alert(activeform);
});
});
另外,您还需要确保在登录和注册按钮上都具有Submit CSS类。
这是一个工作样本
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