我有2个表单登录并在同一个index.php上注册并根据用户需要进行切换。当您在注册表单被隐藏时打开index.php并且仅在您出现时出现登录表单点击这里注册链接。
现在我的问题是,在注册时是否有任何错误(验证,数据库错误),然后当显示点击提交按钮时页面刷新并返回到我打开/刷新index.php时出现的登录表单。我必须点击这里的注册链接来解决错误。
我只想在出现错误时留在注册表单上,这意味着我不希望在点击提交时刷新页面,但希望出现错误并在那里解决它们。
$.ajax({
type: "POST",
url: "phpindex.php",
data: $("#signUpForm").serialize(),
success:function(data)
{
if ("#error".val(data))
{
e.preventDefault();
}
}
});
});
My validation and database code is written in phpindex.php
***************signup form************************
<form method="post" id="signUpForm">
<div class="container" id="signupContainer">
<h1>Sign Up</h1>
<div id="error"><?php if ($error!="") {
echo '<div class="alert alert-danger" role="alert">'.$error.'</div>';
} ?></div>
<div class="form-group row">
<label class="col-sm-2 form-control-label">Name</label>
<div class="col-sm-10">
<input type="text" class="form-control" name="name" placeholder="Your name">
<span class="fa fa-user"></span>
</div>
</div>
<div class="form-group row">
<label class="col-sm-2 form-control-label">Address1</label>
<div class="col-sm-10">
<input type="text" id="address1" class="form-control" name="address1" placeholder="Home address">
<span class="fa fa-map-marker"></span>
</div>
</div>
<div class="form-group row">
<label class="col-sm-2 form-control-label">Address2</label>
<div class="col-sm-10">
<input type="text" id="address2" class="form-control" name="address2" placeholder="City,Pincode....">
<span class="fa fa-map-marker"></span>
</div>
</div>
<div class="form-group row">
<label class="col-sm-2 form-control-label">Email</label>
<div class="col-sm-10">
<input type="text" class="form-control" name="email" placeholder="Email Address">
<span class="fa fa-envelope"></span>
</div>
</div>
<div class="form-group row">
<label class="col-sm-2 form-control-label">Password</label>
<div class="col-sm-10">
<input type="password" class="form-control" name="password" placeholder="Password">
<span class="fa fa-key"></span>
</div>
</div>
<div class="form-group row">
<div class="col-sm-offset-5 col-sm-10">
<input type="submit" id = "submit1"class="btn btn-success btn-lg" value="Sign In" name="submit">
</div>
</div>
<p><a class="toggleForms">Back to Log in</a></p>
</div>
</form>
***************login form*****************
<form method="post" id="logInForm">
<div class="container" id="logInContainer">
<h1>Log In</h1>
<div id="success"><?php if ($success!="") {
echo '<div class="alert alert-success" role="alert">'.$success.'</div>';
} ?></div>
<div class="form-group row">
<label class="col-sm-3 form-control-label">Emai</label>
<div class="col-sm-8">
<input type="email" class="form-control" placeholder="Email" name="email">
<span class="fa fa-envelope"></span>
</div>
</div>
<div class="form-group row">
<label class="col-sm-3 form-control-label">Password</label>
<div class="col-sm-8">
<input type="password" class="form-control" placeholder="Password" name="password">
<span class="fa fa-key"></span>
</div>
</div>
<div class="form-group row">
<div class="col-sm-offset-4 col-sm-10">
<input type="hidden" name="signUp" value="0">
<input type="submit" id = "submit2" class="btn btn-success btn-lg" value="Log In" name="submit">
</div>
</div>
<p><a class="toggleForms">Sign Up Here</a></p>
</div>
</form>
答案 0 :(得分:0)
首先,您提交的代码中存在大量错误。我相信它们主要是由于笨拙的复制和粘贴,但我认为你的代码中只有一个,它是#34;#error&#34;
的缺失jQuery参考。"#error".val(data)
我想你想写这样的东西
$("#error").val(data)
此错误将停止&#34; preventDefault&#34;从发生。
其次,我想知道这些表单的提交处理程序在哪里以及它的外观。如果你没有,那么你有答案。你只是没有抓住提交事件。
最后,我会考虑调用,而不是preventDefault:
return false
请参阅此主题的原因 event.preventDefault() vs. return false