提交表格後,我有一個HTML格式如下:形式disappers用ajax
<form role="form" name="login" id="login" >
<div id="alert-success" class="alert alert-success" style="display: none" ></div>
<div id="alert-danger" class="alert alert-danger" style="display: none" ></div>
<div class="form-group">
<label for="exampleInputEmail1">Name</label>
<input type="name" class="form-control" id="username" placeholder="Enter name" style="width: 30%;">
</div>
<div class="form-group">
<label for="exampleInputEmail1">Email address</label>
<input type="email" class="form-control" id="usermail" placeholder="Enter email" style="width: 30%;">
</div>
<div class="form-group">
<label for="exampleInputPassword1">Username</label>
<input type="password" class="form-control" id="loginname" placeholder="username" style="width: 30%;">
</div>
<div class="form-group">
<label for="exampleInputPassword1">Password</label>
<input type="password" class="form-control" id="loginpassword" placeholder="Password" style="width: 30%;">
</div>
<div class="checkbox">
<label>
<input type="checkbox"> Check me out
</label>
</div>
<button type="submit" id="submit" data-dismiss="alert" name="submit" class="btn btn-default">Join Us</button>
</form>
在同一個文件一個jQuery進行驗證,並提交形式:
$(document).ready(function() {
$("#submit").click(function() {
var username = $("#username").val();
var usermail = $("#usermail").val();
var loginname = $("#loginname").val();
var loginpassword = $("#loginpassword").val();
var emailregex = /^([a-zA-Z0-9_\.\-])+\@(([a-zA-Z0-9\-])+\.)+([a-zA-Z0-9]{2,4})+$/;
if (username == "" || username.length > 20 || !isNaN(username)) {
$('#alert-danger').show();
$("#alert-danger").html("Please Enter Proper Name");
$("#username").focus();
return false;
} else if (usermail == "" || !emailregex.test(usermail)) {
$('#alert-danger').show();
$("#alert-danger").html("Please Enter Your Email Id");
$("#usermail").focus();
return false;
} else if (loginname == "" || loginname.length > 10) {
$('#alert-danger').show();
$("#alert-danger").html("Please Enter Username");
$("#loginname").focus();
return false;
} else if (loginpassword == "" || loginpassword.length > 10) {
$('#alert-danger').show();
$("#alert-danger").html("Please Enter Proper Password");
$("#loginpassword").focus();
return false;
} else {
var datastring = 'name=' + username + '&email=' + usermail + '&loginusername=' + loginname + '&loginuserpassword=' + loginpassword;
$.ajax({
type: "POST",
url: "userdata.php",
data: datastring,
cache: false,
sucess: function (dataitem) {
if (dataitem == 1) {
$("#alert-danger").html("Something Went Wrong");
} else {
$("#alert-danger").html("Something Went Wrong");
}
}
})
}
})
})
現在我遇到的問題在提交表單時,它確實起作用,因爲數據已保存到數據庫表中,但在提交後我無法看到我的表單,但我仍然使用location.reload();
函數,但它仍然不工作,請幫助
寫「返回false 「在功能的最後。 – SeeTheC
來吧人們,當已經使用jQuery時,爲什麼不這樣做的「正確」的方式...使用'event.preventDefault' – CBroe