Javascript Html表单拒绝通过JQuery提交
因此,我在学习AJAX/JQuery教程,学习一个注册脚本,该脚本将由PHP/MySQL执行,并通过JQuery提交 现在我遇到的问题是表单直接提交到操作页面,这不应该是这样,因为它应该提交到script.js以下是表单的html代码Javascript Html表单拒绝通过JQuery提交,javascript,php,jquery,html,ajax,Javascript,Php,Jquery,Html,Ajax,因此,我在学习AJAX/JQuery教程,学习一个注册脚本,该脚本将由PHP/MySQL执行,并通过JQuery提交 现在我遇到的问题是表单直接提交到操作页面,这不应该是这样,因为它应该提交到script.js以下是表单的html代码 <form method="post" id="register-form" action="transact-user.php"> <h2 class="form-signin-heading"></h2>
<form method="post" id="register-form" action="transact-user.php">
<h2 class="form-signin-heading"></h2>
<div class="form-group">
<div class='row'>
<div class='col-sm-6'>
<input type='text' class='form-control' id='fname' name='fname' placeholder="First name">
</div>
<div class='col-sm-6'>
<input type='text' class='form-control' id='lname' name='lname' placeholder="Last name">
</div>
</div>
</div>
<div class="form-group">
<input type="text" class="form-control" placeholder="Phone Number" name = "phone" id = "phone" >
</div>
<div class="form-group">
<input type="email" class="form-control" placeholder="Email address" name = "email" id ="email">
<span id="check-e"></span>
</div>
<div class="form-group">
<input type="password" class="form-control" placeholder="Password" name = "password" id = "password">
</div>
<div class="form-group">
<input type="password" class="form-control" placeholder="Password Again" name = "confirmpassword" id = "confirmpassword">
</div>
<div class="form-group">
<button class="btn btn-primary btn-block btn-apply" type="submit" name="btn-save" id = "btn-submit"><span class="glyphicon glyphicon-log-in"></span> Register</button>
</div>
</form>
</div> <!-- /container -->
<div id = "ack"></div>
最后,这是操作php脚本transact-user.php
<?php
if($_POST)
{
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$email = $_POST['email'];
$phone = $_POST['phone'];
$password = $_POST['password'];
$confirmpassword = $_POST['confirmpassword'];
if($email != '') {
$qry = "SELECT * FROM users WHERE email='$email'";
$result = mysqli_query($mysqli,$qry);
if($result) {
if(mysqli_num_rows($result) > 0) {
echo "Email already in use";
}
@mysqli_free_result($result);
}
else {
die("Query failed");
}
}
$activation = md5(uniqid(rand(), true));
if ($stmt = "INSERT INTO users(firstname, lastname, email, password, verification, phone)"
." values('$fname', '$lname', '$email', '$password', "
. "'$activation','$phone')" or
die("Could not perform query ".mysqli_error($mysqli))) {
$result = mysqli_query($mysqli, $stmt)
or die("The system could not register you"
. "".mysqli_error($mysqli) . "<br>" . $stmt);
if ($result){
echo "sent";
}
//echo "<a href=\"gethotel.php?hid=".$row['hotel_id'].
//<a href = ""></a>
/* close statement */
//$stmt->close();
}
}
?>
之所以向我们发出此问题,是因为您正在挂接提交
按钮的单击
事件,并且没有阻止事件完成
还要注意,用于生成查询字符串的选择器不正确,因为在:
之前应该有一个空格,而在:
和输入之间应该没有空格,例如$(“#注册表单:输入”)
要修复问题,请将钩子连接到表单的提交事件
,并使用preventDefault()
。试试这个:
$("#register-form").submit(function(e) {
e.preventDefault();
if ($("fname").val().trim() == "" || $("lname").val().trim() == "") {
$("div#ack").html("Please enter both your first name and your surname");
} else {
$.post(this.action, $(this).find(':input').serializeArray(), function(data) {
$("div#ack").html(data);
});
}
});
看一看PreventDefault将注册按钮类型从submit
更改为按钮
@ReaganGallant这是个坏主意。这意味着当用户点击相应字段上的“返回”按钮时,表单无法提交,并使屏幕阅读器无法访问表单。@Rorymcrossan对于屏幕阅读器,请添加属性role=“button”。他的表单是使用action方法而不是jquery发布的,因为他的按钮被设置为submit。@ReaganGallant更改了它,它工作得非常好。谢谢
$("#register-form").submit(function(e) {
e.preventDefault();
if ($("fname").val().trim() == "" || $("lname").val().trim() == "") {
$("div#ack").html("Please enter both your first name and your surname");
} else {
$.post(this.action, $(this).find(':input').serializeArray(), function(data) {
$("div#ack").html(data);
});
}
});