Php 如何使用AJAX更新表单?
我正在尝试通过AJAX更新表单,以便在提交表单时不会重定向。现在,我已经遵循了一个教程,但似乎无法使这项工作。当我点击提交按钮时,什么也没发生 HTML:Php 如何使用AJAX更新表单?,php,mysql,ajax,Php,Mysql,Ajax,我正在尝试通过AJAX更新表单,以便在提交表单时不会重定向。现在,我已经遵循了一个教程,但似乎无法使这项工作。当我点击提交按钮时,什么也没发生 HTML: <form class="profile_form" id="form" method="GET" action="include/profile/form/settings.php"> <div class="profileFlex"> <div class="settings-grid"&
<form class="profile_form" id="form" method="GET" action="include/profile/form/settings.php">
<div class="profileFlex">
<div class="settings-grid">
<div class="profile-table">
<div class="input-holder">
<label for="first_name">First name</label>
</div>
<div class="input-holder">
<input type="text" name="first_name" value="<?php echo $firstname[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder">
<label>Last name</label>
</div>
<div class="input-holder">
<input type="text" name="last_name" id="last_name" value="<?php echo $lastname[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder">
<label>Address</label>
</div>
<div class="input-holder">
<input type="text" name="address" value="<?php echo $address[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder">
<label>City</label>
</div>
<div class="input-holder">
<input type="text" name="city" id="city" value="<?php echo $city[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder">
<label>ZIP code</label>
</div>
<div class="input-holder">
<input type="text" name="zip" value="<?php echo $zip[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder">
<label>Country</label>
</div>
<div class="input-holder">
<input type="text" name="country" value="<?php echo $country[0]; ?>">
</div>
</div>
</div>
<div class="settings-grid">
<div class="profile-table">
<div class="input-holder">
<label>Telephone</label>
</div>
<div class="input-holder">
<input type="text" name="phone" value="<?php echo $phone[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder">
<label>Telephone 2</label>
</div>
<div class="input-holder">
<input type="text" name="phone2" value="<?php echo $phone2[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder">
<label>Mobile phone</label>
</div>
<div class="input-holder">
<input type="text" name="mobile" value="<?php echo $mobile[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder">
<label>Mobile phone 2</label>
</div>
<div class="input-holder">
<input type="text" name="mobile2" value="<?php echo $mobile2[0]; ?>">
</div>
</div>
<div class="profile-table">
<div class="input-holder checkbox-holder">
<input type="checkbox" id="sendmails" class="profilebox" name="checkbox" <?php if ($checkbox[0] == 'on') { print 'checked="checked" '; } ?>>
<label for="sendmails">Allow website to send promotional emails</label>
</div>
</div>
<div class="profile-table">
<div class="input-holder checkbox-holder">
<input type="checkbox" class="profilebox" name="checkbox2" id="hide_mail" <?php if ($checkbox2[0] == 'on') { print 'checked="checked" '; } ?>>
<label for="hide_mail">Hide email from listings</label>
</div>
</div>
</div>
</div>
<input type="submit" name="update" id="update" value="Update" />
</form>
<?php
session_start();
$detectUser = $_SESSION['token'];
include "../../../layout/config.php";
if (isset($_GET['update'])) {
$first_name = mysqli_real_escape_string($conn, $_GET['first_name']);
$last_name = mysqli_real_escape_string($conn, $_GET['last_name']);
$address= mysqli_real_escape_string($conn, $_GET['address']);
$city = mysqli_real_escape_string($conn, $_GET['city']);
$zip = mysqli_real_escape_string($conn, $_GET['zip']);
$country = mysqli_real_escape_string($conn, $_GET['country']);
$phone = mysqli_real_escape_string($conn, $_GET['phone']);
$phone2 = mysqli_real_escape_string($conn, $_GET['phone2']);
$mobile = mysqli_real_escape_string($conn, $_GET['mobile']);
$mobile2 = mysqli_real_escape_string($conn, $_GET['mobile2']);
if(isset($_GET['checkbox'])){
$checkbox = $_GET['checkbox'];
}else{
$checkbox = "";
}
if(isset($_GET['checkbox2'])){
$checkbox2 = $_GET['checkbox2'];
}else{
$checkbox2 = "";
}
$sql = "UPDATE users SET user='$first_name', last_name='$last_name', address='$address', city='$city', zip='$zip', country='$country', phone='$phone', phone2='$phone2', mobile='$mobile', mobile2='$mobile2', checkbox='$checkbox', checkbox2='$checkbox2' WHERE `token` = '$detectUser'";
if(mysqli_query($conn, $sql)){
echo "Records inserted successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($conn);
}
}
?>
$(document).ready(function (e) {
$('#update').click(function (event)
{
event.preventDefault();
$.ajax({
data: $('form').serialize(),
url: "include/profile/form/settings.php", //php page URL where we post this data to save in database
type: 'GET',
success: function (strMessage) {
$('#message').text("strMessage");
}
})
});
});
我知道这个问题已经被问过很多次了,但我现在才第一次面对AJAX,所以任何帮助都将意味着很多!谢谢 将
async
转换为false
。我还建议添加一个error()
回调
$(document).ready(function (e) {
$('#update').click(function (event) {
event.preventDefault();
$.ajax({
data: $('form').serialize(),
url: "include/profile/form/settings.php", //php page URL where we post this data to save in database
type: 'GET',
success: function (strMessage) {
$('#message').text(strMessage);
},
error: function(ts) {
console.log(ts)
},
async: false
});
});
});
您还应该确保您正在php脚本中检查
$\u POST
,而不是$\u GET
您正在使用ajax请求(正文中的数据)发送POST数据,但希望在php脚本中获取数据(url中的数据)。替换$('#消息')。文本(“strMessage”)带有$('#message').html(strMessage)的code>
尝试了这两种方法,但都没有修复。当控制台中的async为false时,我发现主线程上的[Deprection]Synchronous XMLHttpRequest已被弃用,因为它会对最终用户的体验产生有害影响。要获得更多帮助,请检查。我还将php文件中的所有GET替换为POST,但当我单击按钮时,仍然没有发生任何事情。您应该尝试从虚拟页面中添加$u POST以包括/profile/form/settings.php,然后查看发生了什么。您通常不会通过脚本中的AJAX调用返回错误消息,只是什么都没有发生。很明显,调用已经发送(因此您有正确的url),它也没有任何错误(因为错误函数没有响应),因此我认为您发布到的脚本中有问题。