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JavaScript Ajax 函数返回错误,状态代码为 'OK'

转载 作者:行者123 更新时间:2023-11-28 08:38:38 25 4
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我不明白为什么这个 ajax 请求返回错误,即使错误状态正常(200),有什么想法吗?似乎我错过了一些非常明显的东西,但我一直在努力寻找答案!我是 ajax 新手,因此几乎没有经验!感谢您的帮助。

这是我的ajax请求:

    var dataString = $('#cform').serialize();
$.ajax({
type: "POST",
url: 'contact.php',
data: dataString,
dataType: 'json',
success: function (data) {
if (data.success == 0) {
var errors = '<ul><li>';
if (data.name_msg != '')
errors += data.name_msg + '</li>';
if (data.email_msg != '')
errors += '<li>' + data.email_msg + '</li>';
if (data.godaddyemail_msg != '')
errors += '<li>' + data.godaddyemail_msg + '</li>';
if (data.godaddyuser_msg != '')
errors += '<li>' + data.godaddyuser_msg + '</li>';

$("div#output").removeClass('alert-success').addClass('alert-error').show().html('<button type="button" class="close" data-dismiss="alert">x</button><p> Could not complete your request. See the errors below!' + errors + '</p>');
}
else if (data.success == 1) {

$("div#output").removeClass('alert-error').addClass('alert-success').show().html('<button type="button" class="close" data-dismiss="alert">x</button><p>You message has been sent successfully!</p>');
}

},

error: function (error) {
$("div#output").removeClass('alert-success').addClass('alert-error').show().html('<button type="button" class="close" data-dismiss="alert">x</button><p> Could not complete your request. The Function returned an error!</p>' + error.statusText); }
});

return false;

}

这是 contact.php(请求发送到的位置):

<?php 

$send_email_to = "My Email";
$message = 'message';
$subject = 'subject';
function send_email($name,$email,$godaddyusername,$godaddyemail)
{
global $send_email_to;
if($message=='message')$message='';
$headers = "MIME-Version: 1.0" . "\r\n";
$headers .= "Content-type:text/html;charset=iso-8859-1" . "\r\n";
$headers .= "From: ".$email. "\r\n";
$message = "<strong>Email = </strong>".$email."<br>";
$message .= "<strong>Name = </strong>".$name."<br>";
$message .= "<strong>GoDaddy Username = </strong>".$godaddyusername."<br>";
$message .= "<strong>GoDaddy Email = </strong>".$godaddyemail."<br>";
mail($send_email_to, $subject, $message,$headers);
return true;
}

function validate($name,$email,$godaddyusername,$godaddyemail)
{
$return_array = array();
$return_array['success'] = '1';
$return_array['name_msg'] = '';
$return_array['email_msg'] = '';
$return_array['godaddyuser_msg'] = '';
$return_array['godaddyemail_msg'] = '';

if($email == '')
{
$return_array['success'] = '0';
$return_array['email_msg'] = 'Email is required';
}
else
{
$email_exp = '/^[A-Za-z0-9._%-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}$/';
if(!preg_match($email_exp,$email)) {
$return_array['success'] = '0';
$return_array['email_msg'] = 'Enter valid emmail.';
}
}

if($name == '')
{
$return_array['success'] = '0';
$return_array['name_msg'] = 'Name is required';
}
else
{
$string_exp = "/^[A-Za-z .'-]+$/";
if (!preg_match($string_exp, $name)) {
$return_array['success'] = '0';
$return_array['name_msg'] = 'Enter valid Name.';
}
}

if($godaddyusername == '')
{
$return_array['success'] = '0';
$return_array['godaddyuser_msg'] = 'GoDaddy Username is required';
}

if($godaddyemail == '')
{
$return_array['success'] = '0';
$return_array['godaddyemail_msg'] = 'GoDaddy Email is required';
}

else
{
$email_exp = '/^[A-Za-z0-9._%-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}$/';
if(!preg_match($email_exp,$godaddyemail)) {
$return_array['success'] = '0';
$return_array['godaddyemail_msg'] = 'Enter a valid GoDaddy email.';
}
}

return $return_array;
}

$name = $_POST['name'];
$email = $_POST['email'];
$godaddyemail = $_POST['gde'];
$godaddyusername = $_POST['gdu'];

$return_array = validate($name,$email,$godaddyusername,$godaddyemail);
if($return_array['success'] == '1')
{
send_email($name,$email,$godaddyusername,$godaddyemail);
}

header('Content-type: text/json');

echo json_encode($return_array);
die();

?>

最佳答案

这是因为您的 send_email() 函数中有两个 undefined variable 。

这里有这个 if 语句..

if($message=='message')$message='';   // $message is not available inside this function.

这里

mail($send_email_to, $subject, $message,$headers);  //$subject is also undefined.

关于JavaScript Ajax 函数返回错误,状态代码为 'OK',我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20773637/

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