javascript - 如何防止此ajax请求丢失数据

Question

我的博文末尾有一个评论表。到目前为止的测试中，绝大多数评论都成功保存到数据库中，但偶尔页面给人的印象是评论已成功发布 - 但在重新加载页面后，评论消失了(并检查数据库表确认它从来没有做到那么远)。有没有办法在某处修改我的代码以捕获这些异常事件？

我知道 $.ajax 有一个错误函数，但我不认为在这种情况下添加它会有帮助。实际的 ajax 请求似乎在工作——因为它总是运行“成功”函数中的内容。那么也许是 postComment.php 需要修改？

表单提交背后的代码:

if( $(".blogPost").length ) {
    $(".commentForm").submit(function(e) {
        e.preventDefault();
    }).validate({
            submitHandler: function (form) {
                var url = window.location.pathname;
                var post_url = url.substring(url.lastIndexOf('/') + 1);
                $("input[name=post_url]").val(post_url);
                var formData = $(form).serialize();
                var post_id = $(".post").attr("id");
                $.ajax({
                    url:"/postComment.php?PostID=" + post_id,
                    type:"POST",
                    data: formData,
                    success:function(data){
                        $(".comments").prepend(data);
                        $("#commentName").val("");
                        $("#commentEmail").val("");
                        $("#commentWebsite").val("");
                        $("#comment").val("");
                        $(".commentForm input[type='submit']").val('Success!').delay(5000).queue(function(){
                            $(".commentForm input[type='submit']").val('Post Comment');
                        });
                    }
                });
            }
        });

}

postComment.php 页面代码:

<?php
include('dbconnect.php');

$name = $_POST['commentName'];
$email = $_POST['commentEmail'];

$website = $_POST['commentWebsite'];
if( $website != ''){
    if  ( $ret = parse_url($website) ) {

          if ( !isset($ret["scheme"]) )
           {
           $website = "http://{$website}";
           }
    }
}

$comment = $_POST['comment'];
$date = date('Y-m-d H:i:s');
$post_id = $_GET['PostID'];

$blogAuthor = '';
if( $name == "Luke Twomey"){
    $blogAuthor = "<span> - Blog Author</span>";
}else{
    $blogAuthor = false;
}

$SQL = "INSERT INTO comments (name, email, website, comment, date, post_id) VALUES ('$name', '$email', '$website', '$comment', '$date', '$post_id')";
mysqli_query($link, $SQL);

echo "<section class='comment'>
            <h3 class='commentAuthor'>$name$blogAuthor</h3>
            <a href='$website'><p class='commentAuthorWebsite'>$website</p></a>
            <p class='postDate'>$date</p>
            <p>$comment</p>
        </section>";

$subject = $name . $_POST['subject'];
$post_url = $_POST['post_url'];
$postedMessage = $_POST['comment'];
$contentForEmail = $postedMessage.'<br><a href="http://www.fakedomainhere.com/blog/'.$post_url.'#comments"><p>View comment on website</p></a>';

$header = "From: fake-email-here\n"
. "Reply-To: fake-email-here\n" . "Content-Type: text/html; charset=ISO-8859-1\r\n";

$email_to = "fake-email-here";

mail($email_to, $subject , $contentForEmail, $header );


?>

Answer 1

首先确保插入成功，然后只返回成功信息。

if(mysqli_query($link, $SQL)){
echo "<section class='comment'>
            <h3 class='commentAuthor'>$name$blogAuthor</h3>
            <a href='$website'><p class='commentAuthorWebsite'>$website</p></a>
            <p class='postDate'>$date</p>
            <p>$comment</p>
        </section>";
}else{
    echo "problem while inserting"; //or return an array with some status to tell the user to submit again.
// or header('HTTP/1.0 500 Internal Server Error'); exit;
    }