我在博客文章的末尾有一个评论表单。到目前为止，在测试中，绝大多数情况下，评论都会成功保存到数据库中，但偶尔页面会给人一种成功发布评论的印象，但在重新加载页面后，评论就消失了（检查数据库表可以确认它从未成功发布过评论）。有没有办法在某个地方修改我的代码来捕捉这些反常事件？
我知道$.ajax有一个错误函数，但我认为在这个例子中添加它不会有帮助。实际的ajax请求似乎正在工作，因为它总是运行"success"函数中的内容。所以也许是postComment.php需要修改？
提交表单背后的代码：
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页面上的代码：
<?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, $header );
?>
首先确保插入成功，然后只返回成功消息。
if(mysqli_query($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;
}
用以下内容确认插入没有问题：
$result = mysqli_query($link, $SQL);
if(!$result) {
printf("Error: %s'n", mysqli_error($link));
} else {
echo "<section ...
您可能应该了解如何正确地转义输入：http://php.net/manual/en/mysqli.real-escape-string.php
服务器端：
-
检查SQL查询是否成功。
-
echotrue或false基于查询的成功或失败。
客户端：
$.ajax({
url: "/postComment.php?PostID=" + post_id,
type: "POST",
data: formData,
success:function(data){
if(data == 'true') { // Prepend only If Successful
$(".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');
});
} else { // Error
alert("There was an issue in submitting your Comment. Please try again.");
}
}
});