大家好，我有一个(可能)简单的问题，但我就是不明白为什么它不起作用。
下面的代码应该相当容易理解:
<?php
require('../db.php');
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN"
"http://www.w3.org/TR/html4/strict.dtd">
<html lang="en">
<head>
<meta http-equiv="content-type" content="text/html; charset=utf-8">
<title>Website Checker</title>
<link href="style.css" rel="stylesheet" type="text/css" media="screen" />
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js"></script>
<script type="text/javascript" src="js/jquery.simpletip-1.3.1.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('#check').click(function(){
$('.http-status').html('<img src="images/spinner.gif"/>').fadeOut('slow').fadeIn('slow');
$('#site-list li.websites').each(function()
{
//var newurl = $(this).find('span.url a').attr('href');
var newurl = $(this).find('span.url').html();
$.ajax(
{
type: "POST",
url: "process.php",
data: ({"a":newurl}),
cache: false,
success: function(message)
{
$(this).find('span.http-status').html(message);
} //End AJAX return
}); //End AJAX call
}); //End li each
}); //End Check
$('#check').click();
});
</script>
<ul id="site-list" class="list">
<li class="title">
<span class="id"></span>
<span class="name">Title</span>
<span class="url">URL</span>
<span class="status">HTTP Status</span>
</li>
<?php
// some PHP to fetch all the gig entries from the shows table
$sql = "SELECT * FROM `check`";
$query = mysql_query($sql) or die(mysql_error());
// a loop to place all the values in the appropriate table cells
while ($row = mysql_fetch_array($query)){
//begin the loop...
$id=$row['id'];
$name=$row['name'];
$url=$row['url'];
?>
<li class="websites">
<span class="id"><?php echo $id; ?></span>
<span class="name"><?php echo $name; ?></span>
<span class="url"><?php echo $url; ?></span>
<span class="status http-status"></span>
</li>
<?php
}
?>
</ul>
<br />
<a href="#" id="check" class="button">Check Now</a>
这基本上从数据库中提取站点数据并将其呈现给用户，我想做的是当您单击要检查和显示的每个站点的http代码的按钮时(并且还首先在页面上运行)正在加载)
这似乎工作正常 - 发送和接收了正确的数据，但之后旋转器仍然在那里!
链接:http://www.4playtheband.co.uk/check/
process.php(如果有帮助):
<?php
require('../db.php');
$url= NULL;
if(isset($_POST['a'])) { $url = mysql_real_escape_string($_POST['a']); }
function Visit($url)
{
$agent = "Mozilla/4.0 (compatible; MSIE 5.01; Windows NT 5.0)";$ch=curl_init();
curl_setopt ($ch, CURLOPT_URL,$url );
curl_setopt($ch, CURLOPT_USERAGENT, $agent);
curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1);
curl_setopt ($ch,CURLOPT_VERBOSE,false);
curl_setopt($ch, CURLOPT_TIMEOUT, 5);
$page=curl_exec($ch);
//echo curl_error($ch);
$httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);
if($httpcode>=200 && $httpcode<300)
{
echo '<span class="up">'.$httpcode.'<span class="icon"><img src="images/info.png" alt="website is up"/></span></span>'; exit;
}
else
{
$httpcode=404;
$date = date("l, j \of F Y \@ H:i");
$to = "someone@domain.com";
$subject = "Urgent: $url is down";
$message = "Hello,\n\nIt appears that on our latest check of $url on $date that the site was down.\n\nRegards,\nWeb Checker";
$headers = 'From: noreply@webchecker.co.uk' . "\r\n" .
'Reply-To: noreply@webchecker.co.uk' . "\r\n";
mail($to, $subject, $message, $headers);
echo '<span class="down">'.$httpcode.'<span class="icon"><img src="images/info.png" alt="website is down"/></span></span>'; exit;
}
}
if(!empty($url)){ Visit($url); exit; }
?>
最佳答案
问题出在这里:
$(this).find('span.http-status').html(message);
根据 [jQuery .ajax 文档][1]:
The this reference within all callbacks is the object in the context option passed to $.ajax in the settings; if context is not specified, this is a reference to the Ajax settings themselves.
要解决此问题，您可以将 $( this ) 的值分配给另一个变量，并在 success 回调正文中访问该变量:
$('#site-list li.websites').each(function()
{
theElement = $( this );
var newurl = $(this).find('span.url').html();
$.ajax(
{
// ...
success: function(message)
{
theElement.html(message);
} //End AJAX return
关于php - 这个 jQuery 选择器的问题，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/7285424/