使用 Smarty & Mysql & Jquery 的问题

Question

我在使用 Samrtm Mysql 和 Jquery 或 Ajax 时遇到问题 如果我使用 div 自动重新加载器 它显示一个php错误.. 这些是我的文件： accept.html（模板）

{literal}
<script type="text/javascript">
var auto_refresh = setInterval(
function ()
{
$('#invite').load('accept.php').fadeIn("slow");
}, 1000);
</script>
{/literal}
<div id="invite">
<fieldset>
<legend>Invites</legend>
{section name=ct loop=$invites}
You have an invite {$sender.user}<br />
<a href="?action=accept_contest&sid={$invites[ct].sendid}">Accept</a> | <a href="?action=cancel_contest&sid={$invites[ct].sendid}">Cancel</a>
{/section}
</fieldset>
</div>

accept.php（主文件）

<?php

$time = date("i:s",time("s"));
$sql_select_invite = mysql_query("SELECT * FROM invite WHERE recid = '$_SESSION[id]' AND accept = '0' AND endtime > '$time'");

while($invite = mysql_fetch_array($sql_select_invite)){
 $sender = mysql_fetch_array(mysql_query("SELECT * FROM member WHERE id = '$invite[sendid]'"));
    $contest->assign("sender", $sender);
    $invites[] = $invite;
}
$contest->assign("invites", $invites);

?>

当我使用 Jquery 或 Ajax 时 它显示此错误：

警告：mysql_fetch_array()：提供 参数不是有效的 MySQL 结果 资源在 C:\AppServ\www\somefo\accept.php 上线6

致命错误：调用成员函数 在非对象上分配（） C:\AppServ\www\somefo\accept.php 上线11

我不知道为什么它会显示这个.. 希望你能回答我 以及如何解决这个问题

谢谢..

Answer 1

对于 php：

<?php

$sql = mysql_connect(......);  // connect to your server
if (!$sql) {
    die('Can\'t connect to database');
}

// select your database
mysql_select_db('your_db', $sql) or die('Can\'t select database');

$time = date("i:s",time("s"));
$sql_select_invite_query = "SELECT * FROM invite WHERE recid = '$_SESSION[id]' AND accept = '0' AND endtime > '$time'";

// print it if you want to see sql query to test it in phpmyadmin
echo $sql_select_invite_query; 
// --------------------------------------------------------------

$sql_select_invite = mysql_query($sql_select_invite_query, $sql);

if (!$sql_select_invite) {    // check for error in query
    die(mysql_error());       // and print it
}

$invites = array();  // don't forget to make an empty array

while($invite = mysql_fetch_array($sql_select_invite, $sql)){
    // this place is not good, just make it as you do before, with checks
    $sender = mysql_fetch_array(mysql_query("SELECT * FROM member WHERE id = '$invite[sendid]'"));

    $contest->assign("sender", $sender);
    $invites[] = $invite;
}
$contest->assign("invites", $invites);   // in your case there is no $invite 
                                         // array cause mysql_fetch_array returned
                                         // false and it is not defined before [while]

?>

供JS测试：

setInterval( function () {
  $('#invite').load('accept.php', function() {
    alert('loaded');  // it is to check content is loaded
  }).fadeIn("slow");
}, 1000);