Php Liking系统声明变量未定义，即使它已定义

我有下面的代码来加载喜欢的特定状态的动态使用，如果用户喜欢他们自己的帖子，它会起作用。但如果一个用户喜欢其他人，它就会失败

我的代码：

<?php

  mysql_connect ('localhost', 'funding9_joeb', 'Brailsf0rdJ0e');
  mysql_select_db ('mingle');
    include('../includes/ts.php');
  $_COOKIE['wds']; 
        $ssid = $_COOKIE['wds'];

        $lu = "SELECT * FROM mingle_sessions WHERE sid = '$ssid' LIMIT 1";
        $luq = mysql_query($lu) or die (mysql_query());
        while($uidr = mysql_fetch_assoc($luq)) {
            $uid = $uidr['uid'];
        }

    $sql = "SELECT * FROM users WHERE uid = '$uid' LIMIT 1";


$s = htmlentities(strip_tags(stripslashes($_GET['sid'])));


$result = mysql_query($sql) or print ("Can't select entry from table mingle_usr.<br />" . $sql . "<br />" . mysql_error());

while($row = mysql_fetch_array($result)) {
    $fname = stripslashes($row['fname']);
    $sname = stripslashes($row['sname']);
    $dp = $row['dp'];
    }
  //--------------------------------------------------------------------------
  // 2) Query database for data
  //--------------------------------------------------------------------------

    $sql = "SELECT * FROM mingle_likes WHERE byuid = '$uid' AND onuid = '$s' AND type = 'status'";
    $sr = mysql_query($sql) or die(mysql_error());

if(mysql_num_rows($sr) >= 1) {

    $re = ("SELECT id as id, status as status, sid as sid, UNIX_TIMESTAMP(timestamp) as timestamp FROM mingle_status WHERE uid = '$uid' AND sid = '$s' ORDER BY timestamp DESC LIMIT 1"); //query

  $result = mysql_query($re) or die (mysql_error());
  while($st = mysql_fetch_assoc($result)) {
    $status = nl2br($st['status']);
    $sid = $st['sid'];  
    $td = $st['timestamp'];
    $id = $st['id'];
    }

    $like = mysql_fetch_array(mysql_query("SELECT COUNT(*) as num FROM mingle_likes WHERE onuid = '$sid' AND type = 'status' AND byuid != '$uid'"));
    $likes = ceil($like['num']);
    $haslike = mysql_fetch_array(mysql_query("SELECT COUNT(*) as nu FROM mingle_likes WHERE onuid = '$sid' AND type = 'status' AND byuid = '$uid'"));
    $hasliked = ceil($haslike['nu']);

?>
<script type="text/javascript" src="js/ld_s.js"></script>
<form action='ld.php' method='post' id='ls' style='display:inline; border:0px; margin: 0 0 0 0; padding: 0 0 0 0;'>
<input type="hidden" class="do_<?php echo $sid; ?>" name="onuid" value="<?php echo $sid; ?>" />
<input type="hidden" class="db_<?php echo $sid; ?>" name="byuid" value="<?php echo $uid; ?>" />
<input type="hidden" class="dp_<?php echo $sid; ?>" name="uid" value="<?php echo $uid; ?>" />
<input type="hidden" class="dt_<?php echo $sid; ?>" name="type" value="status" />
<?php
if($hasliked == 1) {
    ?>
    <label>You <?php if($likes >= 1) { echo "and " . $likes . " other people like this"; } else{ echo "liked this |"; }?></label><input type="submit" id="like" class="<?php echo $sid; ?>" name="submit" value="Unlike" />
    <?php
}
else {
    ?>
    <label><?php if($likes >= 1) { echo $likes . " people like this |"; }?></label><input type="submit" id="like" class="<?php echo $sid; ?>" name="submit" value="Like" />
    <?php
}
?>
</form> 
<a href="#" id="time"><?php echo time_since($td); ?></a>

    <?php
}
?>

然后，对于$sid的每次出现，都是相同的

我完全不知道为什么，就像我说的，它对喜欢他/她自己的帖子的用户来说是完美的，而不是其他人

---

有什么想法吗-/

你确定你不是故意的吗？虽然这堵没有藏宝图或某处X标记的代码墙有点难说。@PeeHaa是的，它来自它上面的while循环，$sid=$st['sid']；您正在while循环中声明它。您确定查询有结果吗？PHP的mysql函数是。有一些既受支持又受支持。

( ! ) Notice: Undefined variable: sid in C:\wamp\www\mingle\ajax\loadslikes.php on line 47