Question

嗨我收到下面的错误，我已经仔细检查了一切，不知道为什么显示它。警告：mysql_num_rows（）：提供的参数不是第6行/home/namebob/public_html/site_reg/statlookup.php中的有效MySQL结果资源

<?php
    include_once('common.php');
    $cid        =   $_REQUEST['cid'];
    $_elmID     =   $_REQUEST['elmID'];
    $scan = mysql_query("SELECT  `id`, `state` FROM `mast_state` WHERE `countryid` = $cid");
    if(mysql_num_rows($scan)>0)
    {
...

Answer 1

值得尝试var_dump($scan)，如果你得到类似null / false的东西，那就意味着查询失败了。如果是这种情况，请尝试echo mysql_error()检查是否有任何sql错误（缺少表格等）

Answer 2

很可能您的查询失败了。尝试：

if($scan) {
  if(mysql_num_rows($scan) > 0) {
    //...
  }
} else {
   trigger_error(mysql_error()); 
};

Answer 3

$scan = mysql_query("SELECT  `id`, `state` FROM `mast_state` WHERE `countryid` = $cid");
    $stock_num = '';
    if($scan )
    {
        while($row=mysql_fetch_object($scan))
        {
            if($row->state)
            $state  = $row->state;
        }   
    }

Answer 4

$scan = mysql_query("SELECT  `id`, `state` FROM `mast_state` WHERE `countryid` = '".mysql_real_escape_string($cid)."'");

但实际上，请使用PDO。

$stm = $db->prepare("SELECT  `id`, `state` FROM `mast_state` WHERE `countryid` = ?");
if( $stm && $stm->execute(array($cid)) ) {
  while( $data = $stm->fetch(PDO::FETCH_ASSOC) ) {

  }
}

警告：mysql_num_rows（）：提供的参数无效

4 个答案: