我似乎无法摆脱这个错误。

警告:mysql_fetch_array():第47行/windsor.php中提供的参数不是有效的MySQL结果资源

if ($insertdb == NULL)
  mysql_query("INSERT INTO `asd` (`id`, `1`, `2`,`3`, `4`, `5`, `pubdate`) VALUES (296, maddeal', 'Windsor', 'ON', '', '', '') ON DUPLICATE KEY UPDATE `1`=VALUES(`1`),`2`=VALUES(`2`),`3`=VALUES(`3`),`4`=VALUES(`4`),`5`=VALUES(`5`)") or die(mysql_error());
 else
//Check and see if value has changed...

$checksql = mysql_query("SELECT `1` FROM deal WHERE `id`= 296") or die(mysql_error());
while($row = mysql_fetch_array($checksql))
  {
  $checksqlver = $row['deal'];
  }

$checksqlver = mysql_real_escape_string($checksqlver);
//$checksqlver = stripslashes($checksqlver);
echo "$checksqlver<br>";

if ($checksqlver == $insertdb)
  exit();
 else
  echo "No Match<br>";

//percent

最佳答案

您是否使用mysql_connect连接到数据库?

那会导致您描述的错误。

10-05 20:33