2011-01-28 81 views
0

我似乎無法超越這個錯誤。PHP mysql_fetch_array問題

警告:mysql_fetch_array():提供的參數不是在/windsor.php一個有效的MySQL結果資源上線

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 

回答

1

你連接到使用mysql_connect數據庫?

這會導致您描述的錯誤。

+0

是的,我有指定的頂部,只是沒有發佈。其他腳本的工作,只是這一個讓我悲傷... – mrlayance 2011-01-28 01:05:45