2013-01-25 90 views
-1

我被困在一個非常基本的sql查詢腳本中。有人會注意到我無法看到的東西。我已經檢查了SQL查詢工作正常時,它是從中mysqladmin執行:PHP錯誤mysql_fetch_array():提供的參數不是有效的MySQL結果資源

<? 
include ('gps_db_connect.php'); 
$query = "SELECT * from gps WHERE server_time > '20130124'"; 
echo $query; 
$result = mysqli_query($connection, $query) or die(' Error getting data'); 
echo ' After query'; 
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) 
    { 
    echo $row['server_time']; 
    } 

?> 

這裏是屏幕顯示/輸出:

SELECT * from gps WHERE server_time > '20130124' After      
query 
PHP Error Message 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9440109/public_html/test.php on line 7 
+0

看看你的right - > –

回答

0

試試這個

$result = mysql_query($query, $connection) or die('Error'); 
+0

非常感謝!不知何故,我沒有看到。 – user1834682

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