PHP-无法跳转到MySQL结果索引13的第0行

标题错误的原因是什么？我在MysqL Front测试了mysql查询,没有问题.我的脚本在下面.谢谢.

$fbFoodPerma = $fbSiteID."-".$fbFoodPerma;  
    $sql = MysqL_query("  
        SELECT fbFoodPerma   
        FROM fbFoods   
        WHERE fbFoodPerma   
        LIKE '$fbFoodPerma'") or die(MysqL_error());  
    $isstored = MysqL_result($sql,0,"fbFoodPerma");  
    while ($isstored == "$fbFoodPerma") {  
    $fbFoodPerma = $fbFoodPerma."-";  
    $sql = MysqL_query("  
        SELECT fbFoodPerma   
        FROM fbFoods   
        WHERE fbFoodPerma   
        LIKE '$fbFoodPerma'") or die(MysqL_error());  
        $isstored = MysqL_result($sql,0,"fbFoodPerma");  
        }  

解决方法:

嗯,试试看看它是否正在收集任何数据：

$sql="SELECT * FROM fbFoods";
$query = MysqL_query($sql) or die ("Could not match data because ".MysqL_error());
$num_rows = MysqL_num_rows($query);

if ($num_rows > 0) {
    echo "rows found";
    } else {
    echo "none found";
}

还要确保$fbFoodPerma实际上包含某些内容-尝试在查询末尾回显它.

echo $fbFoodPerma;