Question

• 支付费用 GET将数字从正确来源得出。

My url 看像:指数.php?page=items&catID=5

当用户提出像3类的东西时,在散射电离层上就不存在。我希望它能够显示错误信息。

$catID = intval($_GET["catID"]);

if($catID) {
    $checkSQL = mysql_query("SELECT * FROM category WHERE category_type= 2 ");
    while($checkROW = mysql_fetch_array($checkSQL)) {
    $checkCAT != $checkROW["categoryID"];
    echo "err msg";
    }

这一点迄今为止可以出现,但是,即便在正确的网页上发生火灾错误,也无所作为。

谢谢。

Answer 1

隔离墙是好的,如果是的话,就存在失踪人员。但是,如果正确读到这一点,那么你会做些什么呢? 现在是:

if($catID) {

实际上,只有卡迪卡(或,从_GET美元中支离破碎)是非零的(不是假的)。我的猜测是,如果要检查一下信息数据库是从KQ返回的类别?

$catID = intval($_GET["catID"]);

checkcat($catID);

function checkcat($check_category) {
    $checkSQL = mysql_query("SELECT * FROM category WHERE category_type= 2 ");
    while($checkROW = mysql_fetch_array($checkSQL)) {
        if ( $check_category != $checkROW["categoryID"] ) {
            echo "err msg";
        } else { 
            echo "not an error message";
        }
    }
}

了解你所期待的是什么,为什么像现在这样?

$catID = ($_GET["catID");

if ( !is_numeric($catID) ) {
    echo "Not a numeric category!"
} else {
    $checkSQLQuery = "SELECT * FROM category WHERE categoryID =  {$catID}  AND category_type= 2 "
    $resultSQL = mysql_query($checkSQLQuery, $db);
    /* NOTE!:  Guessing on what your database resouce 
    pointer is - it isn t included in the origin snippet.
    Although, the last opened should be used by default if 
    this is left out. */
    if ( mysql_num_rows($resultSQL) < 1 ) {
        echo "Error message, category ID not found" 
    } else {
        echo "Found it!"
    }
}

Answer 2

见Oh先生。第1行“如果”

    while ($checkROW = mysql_fetch_array($checkSQL)) {
        if ($checkCAT != $checkROW["categoryID"])
            echo "err msg";

Answer 3

它想像你想要使用我的sql_fetch_assoc,而不是我的sql_fetch_array()。

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