從選擇查詢抓取結果我從MySQL數據庫選擇的PHP函數:在PHP函數
function Master_file($name, $latin){
$HOST_DB ="localhost";
$NAME_DB="nom";
$USER_DB ="utilisaeur";
$PWD_DB="K3Pud1";
$connect = mysql_connect($HOST_DB,$USER_DB,$PWD_DB);
$db=mysql_select_db($NAME_DB);
$qry = "SELECT tax_id FROM master where name =".$name." and latin =".$latin;
echo $qry;
$result = mysql_query($qry);
while ($Res_user = mysql_fetch_assoc($result)) {
return $Res_user['tax_id'];
}
}
顯示Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/admin/public_html/hitlist/include/fg_membersite.php on line 446
一個錯誤,該行是while ($Res_user = mysql_fetch_assoc($result)
那麼是什麼問題?我該如何解決它?
呦你的查詢失敗。我假設你在變量周圍缺少引號。 –
數據庫連接是否有效?它可能是你沒有真正連接到數據庫,因爲你沒有任何錯誤捕捉,你可能不知道它 –
@FabrícioMatté我改變查詢到'$ qry =「選擇'tax_id'從主'where'名稱'=「。$ name。」和'latin' =「。$ latin;'但是結果相同 –