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我試圖使用IFNULL()
函數來防止查詢結果爲空,因爲當結果轉換爲JSON數組時,空值會導致錯誤。處理空值和JSON
$sql = mysql_query("select IFNULL(status,'nothing'),
foodname from disease_food,
food where disease_food.Disease_ID=$d1 or disease_food.Disease_ID=$d2 and
Food_ID=$res1 and disease_food.Food_ID=food.ID");
while($row=mysql_fetch_assoc($sql)) {
$output[] = $row;
}
$data = json_encode($output);
print($data);
mysql_close();
的錯誤:(當結果爲空)
Undefined variable: output
您必須使用準備好的語句,這些天,這種查詢是危險的,過時的:http://php.net/manual/en/pdo.prepared-statements.php – freshnode
嘗試'$ SQL =請求mysql_query( 「選擇IFNULL(status,'nothing'),來自disease_food的foodname,食物,其中disease_food.Disease_ID = $ d1或disease_food.Disease_ID = $ d2和 Food_ID = $ res1 and disease_food.Food_ID = food.ID」)或者die(mysql_error ());'並告訴我什麼錯誤(如果有的話) – freshnode
如何定位並修復該錯誤? –