我正在嘗試從數據庫中選擇多個產品并將 id 插入到另一個表中。我嘗試如下執行，如果我只是 echo $name，$_POST["txtMaterial"][$key]并且$_POST["txtSize"][$key]在查詢之前，我會獲取所有選定的名稱、材料和尺寸，但是當我綁定查詢中的值時，我只得到第一個。 foreach($_POST["txtName"] as $key => $name){ $sQuery = $db->prepare('SELECT id FROM products WHERE name = :sName AND material = :sMaterial AND size = :sSize'); $sQuery->bindValue(':sName', $name); $sQuery->bindValue(':sMaterial', $_POST["txtMaterial"][$key]); $sQuery->bindValue(':sSize', $_POST["txtSize"][$key]); $sQuery->execute(); $aOrders = $sQuery->fetchAll(); foreach($aOrders as $aOrder){ echo print_r($aOrders); }}的 print_r$aOrders是這樣的："Array( [id] => 174)1"這是我需要的第一個 id。誰能幫我看看如何獲取所有的 id？$_POST["txtName"]foreach之前的var_dump ："array(50) { [0]=> string(0) "" [1]=> string(6) "BACURI" [2]=> string(0) "" [3]=> string(0) "" [4]=> string(0) "" [5]=> string(0) "" [6]=> string(0) "" [7]=> string(0) "" [8]=> string(0) "" [9]=> string(0) "" [10]=> string(0) "" [11]=> string(0) "" [12]=> string(0) "" [13]=> string(0) "" [14]=> string(0) "" [15]=> string(0) "" [16]=> string(0) "" [17]=> string(0) "" [18]=> string(0) "" [19]=> string(0) "" [20]=> string(0) "" [21]=> string(0) "" [22]=> string(0) "" [23]=> string(0) "" [24]=> string(0) "" [25]=> string(0) "" [26]=> string(0) "" [27]=> string(0) "" [28]=> string(0) "" [29]=> string(6) "CAJARI" [30]=> string(0) "" [31]=> string(0) "" [32]=> string(0) "" [33]=> string(0) "" [34]=> string(0) "" [35]=> string(0) "" [36]=> string(0) "" [37]=> string(0) "" [38]=> string(0) "" [39]=> string(0) "" [40]=> string(0) "" [41]=> string(0) "" [42]=> string(0) "" [43]=> string(0) "" [44]=> string(0) "" [45]=> string(0) "" [46]=> string(0) "" [47]=> string(0) "" [48]=> string(0) "" [49]=> string(0) ""}"foreach之前的var_dump $_POST["txtMaterial"]："array(2) { [0]=> string(1) "3" [1]=> string(1) "2"}
PDO SELECT 僅在 foreach 循環中選擇第一個
慕婉清6462132
2023-06-24 15:39:26
