php - 将数据添加到两个以id连接的表中

Question

嗨，我有这段代码将数据插入 mysql 数据库到表 radnici 中:

 try {        
                $STH = $db->prepare("INSERT INTO radnici (ime_prezime, jmbg, grad, ulica, posta, telefon, uloga, email, user_id) VALUES (:1,:2,:3,:4,:5,:6,:7,:8,:9)");

                $STH->bindParam(':1', $_POST['ime']);
                $STH->bindParam(':2', $_POST['jmbg']);
                $STH->bindParam(':3', $_POST['grad']);
                $STH->bindParam(':4', $_POST['ulica']);
                $STH->bindParam(':5', $_POST['posta']);
                $STH->bindParam(':6', $_POST['telefon']);
                $STH->bindParam(':7', $_POST['pozicija']);
                $STH->bindParam(':8', $_POST['email']);
                $STH->bindParam(':9', $user_id);

//HERE I NEED TO INSERT DATA TO TABLE WORKHOURS BUT CONNECTED WITH JUST ADDEDED ID
//SO here I NEED query something like this // INSERT INTO radnici (ID, ID_radnici, ime_prezime, jmbg,... here ID_radnici is =with ID from table radnici of added data


                $STH->execute();

            } catch (PDOException $e) {
                echo $e->getMessage();
            }
            echo "<p>Data submitted successfully</p>";

所以现在当我添加这个时，我需要添加到这个表workhours数据中，但与刚刚添加的数据的ID连接......

所以在第二个表中我有:ID，ID_radnici(这里我必须添加ID_radnici与表radnici上的ID相同的值)，value，user_id

我写了这个但不起作用:更新:

 try {        
                $STH = $db->prepare("INSERT INTO radnici (ime_prezime, jmbg, grad, ulica, posta, telefon, uloga, email, user_id) VALUES (:1,:2,:3,:4,:5,:6,:7,:8,:9)");

                $STH->bindParam(':1', $_POST['ime']);
                $STH->bindParam(':2', $_POST['jmbg']);
                $STH->bindParam(':3', $_POST['grad']);
                $STH->bindParam(':4', $_POST['ulica']);
                $STH->bindParam(':5', $_POST['posta']);
                $STH->bindParam(':6', $_POST['telefon']);
                $STH->bindParam(':7', $_POST['pozicija']);
                $STH->bindParam(':8', $_POST['email']);
                $STH->bindParam(':9', $user_id);

                $orderID = $db -> lastInsertId();

   $STH1 = $db->prepare("INSERT INTO workhours (ID_radnika) values($orderID)");  // where $value is the value you want to insert...
   $STH->execute();
      $STH1->execute();






            } catch (PDOException $e) {
                echo $e->getMessage();
        }
        echo "<p>Data submitted successfully</p>";

Answer 1

使用lastInsertId将执行您想要的操作:

$orderID = $db -> lastInsertId();