php - 尝试从一个页面获取变量以将其重新填充到另一页面上

Question

我在第一页上使用下拉框，该下拉框应该将变量传递到链接页面，并在第二页上使用mysql查询中第二个的变量“$id”。

选择要查找的人的首页代码

客户查询

                        require ('dbconnect.php');
                        $result = $con->query("select id, lastname, firstname from customer");

                        while ($row = $result->fetch_assoc()) {

                        unset($id, $name);
                        $id = $row['id'];
                        $name = $row['lastname'];
                        $firstname = $row['firstname']; 

                        echo '<option value="/customerpage.php?='.$id.'">'.$name.','.$firstname.'</option>';

                        }

                        echo "</select>";
                        mysqli_close($con);
                        ?> 

第二页，即接收页面

$id = $_GET['id'];
echo $id;


require ('dbconnect.php');

                            if (mysqli_connect_errno()) {
                              echo "Failed to connect to MySQL: " . mysqli_connect_error();
                            }
                                 $result = mysqli_query($con,"SELECT * FROM customer WHERE id='$id'");
                            while($row = mysqli_fetch_array($result)) {
                              echo $row['firstname'];

                              echo "<br>";
                                    }

?>

第二页网址与/customerpage.php?=1 一起出现，效果很好

Answer 1

试试这个:

 $result = mysqli_query($con,"SELECT * FROM customer WHERE id=' . $id . '");