Mysql 如何编写组合3个表数据的sql语句？

这是我从shipping infos和order这两个表中获取数据后编写的代码。在我得到的数据中，有一个是product_id，后面的product_id是第三个表名product的主键。从第三个产品表中，我想检索产品名称和产品价格。我应该如何编写sql？下面显示了从两个表检索数据的sql

 <?php
 include ('classes/functions.php');

 if(isset($_POST['user_id'])){
$user_id = $_POST['user_id'];
$check_receipt = "select  si.shipping_name,
    si.shipping_address,
    si.shipping_contact,
    si.shipping_email,
    o.order_date,
    o.trx_id,
    o.tracking_num,
    o.quantity,
    o.product_id
   from shipping_infos si
   inner join orders o
on si.user_id = o.user_id    
where si.user_id='".$user_id."';";

        $run_receipt_checking = mysqli_query($con, $check_receipt);
        $result = array();
    while($row = mysqli_fetch_array($run_receipt_checking)){
    array_push($result,
    array(
          'shipping_name'=>$row[0],
          'shipping_address'=>$row[1],
          'shipping_contact'=>$row[2],
          'shipping_email'=>$row[3],
          'order_date'=>$row[4],
          'trx_id'=>$row[5],
          'tracking_num'=>$row[6],  
          'quantity'=>$row[7],
          'product_id'=>$row[8]       

));
}
echo json_encode(array("result"=>$result));
 }
?>

---

您可以将三个表与
联接组合起来

试试下面这样

 <?php
 include ('classes/functions.php');

 if(isset($_POST['user_id'])){
$user_id = $_POST['user_id'];
$check_receipt = "select  si.shipping_name,
    si.shipping_address,
    si.shipping_contact,
    si.shipping_email,
    o.order_date,
    o.trx_id,
    o.tracking_num,
    o.quantity,
    o.product_id
   from shipping_infos si
   inner join orders o   
on si.user_id = o.user_id    
inner join products p on p.product_id  = o.product_id 
where si.user_id='".$user_id."';";

        $run_receipt_checking = mysqli_query($con, $check_receipt);
        $result = array();
    while($row = mysqli_fetch_array($run_receipt_checking)){
    array_push($result,
    array(
          'shipping_name'=>$row[0],
          'shipping_address'=>$row[1],
          'shipping_contact'=>$row[2],
          'shipping_email'=>$row[3],
          'order_date'=>$row[4],
          'trx_id'=>$row[5],
          'tracking_num'=>$row[6],  
          'quantity'=>$row[7],
          'product_id'=>$row[8]       

));
}
echo json_encode(array("result"=>$result));
 }

为什么不将第三个表与其他表连接起来呢？这是因为表1和表2共享同一个变量，即用户id，而第三个表没有用户id，而是有表2也有的产品id。我可以把这三张桌子和那个组合起来吗？@CheongCharlene是的，你可以。请参考下面我的建议answer@Cheong，使用产品id加入第三和第二个表