Question

我现在是php和sql的新手，所以我很困惑我的代码中的错误。（需要做的）cart_tbl（order_id）food_name，special_request，quantity，amount等于我的order_tbl的order_id。当我的order_tbl和cart_tbl的order_id相同时。我的输出将是该2表的值。这是我现在的代码。

     <?php
$connect = mysqli_connect ("localhost", "root", "" , "db");
if(isset($_POST['order_id'])){ 
$asd = ($_POST['order_id']);
$sql = "SELECT food_name, special_request, quantity, amount 
FROM cart_tbl
WHERE order_id='$asd'";
$result = mysqli_query($connect, $sql);
}

?>

 <table class="table table-hover table-bordered">
 <thead>
 <tr> 
 <th>Food</th>
 <th>Special Request</th>
 <th>Quantity</th>
 <th>Amount</th> 
 </tr>
 </thead> 
 <?php
if(mysqli_num_rows($result)>0)
{
    while($row = mysqli_fetch_array($result))
    {
?>
<tr>
<td><?php echo $row["food_name"];?></td>
<td><?php echo $row["special_request"];?></td>
<td><?php echo $row["quantity"];?></td>
<td><?php echo $row["amount"];?></td>
</tr>
<?php
    }
}
?>

</table>

Answer 1

使用INNER JOIN。

SELECT *
FROM cart_tbl
INNER JOIN order_tbl
ON cart_tbl.order_id = order_tbl.order_id
WHERE order_id='$asd'

或者，如果关联的表具有相同的列名NATURAL JOIN并且列具有相同的数据类型，请使用order_id。

SELECT *
FROM cart_tbl
NATURAL JOIN order_tbl
WHERE order_id='$asd'

如何从db中选择一个值

1 个答案: