Question

大家好，我想显示一个表，以便该表具有2个外键。我的 food 和 restaurant 表具有 name 列。尝试遵循以下代码，但我不知道如何显示不同的餐厅名称和食物名称。

$sql_select_food = "SELECT food.id,restaurant.name,food.name,food.restaurant_id,foodcategory.title from food INNER join foodcategory ON food.foodcategory_id=foodcategory.id INNER join restaurant ON food.restaurant_id=restaurant.id WHERE food.isActive = 1 ORDER BY food.updateDate DESC";
$result_select_food = mysqli_query($connection, $sql_select_food);
if (mysqli_num_rows($result_select_food) > 0) {
      while ($row_select_food = mysqli_fetch_assoc($result_select_food)) {
                        echo "
                            <tr>
                                <td>$row_select_food[name]</td>
                                <td>$row_select_food[name]</td>
                                <td>$row_select_food[title]</td>
                            </tr>
                ";
                    }
                }

Answer 1

使用列别名：

SELECT food.id AS food_id, restaurant.name AS restaurant_name, food.name AS food_name ...

然后

$row_select_food['restaurant_name']是餐厅的名称，
$row_select_food['food_name']是食物的名称，
等

注意：AS关键字是可选的。

Answer 2

为列名添加别名，然后在代码中稍后将其作为列名调用：

$sql_select_food = "SELECT food.id,restaurant.name as rname, food.name as fname, food.restaurant_id,foodcategory.title from food INNER join foodcategory ON food.foodcategory_id=foodcategory.id INNER join restaurant ON food.restaurant_id=restaurant.id WHERE food.isActive = 1 ORDER BY food.updateDate DESC";
$result_select_food = mysqli_query($connection, $sql_select_food);
if (mysqli_num_rows($result_select_food) > 0) {
      while ($row_select_food = mysqli_fetch_assoc($result_select_food)) {
                        echo "
                            <tr>
                                <td>$row_select_food[rname]</td>
                                <td>$row_select_food[fname]</td>
                                <td>$row_select_food[title]</td>
                            </tr>
                ";
                    }
                }

使用内部联接查询显示2合1表

2 个答案: