对于未来的项目，我计划我有三张桌子。
第1个跟踪配方信息，第2个跟踪从配方表中引用recipe_id的跟踪标签，第3个表包含所有成分的列表。
recipe:
+---------+-----------------+-------------+
| id [AI] | title | category |
+---------+-----------------+-------------+
| 20 | Italian Pizza | dinner |
+---------+-----------------+-------------+
| 21 | Greek Soup | lunch |
+---------+-----------------+-------------+
recipe_ingredients_label:
+---------------+------------+-------------+
| label_id [AI] | recipe_id | label |
+---------------+------------+-------------+
| 1 | 20 | Pizza Base |
+---------------+------------+-------------+
| 2 | 20 | Topping |
+---------------+------------+-------------+
recipe_ingredients:
+----------+------------+----------------------+
| label_id | amount | ingredients |
+----------+------------+----------------------+
| 1 | 4 cups | Cheese |
+----------+------------+----------------------+
| 1 | 3/4 cup | White Flour |
+----------+------------+----------------------+
| 2 | 1 cup | Mozzarella Cheese |
+----------+------------+----------------------+
| 2 | 1/2 cup | Tomato Sauce |
+----------+------------+----------------------+
recipe_ingredients引用recipe_ingredients_label表，每个成分都在它自己的标签下面，如下所示：
披萨基地包含 4杯奶酪，3/4杯白面粉。
Topping 包含 1杯莫扎里拉奶酪，1/2杯番茄酱。
问题是并非所有食谱都包含标签，因此我想根据食谱使标签可选。
我想到了这个：
在recipe_ingredients表中创建一个额外的recipe_id列，如果没有找到标签，在插入数据库期间，只需引用recipe_id，如下所示：
recipe_ingredients:
+----------+------------+-----------------------------------+
| label_id | recipe_id | amount | ingredients |
+----------+------------+-----------------------------------+
| 1 | NULL | 4 cups | Cheese |
+----------+------------+-----------------------------------+
| 1 | NULL | 3/4 cup | White Flour |
+----------+------------+-----------------------------------+
| 2 | NULL | 1 cup | Mozzarella Cheese |
+----------+------------+-----------------------------------+
| 2 | NULL | 1/2 cup | Tomato Sauce |
+----------+------------+-----------------------------------+
| NULL | 21 | 2 cups | Chicken |
+----------+------------+-----------------------------------+
| NULL | 21 | 1/2 cup | Half carrots |
+----------+------------+-----------------------------------+
| NULL | 21 | 2 | Celery stalks |
+----------+------------+-----------------------------------+
这是组织/优化此表的最佳方式，以提高性能，减少整个数据库中丢失的数据吗？你会推荐什么？