mysql - 简化我的 Django 查询逻辑

Question

我有一个看起来像这样的表格。

+----+--------+---------+-----------+------------------------+
| id | parent | type    | libTypeId | name                   |
+----+--------+---------+-----------+------------------------+
|  2 |      1 | project |         2 | 1p6m4x0y1z_1.2-1.8_hvt |
|  5 |      1 | project |         6 | 1p6m4x0y1z_1.2-1.8_hvt |
|  8 |      1 | project |        16 | 1p6m4x0y1z_1.2-1.8_hvt |
| 11 |      1 | project |        21 | 1p6m4x0y1z_1.2-1.8_hvt |
|  3 |      1 | project |         2 | 1p6m4x0y1z_1.2-1.8_lvt |
|  6 |      1 | project |         6 | 1p6m4x0y1z_1.2-1.8_lvt |
|  9 |      1 | project |        16 | 1p6m4x0y1z_1.2-1.8_lvt |
| 12 |      1 | project |        21 | 1p6m4x0y1z_1.2-1.8_lvt |
|  1 |      1 | project |         2 | 1p6m4x0y1z_1.2-1.8_svt |
|  4 |      1 | project |         6 | 1p6m4x0y1z_1.2-1.8_svt |
|  7 |      1 | project |        16 | 1p6m4x0y1z_1.2-1.8_svt |
| 10 |      1 | project |        21 | 1p6m4x0y1z_1.2-1.8_svt |
| 13 |      2 | project |         2 | 065nm_GPIO             |
| 17 |      2 | project |         4 | 065nm_GPIO             |
| 14 |      2 | project |         6 | 065nm_GPIO             |
| 18 |      2 | project |        12 | 065nm_GPIO             |
| 15 |      2 | project |        16 | 065nm_GPIO             |
| 16 |      2 | project |        21 | 065nm_GPIO             |
| 19 |      2 | project |         2 | 065nm_Specialized      |
+----+--------+---------+-----------+------------------------+

我正在寻找一个查询，该查询会产生一个列表，其中我们获取 id = 1 的所有项目，该项目仅按名称、libtypeid 和第一个 libtypeid 排序。

换句话说，我应该得到这样的结果:

+----+--------+---------+-----------+------------------------+
| id | parent | type    | libTypeId | name                   |
+----+--------+---------+-----------+------------------------+
|  2 |      1 | project |         2 | 1p6m4x0y1z_1.2-1.8_hvt |
|  3 |      1 | project |         2 | 1p6m4x0y1z_1.2-1.8_lvt |
|  1 |      1 | project |         2 | 1p6m4x0y1z_1.2-1.8_svt |
+----+--------+---------+-----------+------------------------+

现在我可以走到这一步了，但是我怎样才能只得到第一个？

Variant.objects.filter(parent=self.id).order_by('name', 'libtype_id')

然后我进一步这样做..

full = Variant.objects.filter(parent=self.id).order_by('name', 'libtype_id')
names, out = [], []
for v in full:
    if v.name not in names:
       out.append(v)
       names.append(v.name)
return out

如果有人能稍微清理一下，我将不胜感激..

Answer 1

如果我正确理解了这个问题，您需要对您的集合进行“分组”并返回每组的第一行。您需要查看聚合:https://docs.djangoproject.com/en/dev/topics/db/aggregation/

类似的东西

Variant.objects
    .values('name', 'libtype_id')
    .annotate(min_libtype_id=Min('libtype_id'))
    .filter(parent=self.id)

未经测试。