1

我有一个模型Level

class Level:
   level1_id = models.IntegerField()
   level2_id = models.IntegerField()
   level3_id = models.IntegerField()
   level4_id = models.IntegerField()
   level5_id = models.IntegerField()
   level6_id = models.IntegerField()
   level7_id = models.IntegerField()
   level_name = models.CharField()

我正在传递 1-7 范围内的整数 id 和来自 AJAX 的名称。现在我想获得具有相应 id 和 name 的 levelX_id，X 为 id(1-7)。

这就是我正在做的。

id = request.POST['id']
name = request.POST['name']  


if id == 1:
    level_name = Level.objects.all(level_name = name)[0].level1_id


if id == 2:
    level_name = Level.objects.all(level_name = name)[0].level2_id


if id == 3:
    level_name = Level.objects.all(level_name = name)[0].level3_id


if id == 4:
    level_name = Level.objects.all(level_name = name)[0].level4_id


if id == 5:
    level_name = Level.objects.all(level_name = name)[0].level5_id


if id == 6:
    level_name = Level.objects.all(level_name = name)[0].level6_id


if id == 7:
    level_name = Level.objects.all(level_name = name)[0].level7_id

我可以让它更通用。就像是。

level_X_id = "level"+id+"_id"
level_name = Level.objects.all(level_name = name)[0].level_X_id
4

1 回答 1

1

我想getattr这就是你想要的。你可以这样做：

level_X_id = "level"+id+"_id"
level_name = getattr(Level.objects.all(level_name = name)[0], level_X_id)

希望这可以帮助！

于 2013-11-04T17:33:27.343 回答