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