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()
我传递的是1-7范围内的整数id和来自AJAX的名称。现在我想得到levelXïid,它有各自的id和名称,X是id(1-7)。你知道吗
我就是这么做的。你知道吗
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
我能把它做得更普通些吗。差不多吧。你知道吗
level_X_id = "level"+id+"_id"
level_name = Level.objects.all(level_name = name)[0].level_X_id
我认为^{} 是你想要的。你可以这样做:
希望这有帮助!你知道吗
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