使用字段参数列表过滤 Django 模型
Filter Django Models using lists of field parameters
我希望能够将任意数量的字段和值传递到函数中以识别相关行:
models.py
class Player(models.Model):
name = CharField(max_length = 50, default = 'Ronaldo')
defender = BooleanField(default = False)
midfielder = BooleanField(default = False)
attacker = BooleanField(default = False)
goalkeeper = BooleanField(default = False)
views.py
def find_player(**kwargs):#to be used in some view
players = Player.objects.filters(kwargs).all()
for player in players:
#do something with player...
find_player({defender:True, goalkeeper:True})#selects rows with players who defend and play in goal
find_player({defender:True, attacker:False})#...
find_player({defender:False})
我在上面尝试做的显然是行不通的!我知道我也可以使用 exec() 来获得我想要的东西:
def find_player(string_of_params):
players = exec(Player.objects.filters(string_of_params).all())
for player in players:
print(player)
find_player('defender=True, goalkeeper=True')#prints rows with players who can defend and go in goal
find_player('defender=True, attacker=False')#...
find_player('defender=False'})
但我认为必须有一种更自然的方式将字典的内容直接解包到 filter()
。
欢迎任何指导。
你应该解压函数中的字典,这样**kwargs
而不是kwargs
:
# ↓ no double asterisk
def find_player(kwargs):
# ↓↓ double asterisk
players = Player.objects.filters(<strong>**kwargs</strong>).all()
# …
然后你可以通过传递字典来使用它:
find_player({defender:True, goalkeeper:True})
find_player({defender:True, attacker:False})
find_player({defender:False})
我希望能够将任意数量的字段和值传递到函数中以识别相关行:
models.py
class Player(models.Model):
name = CharField(max_length = 50, default = 'Ronaldo')
defender = BooleanField(default = False)
midfielder = BooleanField(default = False)
attacker = BooleanField(default = False)
goalkeeper = BooleanField(default = False)
views.py
def find_player(**kwargs):#to be used in some view
players = Player.objects.filters(kwargs).all()
for player in players:
#do something with player...
find_player({defender:True, goalkeeper:True})#selects rows with players who defend and play in goal
find_player({defender:True, attacker:False})#...
find_player({defender:False})
我在上面尝试做的显然是行不通的!我知道我也可以使用 exec() 来获得我想要的东西:
def find_player(string_of_params):
players = exec(Player.objects.filters(string_of_params).all())
for player in players:
print(player)
find_player('defender=True, goalkeeper=True')#prints rows with players who can defend and go in goal
find_player('defender=True, attacker=False')#...
find_player('defender=False'})
但我认为必须有一种更自然的方式将字典的内容直接解包到 filter()
。
欢迎任何指导。
你应该解压函数中的字典,这样**kwargs
而不是kwargs
:
# ↓ no double asterisk
def find_player(kwargs):
# ↓↓ double asterisk
players = Player.objects.filters(<strong>**kwargs</strong>).all()
# …
然后你可以通过传递字典来使用它:
find_player({defender:True, goalkeeper:True})
find_player({defender:True, attacker:False})
find_player({defender:False})