获取“first_name”的Django是此函数的无效关键字参数“创建模型类实例时的TypeError”

Posted 2021-05-03

如果相关，请使用Django 1.11和django-pyodbc-azure最新版本。

我是Django的新手，并一直关注1.11教程，没有任何问题，直到这一点，我非常困惑。

这是我的models.py：

from django.db import models

# Create your models here.


class Player(models.Model):
    first_name = models.CharField(max_length=20, name='First Name')
    last_name = models.CharField(max_length=20, name='Last Name')

    def __str__(self):
        return '{}, {} ({})'.format(self.last_name, self.first_name, self.id)


class Game(models.Model):
    players = models.ManyToManyField(Player, name='Players')

    def __str__(self):
        return ' vs. '.join(self.players)


class Round(models.Model):
    GIN = 'GI'
    UNDERCUT = 'UN'
    KNOCK = 'KN'
    ENDING_ACITONS = (
        (GIN, 'Gin'),
        (UNDERCUT, 'Undercut'),
        (KNOCK, 'Knock'),
    )

    game = models.ForeignKey(Game, on_delete=models.CASCADE, name='Parent Game')
    winner = models.ForeignKey(Player, on_delete=models.CASCADE, name='Winning Player')
    points = models.IntegerField(name='Points Awarded')
    end = models.CharField(max_length=2, choices=ENDING_ACITONS)

    def __str__(self):
        return '{} awarded {} points via {}'.format(self.winner, self.points, self.end.name)

现在当我运行manage.py shell并输入：

from game.models import *
bobby = Player(first_name='Bobby', last_name='Fisher')

我遇到了这个错误：

Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "C:UsersmteperPycharmProjectsGinRummyvenvlibsite-packagesdjangodbmodelsase.py", line 571, in __init__
    raise TypeError("'%s' is an invalid keyword argument for this function" % list(kwargs)[0])
TypeError: 'first_name' is an invalid keyword argument for this function

任何见解都会非常感激，因为，正如我之前所说，我是Django和Python的新手。

答案

这是导致问题的name论点。你应该设置verbose_name。

class Player(models.Model):
    first_name = models.CharField(max_length=20, verbose_name='First Name')
    last_name = models.CharField(max_length=20, verbose_name='Last Name')