Django：附加列表，然后分页

Question

在我的网站上，我有一个供用户玩的 public 游戏列表。我想对该列表进行分页，使其每页仅显示 20 个游戏。同时，我想保留游戏统计数据，以便用户可以知道他们是否玩过该游戏，以及他们是否喜欢该游戏（这样 URL 就可以更改）。目前它给我的网站带来了巨大的痛苦。我正在使用内置的分页器。

我的代码：

def games(request):
    thisuser = request.user
    games_list = Game.objects.filter(private_game=False).order_by('-pubdate')
    paginator = Paginator(games_list, 20)
    next = request.path
    page = request.GET.get('page')
    try:
        games = paginator.page(page)
    except PageNotAnInteger:
        games = paginator.page(1)
    except EmptyPage:
        games = paginator.page(paginator.num_pages)

    args = {'games': games}
    args.update(csrf(request))

    for game in games:
        played = (game, Result.objects.filter(player=thisuser, game=game).exists())
        liked = (game, Game.objects.filter(id=game.id, gamelikes=thisuser).exists())
        args['games'].append(played, liked)

return render(request, 'games.html', args)

错误是“'Page' 对象没有属性 'append'”

游戏模型：

class Game(models.Model):
    name            = models.CharField(max_length=200, null=True, blank=True)
    description     = models.TextField(null=True, blank=True)
    pubdate         = models.DateTimeField(default=timezone.now)
    createdby        = models.ForeignKey(User, related_name='game_creator')
    lastedit         = models.DateTimeField(default=timezone.now)
    private_game     = models.BooleanField(default=False)
    private          = models.ManyToManyField(User, related_name='private_game_players')
    item01           = models.CharField(max_length=200, null=True, blank=True)
    item01photocount = models.IntegerField(default=0)
    item02           = models.CharField(max_length=200, null=True, blank=True)
    item02photocount = models.IntegerField(default=0)
    item03           = models.CharField(max_length=200, null=True, blank=True)
    item03photocount = models.IntegerField(default=0)
    item04           = models.CharField(max_length=200, null=True, blank=True)
    item04photocount = models.IntegerField(default=0)
    value            = models.CharField(max_length=200, null=True, blank=True)
    endby_date       = models.DateField(null=True, blank=True)
    game_closed      = models.BooleanField(default=False)
    location         = models.CharField(max_length=200, null=True, blank=True)
    gamelikes        = models.ManyToManyField(User, related_name='game_likes')
    photocount       = models.IntegerField(default=0)

Answer 1

您可以通过将属性直接粘贴到对象上来简化代码。

# ...
for game in games:
    game.played = Result.objects.filter(player=thisuser, game=game).exists()
    game.liked = Game.objects.filter(id=game.id, gamelikes=thisuser).exists()

ctx = {'games': games}
ctx.update(csrf(request))

return render(request, 'games.html', ctx)

当您遍历模板中的 games 对象时，您可以访问属性：

{% for game in games %}
{{ game.played }}, {{ game.liked }}
{% endfor %}

唯一要注意的是确保您没有覆盖 game 对象上可能存在的属性，例如通过在属性前加上 extra_has_played 或 extra_has_liked.

之类的前缀

Django：附加列表，然后分页

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