有没有办法在 django 的 if 语句中呈现请求？

Question

我正在尝试为黑白棋游戏编写一个 django web 应用程序。我在向网站呈现新 table 时遇到问题。

views.py

def table(request):
    if request.method == "POST":
        coord = request.POST.keys()
        crd = list(coord)[1]
        x = crd.split("_")
        r = int(x[0]) - 1
        c = int(x[1]) - 1
        reversi = ReversiGame()
        grid = draw_grid(reversi)
        ctxt = {"table": grid}
        return render(request, 'table.html', context=ctxt)

模板

{% extends 'base.html' %}
{% block main_content %}
    <div style="text-align: center; width: auto;
    height:auto; margin-right:auto; margin-left:200px;">
    {% for r in table %}
        <div style="float:left">
        {% for c in r %}
             <form action="" method="post">
            {% csrf_token %}
            {% if c == 2 %}
                <input type="submit" style="background: #000000; width:50px; height:50px;
                            color:white;"
                       name="{{forloop.counter}}_{{forloop.parentloop.counter}}" value="2">
            {% elif c == 1 %}
                <input type="submit" style="background: #ffffff;  width:50px; height:50px;"
                       name="{{forloop.counter}}_{{forloop.parentloop.counter}}" value="1">
            {% else %}
                <input type='submit' style="background: #c1c1c1;  width:50px; height:50px;"
                       name="{{forloop.counter}}_{{forloop.parentloop.counter}}" value="o">
            {% endif %}
             </form>
        {% endfor %}
        </div>
    {% endfor %}
    </div>
{% endblock %}

urls.py

urlpatterns = [
    path('', views.HomeView.as_view(), name="home"),
    path('signup/', views.SignUpView.as_view(), name='signup'),
    path('profile/<int:pk>/', views.ProfileView.as_view(), name='profile'),
    path('table/', views.table, name='table')
]

当我尝试 return request.method 中的 HttpResponse 时，出现以下错误：The view GameConfiguration.views.table didn't return an HttpResponse object. It returned None instead.

如果我将一个tab移到return render(request, 'table.html', context=ctxt)的左边，那么ctxt变量，也就是新的板子，就不会被识别（它说它在赋值之前被使用），这意味着我没有可以访问新绘制的 table.

我需要 POST 方法中的行和列来翻转棋盘和切换播放器。

非常感谢您的宝贵时间！谢谢！

Answer 1

您的视图函数仅在 return 时响应 request.method == "POST"。当您在浏览器中访问页面并收到错误 The view ... didn't return an HttpResponse object. 时，那是因为通过浏览器发出的请求有 request.method == "GET".

您可以通过在 if 语句之外添加 return 方法来修复视图方法：

def table(request):
    if request.method == "POST":
        # Here is where you capture the POST parameters submitted by the
        # user as part of the request.POST[...] dictionary.
        ...
        return render(request, 'table.html', context=ctxt)

    # When the request.method != "POST", a response still must be returned.
    # I'm guessing that this should return an empty board.
    reversi = ReversiGame()
    grid = draw_grid(reversi)
    return render(request, 'table.html', context={"table": grid})

有没有办法在 django 的 if 语句中呈现请求？

Is there a way to render a request within an if Statement in django?

python

django

django-views

httpresponse

reversi