Question

我正在尝试将数据从我的django模型表单发布到我的数据库，但我没有运气。我可以从管理员创建并在页面上显示，但似乎无法正确传递我的信息。该评论与学校的外键相关联，我知道这有效。这就是我的模型，视图和HTML的内容。

models.py

from django.forms import ModelForm

class Comment(models.Model):
    school = models.ForeignKey(Schools)
    created = models.DateTimeField(auto_now_add=True)
    author = models.CharField(max_length=60)
    body = models.TextField()

class CommentForm(ModelForm):
    class Meta:
        model = Comment
        fields = ['author', 'body', 'school']

views.py

EDITED

我还添加了评论对象create，就像尝试另一件事一样。

from my_app.models import Comment, CommentForm

if request.method == 'POST':
    form = CommentForm(request.POST) # A form bound to the POST data
    if form.is_valid(): # All validation rules pass
        author = form.cleaned_data['author']
        body = form.cleaned_data['body']
        school = form.cleaned_data['school']
        form.save()
        content = Comment.objects.create(school = school, author = author, body = body)

我在这个视图上尝试了很多变化，但还没有运气。

HTML

<form action="/" method="POST">{% csrf_token %}
     <p>{{ form.body }}</p>
     <div id="submit"><input type="submit" value="Submit"></div>
</form>

Answer 1

一个。您必须指定{{form.as_p}}，{{form.as_table}}或仅{{form}}而不是{{form.body}}。查看文档以获取更多详细信息：https://docs.djangoproject.com/en/dev/topics/forms/

B中。 cleaned_data需要Comment类的属性。因此，使用body = form.cleaned_data['comment']

更改：body = form.cleaned_data['body']

℃。和Django一起玩。

修改

将if request.method == 'POST':移到新方法中，即register_comment：

def register_comment: if request.method == 'POST': form = CommentForm(request.POST) # A form bound to the POST data if form.is_valid(): # All validation rules pass author = form.cleaned_data['author'] body = form.cleaned_data['body'] school = form.cleaned_data['school'] form.save()

修改urls.py：

urlpatters = patterns('', # Some others views here url(r'^new-comment/$', 'app_name.views.register_comment', name="new_comment"), # Maybe Some others views here )

打开网络浏览器，然后转到：http://127.0.0.1:8000/new-comment/

无法获取django模型表单将信息传递给数据库

1 个答案: