success_url在UpdateView中，基于传递的值 [英] success_url in UpdateView, based on passed value

问题描述

如何根据参数设置 success_url ？

我真的想回到我来的地方，而不是一些静态的地方。在伪代码中：

How can I set success_url based on a parameter?
I really want to go back to where I came from, not some static place. In pseudo code:

url(r'^entry/(?P<pk>\d+)/edit/(?P<category>\d+)',
    UpdateView.as_view(model=Entry, 
                       template_name='generic_form_popup.html',
                       success_url='/category/%(category)')),

这将意味着：编辑条目 pk 然后返回'category'。

Which would mean: edit entry pk and then return to 'category'. Here an entry can be part of multiple categories.

推荐答案

创建一个MyUpdateView类，从UpdateView继承并覆盖get_success_url方法：

Create a class MyUpdateView inheritted from UpdateView and override get_success_url method:

class MyUpdateView(UpdateView):
    def get_success_url(self):
        pass #return the appropriate success url

此外，我喜欢传递类似于继承类视图中的template_name和model的参数，但不是在.as_view ）in urls.py

Also i like to pass such parameters like template_name and model inside of inheritted class view, but not in .as_view() in urls.py

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