如何从django视图的url中删除get参数

I have the following view:

我有以下观点:

def copy_group(request):
    copy = request.GET.get('copy','')

    if copy:
        #do my copy process

    context = {'view':'copy-view'}
    return render(request, 'groups/copy-view.html', context)

This invoked with the following url mysite.com/groups/?copy=1

使用以下url mysite.com/groups/?copy=1调用

The problem is that if I refresh the page, my process keeps copying over and over.

问题是，如果我刷新页面，我的进程就会不断地复制。

How can I remove the ge parameter so the url returns after it copies: mysite.com/groups/

如何删除ge参数，以便url在它复制之后返回:mysite.com/groups/

I tried inserting this in my view code:

我尝试把这个插入到我的视图代码中:

request.GET.pop('copy')

But I get the error: This QueryDict instance is immutable

但是我得到了一个错误:这个QueryDict实例是不可变的

2 个解决方案

#1

If I understand it well, once you've done the copy you want the page to be redirected so no more copy will be made. Do like this then:

如果我理解的很好，一旦你完成了复制，你希望页面被重定向，这样就不会再复制了。这样做:

...
if copy:
    the actual copy and then...
    return HttpResponseRedirect(redirect_to='/the-path-without-copy-parameter/')
...

Then you can improve your code:

然后你可以改进你的代码:

add copy as an actual route parameter, def copy_group(request, copy=None)
添加copy作为实际的路由参数def copy_group(request, copy=None)
generate the path in redirect_to instead of hardcoding it
在redirect_to中生成路径，而不是硬编码它
Out of topic: add a functional test for it if there is none :)
脱离主题:如果没有的话，为它添加一个功能测试:)

BTW, I'm not sure why you use a Get parameter and not just a different URL for it??

顺便说一句，我不知道为什么要使用Get参数而不是不同的URL ?

#2

I ended up doing a redirect instead. That removes the get parameter.

结果我重定向了。这将删除get参数。

#1