将Django StreamingHttpResponse转换成模板 [英] Django StreamingHttpResponse into a Template
问题描述
现在,我阅读了这个讨论和第二个答案实际上打印出一个页面中的流(实际上只是数据)。
我想做的是将流响应的输出打印到模板中,不仅仅打印出讨论中的数据。
我该怎么办?我必须使用javascript并调用实现StreamingHttpResponse的视图,或者有一种方法来告诉django渲染模板,然后将StreamingHttpResponse数据发送到模板(然后我需要知道数据存储的变量是什么) ?
编辑:到目前为止我发现的解决方案是将最后的html页面的片段写入生成器(yield)。这个解决方案的问题是,我不能像例如一个随着数据流传播的一个吧(像一个加载栏)。
ciao
是的,但可能不是你真的想要,因为整个模板将被迭代。然而,对于什么是值得的,您可以为模板传送一个重新渲染的上下文。
from django.http import StreamingHttpResponse
从django.template导入上下文,模板
#模板代码在创建时解析一次,所以
#在模块范围内创建更好的性能
t =模板('{{mydata}}< br /> \\\
')
def gen_rendered():
为范围(1,11)中的x:
c =上下文({'mydata':x})
yield t.render(c)
def stream_view(request):
response = StreamingHttpResponse(gen_rendered())
返回响应
编辑:
您还可以呈现模板,只是追加< p>
或< tr>
标签,但与目的相反的模板。 (即将代码分开)
from django.template import loader,Context
from django.http import StreamingHttpResponse
t = loader.get_template('admin / base_site.html')#或任何
buffer =''* 1024
def gen_rendered():
yield t.render(Context({'varname':'some value','buffer':buffer}))
#^^^^^
#embed {{buffer}}你的模板
#(除非已经足够长了)强制显示
对于范围(1,11)中的x:
yield'< p> x = {}< ; / p> {} \\\
'.format(x,buffer)
Django 1.5 is just came out and it ships the StreamingHttpResponse. Now, I've read this discussion and the second answer actually prints out the stream in a page (actually just data).
What I want to do is to print the output of a stream response into a template, not just print the data as in the discussion.
What should I do? do I've to use javascript and call the view that implements the StreamingHttpResponse, or there's a way to tell django to render the template and later send the StreamingHttpResponse data to the template (then I need to know what's the variables where data are stored)?
Edit: the solution I found so far is to write the pieces of the final html page into the generator (yield). The problem of this solution is that I can't have, for example, a bar that grows with the data streamed (like a loading bar).
ciao
Yes, but it may not be what you really want, as the entire template will be iterated. However, for what it's worth, you can stream a re-rendered context for a template.
from django.http import StreamingHttpResponse
from django.template import Context, Template
# Template code parsed once upon creation, so
# create in module scope for better performance
t = Template('{{ mydata }} <br />\n')
def gen_rendered():
for x in range(1,11):
c = Context({'mydata': x})
yield t.render(c)
def stream_view(request):
response = StreamingHttpResponse(gen_rendered())
return response
EDIT:
You can also render a template and just append <p>
or <tr>
tags to it, but it's quite contrary to the purpose of templates in the first place. (i.e. separating presentation from code)
from django.template import loader, Context
from django.http import StreamingHttpResponse
t = loader.get_template('admin/base_site.html') # or whatever
buffer = ' ' * 1024
def gen_rendered():
yield t.render(Context({'varname': 'some value', 'buffer': buffer}))
# ^^^^^^
# embed that {{ buffer }} somewhere in your template
# (unless it's already long enough) to force display
for x in range(1,11):
yield '<p>x = {}</p>{}\n'.format(x, buffer)
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