How to print the output from StreamingHttpResponse to an html template in django?

Question

I have realtime printed the output of my subprocess command onto an an html page using the below code in my views.py . However i want to print this output onto my html template (results.html). how do i do that ?

from django.shortcuts import redirect, render
from django.http import HttpResponse,HttpResponseRedirect,request
import subprocess

# Create your views here.

def home(request):
    return render(request,"home.html")

def about(request):
    return render (request,"About.html")

def contact(request):
    return render (request,"Contact.html")

def process(request):
    ip=request.POST.get('ip')
    with subprocess.Popen(['ping', '-c5',ip], stdout=subprocess.PIPE, bufsize=1, universal_newlines=True) as p:
        for line in p.stdout:
            yield("<html><title>Fetcher Results</title><body><div style='background-color:black;padding:10px'><pre  style='font-size:1.0rem;color:#9bee9b;text-align: center;'><center>"+line+"<center></div></html> ") # process line here
            
    if p.returncode != 0:
        raise subprocess.CalledProcessError(p.returncode, p.args)


def busy(request):
    from django.http import StreamingHttpResponse
    return StreamingHttpResponse(process(request))

Do you want to provide updated lines from the subprocess in a new view or just replace the string in yield with a template? — jan_w
– jan_w, Commented Jul 22, 2022 at 7:40
yes, yield with a template . the output to be printed in a template html — shreesh
– shreesh, Commented Jul 22, 2022 at 8:00

jan_w · Accepted Answer · 2022-07-25 07:38:33Z

You can get a template and render it with get_template [1]. You can then yield your rendered template with:

def process(request):
  ip=request.POST.get('ip')
  template = get_template("result.html")
  with subprocess.Popen(['ping', '-c5',ip], stdout=subprocess.PIPE, bufsize=1, universal_newlines=True) as p:
      for line in p.stdout:
          yield(template.render({"line": line})
  if p.returncode != 0:
      raise subprocess.CalledProcessError(p.returncode, p.args)

For a template result.html:

<html>
    <title>Fetcher Results</title>
    <body>
      <div style='background-color:black;padding:10px'>
        <pre  style='font-size:1.0rem;color:#9bee9b;text-align: center;'>       
          <center>{{ line }}</center>
        </pre>
      </div>
    </body>
 </html>

However, do be aware that the StreamingHttpResponse will concat the template results and you will have a effectively several full <html>...</html> in your result.

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