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Home/ Questions/Q 8176313
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T23:07:26+00:00 2026-06-06T23:07:26+00:00

My views.py file code: #!/usr/bin/python from django.template import loader, RequestContext from django.http import HttpResponse

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My views.py file code:

#!/usr/bin/python 

from django.template import loader, RequestContext
from django.http import HttpResponse
#from skey import find_root_tags, count, sorting_list
from search.models import Keywords
from django.shortcuts import render_to_response as rr

def front_page(request):

    if request.method == 'POST' :
        from skey import find_root_tags, count, sorting_list
        str1 = request.POST['word'] 
        fo = open("/home/pooja/Desktop/xml.txt","r")

        for i in range(count.__len__()):

            file = fo.readline()
            file = file.rstrip('\n')
            find_root_tags(file,str1,i) 

            list.append((file,count[i]))

        sorting_list(list)

        for name, count1 in list:
            s = Keywords(file_name=name,frequency_count=count1)
            s.save()

        fo.close()

        list1 = Keywords.objects.all()
        t = loader.get_template('search/results.html')
        c = RequestContext({'list1':list1,
        })

        return HttpResponse(t.render(c))

    else :  
        str1 = ''
        list = []
        template = loader.get_template('search/front_page.html')
        c = RequestContext(request)
        response = template.render(c)
        return HttpResponse(response)

the variable “file” that I’m sending in function find_root_tags(file,str1,i) has a xml file’s name. This file is present on my desktop, and this code is written in views.py file of my django app, so it is unable to open that file. How can I open that file because xml.txt contains similar filenames that are to be read and then opened. In short, how can I send the file argument as:

file1 = '/home/pooja/Desktop/<filename>'

here <filename> equals the value stored in variable file and finall able to call it as:

find_root_tags(file1, str1, i)

////////////////////////////////////////////////////////////////////////////////////////////

clarification :

1) please refer to the variable “file” in which you can see that I’m storing the read content of xml.txt.

2)xml.txt contains xml filenames. The views.py file is the part of django app’s which is unable to open these files since they are present on desktop.

3) my question is how to modify and send the file variable containing filename appended with it’s absolute path which is:

'/home/pooja/Desktop/filename'

by doing this it will open the files present on the Desktop.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-06T23:07:28+00:00Added an answer on June 6, 2026 at 11:07 pm

    try this:

    pathh='/home/pooja/Desktop/'      #set the base path       
    fo = open("/home/pooja/Desktop/xml.txt")
    for i in range(len(count)):     #use len(count)
        file = fo.readline()
        file = file.strip()          #use strip()
        find_root_tags(pathh+file,str1,i) #base path+file
        mylist.append((file,count[i]))   #using 'list' as a variable name is not good
    
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