Download HTTP Content Disposition python

Question

Im stuck and in need of some help. I cant to seem to find any reference on how to download the file. Ive be using requests to find the file name but I dont know where to go from there.

'Content-disposition': 'attachment;filename=MT0376_DealerPrice.zip'

Any help would be greatly appreciated 

import requests
import csv
import urllib
from urllib.request import urlretrieve
import sys, os, base64, datetime, hashlib, hmac, urllib
from time import gmtime, strftime
import xml.etree.ElementTree as ET
import math
import time
import zipfile


dealerCode = 'Code'
userName ='User'
password = 'Pass'
loginUrl = 'https://www.lemansnet.com/login'

payload = {'rememberMe': 'on', 
           'dealerCode': dealerCode, 
           'dm': 4, 
           'userName': userName, 
           'password': password}


r = requests.post(loginUrl, params=payload)


token = r.headers['loginToken']


with open ("parts.xml", "r") as myfile:
    requestBody = myfile.read().replace('\n', '')

serviceURL = 'https://www.lemansnet.com/pricing/2013/pos'
ContentType = 'Content-Type:text/xml'
ContentLength = len(requestBody)
loginToken = 'loginToken:' + token

httparray = {'content': requestBody}

xml = """<pricing>
<whoForDealer>
<dealerCode>MT0376</dealerCode>
</whoForDealer>
<rememberPreferences>1</rememberPreferences>
</pricing>"""

params = {'http': httparray}

requestHeaders = {'Content-Type': 'text/xml',
                  'Content-Length': ContentLength, 
                  'charset': 'utf-8',
                  'loginToken': token,
                  'Cache-Control': 'no-cache',
                  'Pragma': 'no-cache',
                  'Connection': 'keep-alive'}
files ={'file': ('parts.xml')}


t = requests.post(serviceURL, data=xml, headers=requestHeaders)

#print(r.url)
#print(r.headers)
#print(r.text)
#print(r.status_code)
#print(r.content)
#print(t.url)
#print(t.headers)
#print(t.text)
print(t.status_code)
print(t.content)
with open("MT0376_DealerPrice1.zip", "wb") as code:
    code.write(t.content)
zip = t.headers['Content-disposition']
#print(zip)

#localName = zip.split('filename=')[1]
#print(localName)



print('end')

This is my current response from above code.

200
b'PK\x03\x04\x14\x00\x08\x08\x08\x00H\x8aOF\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x1b\x00\x00\x00PriceFile_system_errors.txt\xe3\xf5M\xac\xc8\xcc-\xcdUHI\xcc\xcc\xa9T\xc8\xc9\xcc\xcd,QH\xadHNMMIM\xd1\xe3\xc5+\x0b\x00PK\x07\x08\xcdO\x92K#\x00\x00\x00<\x00\x00\x00PK\x01\x02\x14\x00\x14\x00\x08\x08\x08\x00H\x8aOF\xcdO\x92K#\x00\x00\x00<\x00\x00\x00\x1b\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00PriceFile_system_errors.txtPK\x05\x06\x00\x00\x00\x00\x01\x00\x01\x00I\x00\x00\x00l\x00\x00\x00\x00\x00'
end

Thanks for the help everyone, I think I managed to figure this one out. I edited the my code above to what I think is working correctly. Unfortunately the website im gathering my info from only allows me to download the file 3 times a day so I wont be able to really test it until tomorrow.

can you share the url and your code? – Padraic Cunningham Feb 15 '15 at 17:57 — Padraic Cunningham, Feb 15 '15 at 17:57
Its a little messy but this is what I got as far. – scate636 Feb 15 '15 at 18:58 — scate636, Feb 15 '15 at 18:58
Ok, what do you want specifically? – felipsmartins Feb 15 '15 at 22:19 — felipsmartins, Feb 15 '15 at 22:19

felipsmartins · Answer 1 · 2015-02-15T18:13:45.497

-1

The python built-in urllib.urlretrieve would be work:

import urllib

output_filename = '/tmp/my_downloaded_file.zip'
url = 'http://www.blog.pythonlibrary.org/wp-content/uploads/2012/06/'
url += 'wxDbViewer.zip'
# download...
urllib.urlretrieve(url, output_filename)

In python 3:

from urllib.request import urlretrieve
...
urlretrieve(url, output_filename)

edited Feb 15 '15 at 18:13

answered Feb 15 '15 at 18:05

felipsmartins

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Download HTTP Content Disposition python

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