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Home/ Questions/Q 8816115
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T04:32:32+00:00 2026-06-14T04:32:32+00:00

I am trying to automate download of historic stock data using python. The URL

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I am trying to automate download of historic stock data using python. The URL I am trying to open responds with a CSV file, but I am unable to open using urllib2. I have tried changing user agent as specified in few questions earlier, I even tried to accept response cookies, with no luck. Can you please help.

Note: The same method works for yahoo Finance.

Code:

import urllib2,cookielib

site= "http://www.nseindia.com/live_market/dynaContent/live_watch/get_quote/getHistoricalData.jsp?symbol=JPASSOCIAT&fromDate=1-JAN-2012&toDate=1-AUG-2012&datePeriod=unselected&hiddDwnld=true"

hdr = {'User-Agent':'Mozilla/5.0'}

req = urllib2.Request(site,headers=hdr)

page = urllib2.urlopen(req)

Error

File “C:\Python27\lib\urllib2.py”, line 527, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) urllib2.HTTPError: HTTP Error 403: Forbidden

Thanks for your assistance

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T04:32:33+00:00Added an answer on June 14, 2026 at 4:32 am

    By adding a few more headers I was able to get the data:

    import urllib2,cookielib
    
    site= "http://www.nseindia.com/live_market/dynaContent/live_watch/get_quote/getHistoricalData.jsp?symbol=JPASSOCIAT&fromDate=1-JAN-2012&toDate=1-AUG-2012&datePeriod=unselected&hiddDwnld=true"
    hdr = {'User-Agent': 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.64 Safari/537.11',
           'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
           'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
           'Accept-Encoding': 'none',
           'Accept-Language': 'en-US,en;q=0.8',
           'Connection': 'keep-alive'}
    
    req = urllib2.Request(site, headers=hdr)
    
    try:
        page = urllib2.urlopen(req)
    except urllib2.HTTPError, e:
        print e.fp.read()
    
    content = page.read()
    print content
    

    Actually, it works with just this one additional header:

    'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
    
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