回答此问题可获得 20 贡献值,回答如果被采纳可获得 50 分。
<p>嘿,我正试着从印度nse网站下载股票数据</p>
<p>所以我用python来做这个</p>
<p>链接是</p>
<pre><code> import urllib
urllib.urlretrieve("https://www.nseindia.com/content/historical/DERIVATIVES/2016/JAN/fo01JAN2016bhav.csv.zip","fo01JAN2016bhav.csv.zip")
</code></pre>
<p>但是当我试图打开下载的文件时,它显示</p>
^{pr2}$
<p>当我尝试从网站正常下载时,只要粘贴链接,下载的文件就会打开</p>
<p>链接</p>
<p><a href="https://www.nseindia.com/content/historical/DERIVATIVES/2016/JAN/fo01JAN2016bhav.csv.zip" rel="nofollow noreferrer">https://www.nseindia.com/content/historical/DERIVATIVES/2016/JAN/fo01JAN2016bhav.csv.zip</a></p>
<p>所以如果我尝试使用urllib 2
我明白了</p>
<pre><code>f=urllib2.urlopen('https://www.nseindia.com/content/historical/DERIVATIVES/2016/JAN/fo01JAN2016bhav.csv.zip')
Traceback (most recent call last):
File "<pyshell#6>", line 1, in <module>
f=urllib2.urlopen('https://www.nseindia.com/content/historical/DERIVATIVES/2016/JAN/fo01JAN2016bhav.csv.zip')
File "C:\Python27\lib\urllib2.py", line 127, in urlopen
return _opener.open(url, data, timeout)
File "C:\Python27\lib\urllib2.py", line 410, in open
response = meth(req, response)
File "C:\Python27\lib\urllib2.py", line 523, in http_response
'http', request, response, code, msg, hdrs)
File "C:\Python27\lib\urllib2.py", line 448, in error
return self._call_chain(*args)
File "C:\Python27\lib\urllib2.py", line 382, in _call_chain
result = func(*args)
File "C:\Python27\lib\urllib2.py", line 531, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
HTTPError: HTTP Error 403: Forbidden
</code></pre>
<p>我怎么解决这个问题?在</p>
<p>只有我试过从imgur下载图片,代码运行良好</p>
<p>当我可以通过浏览器正常访问时,为什么会出现http403错误?在</p>