get image from website
質問
I try to improve a recipe for calibre and replace the default cover image with the cover image of the current newspaper issue.
The way to go has something to do with get_cover_url
(link).
There are two problems:
- The URL of the cover image changes every day.
- I know virtually nothing about python.
I hope for a solution like this (in pseudo-code):
OPEN URL "http://epaper.derstandarddigital.at/";
coverElement = (SEARCH HTML-ELEMENT "<img>" WITH ID "imgPage2" AND CLASS "page");
coverUrl = (GET HTML-ATTRIBUTE "src" FROM coverElement);
RETURN coverUrl;
Would there be a way to achieve this in python*) (using only python standard libraries)?
*) Calibre-Recipes seem to be python code
[edit] here's the solution a friend of mine offered:
#!/usr/bin/env python
import urllib
from time import strftime
def get_cover_url(self):
highResolution = True
date = strftime("%Y/%Y%m%d")
# it is also possible for the past
#date = '2012/20120503'
urlP1 = 'http://epaper.derstandarddigital.at/'
urlP2 = 'data_ep/STAN/' + date
urlP3 = '/V.B1/'
urlP4 = 'paper.htm'
urlHTML = urlP1 + urlP2 + urlP3 + urlP4
htmlF = urllib.urlopen(urlHTML)
htmlC = htmlF.read()
# URL EXAMPLE: data_ep/STAN/2012/20120504/V.B1/pages/A3B6798F-2751-4D8D-A103-C5EF22F7ACBE.htm
# consists of part2 + part3 + 'pages/' + code
# 'pages/' has length 6, code has lenght 36
index = htmlC.find(urlP2) + len(urlP2 + urlP3) + 6
code = htmlC[index:index + 36]
# URL EXAMPLE HIGH RESOLUTION: http://epaper.derstandarddigital.at/data_ep/STAN/2012/20120504/pagejpg/A3B6798F-2751-4D8D-A103-C5EF22F7ACBE_b.png
# URL EXAMPLE LOW RESOLUTION: http://epaper.derstandarddigital.at/data_ep/STAN/2012/20120504/pagejpg/2AB52F71-11C1-4859-9114-CDCD79BEFDCB.png
urlPic = urlP1 + urlP2 + '/pagejpg/' + code
if highResolution:
urlPic = urlPic + '_b'
urlPic = urlPic + '.png'
return urlPic
if __name__ == '__main__':
print get_cover_url(None)
解決
You can use the lxml
library to extract elements from an HTML document. The basic framework will look something like this:
import urllib
import lxml.html
fd = urllib.urlopen('http://epaper.derstandarddigital.at/')
doc = lxml.html.parse(fd)
matches = doc.xpath('//img[@id="imgPage2]')
if matches:
print matches[0].get('src')
...however, the URL you've provided does not actually result in a document with any <img>
tags. It is almost entirely a bunch of Javascript.
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