将 URL 中的 XML 解析为 python 对象
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Parse XML from URL into python object
提问by smilebomb
The goodreads website has this API for accessing a user's 'shelves:' https://www.goodreads.com/review/list/20990068.xml?key=nGvCqaQ6tn9w4HNpW8kquw&v=2&shelf=toread
goodreads 网站具有用于访问用户“货架”的 API：https: //www.goodreads.com/review/list/20990068.xml?key=nGvCqaQ6tn9w4HNpW8kquw &v=2&shelf=toread
It returns XML. I'm trying to create a django project that shows books on a shelf from this API. I'm looking to find out how (or if there is a better way than) to write my view so I can pass an object to my template. Currently, this is what I'm doing:
它返回 XML。我正在尝试创建一个 django 项目，该项目通过此 API 在书架上显示书籍。我正在寻找如何（或者是否有更好的方法）来编写我的视图，以便我可以将对象传递给我的模板。目前，这就是我正在做的事情：
import urllib2
def homepage(request):
file = urllib2.urlopen('https://www.goodreads.com/review/list/20990068.xml?key=nGvCqaQ6tn9w4HNpW8kquw&v=2&shelf=toread')
data = file.read()
file.close()
dom = parseString(data)
I'm not entirely sure how to manipulate this object if I'm doing this correctly. I'm following this tutorial.
如果我正确地执行此操作，我不完全确定如何操作此对象。我正在关注本教程。
采纳答案by alecxe
I'd use xmltodictto make a python dictionary out of the XMLdata structure and pass this dictionary to the template inside the context:
我会使用数据结构xmltodict制作一个 python 字典XML，并将这个字典传递给上下文中的模板：
import urllib2
import xmltodict
def homepage(request):
file = urllib2.urlopen('https://www.goodreads.com/review/list/20990068.xml?key=nGvCqaQ6tn9w4HNpW8kquw&v=2&shelf=toread')
data = file.read()
file.close()
data = xmltodict.parse(data)
return render_to_response('my_template.html', {'data': data})
回答by gies0r
xmltodictusing urllib3
xmltodict使用 urllib3
import traceback
import urllib3
import xmltodict
def getxml()
url = "https://yoursite/your.xml"
import urllib3
import xmltodict
http = urllib3.PoolManager()
response = http.request('GET', url)
try:
data = xmltodict.parse(response.data)
except:
print("Failed to parse xml from response (%s)" % traceback.format_exc())
return data
回答by Vincent
xmltodictusing requests
xmltodict使用 requests
import requests
import xmltodict
url = "https://yoursite/your.xml"
response = requests.get(url)
data = xmltodict.parse(response.content)
