使用 Python 进行基本的 xml 解析

Question

我正在通过他们的 API 从 Flickr 提取图像数据，我打印的是几千个看起来像这样的 xml 对象：

<photo accuracy="15" context="0" dateupload="1398279194" farm="8" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="320" id="13986079375" isfamily="0" isfriend="0" ispublic="1" latitude="41.828482" license="0" longitude="-87.624506" owner="100231432@N02" pathalias="perspectivesschools" place_id="cF8n.mJTWrhYf0uBEw" secret="f46eef0b1d" server="7308" title="Sean Gallagher, Pulitzer Photojournalist visits MSA" url_n="https://farm8.staticflickr.com/7308/13986079375_f46eef0b1d_n.jpg" width_n="213" woeid="28297331" />
<photo accuracy="12" context="0" dateupload="1394558054" farm="4" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="213" id="13086071753" isfamily="0" isfriend="0" ispublic="1" latitude="51.451914" license="2" longitude="-0.122882" owner="96189004@N04" pathalias="" place_id="JYdWRftQUbMvFA" secret="265103ac38" server="3040" title="" url_n="https://farm4.staticflickr.com/3040/13086071753_265103ac38_n.jpg" width_n="320" woeid="13978" />
<photo accuracy="12" context="0" dateupload="1394558019" farm="8" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="213" id="13086343854" isfamily="0" isfriend="0" ispublic="1" latitude="51.451914" license="2" longitude="-0.122882" owner="96189004@N04" pathalias="" place_id="JYdWRftQUbMvFA" secret="a6858f84d2" server="7451" title="" url_n="https://farm8.staticflickr.com/7451/13086343854_a6858f84d2_n.jpg" width_n="320" woeid="13978" />

现在我想一次性提取属性“lat”和“long”的数据。以及另一个中属性“url_n”的数据。我怎样才能在 Python 中做到这一点？我没有解析 xml 数据的经验，也不知道从哪里开始。

非常感谢！

Answer 1

使用 lxml

虽然 Python 中有多个与 XML 相关的包，包括。 stdlib 一，我更喜欢使用lxml，作为 它提供了我需要的所有东西（良好的 XPath 支持、模式验证等），我更喜欢保留数字 我使用的包很小。

对于来自 Flickr 的 xml 文档，解决方案可能如下所示

脚本flickr.py

from lxml import etree
xmllines = """
<photo accuracy="15" context="0" dateupload="1398279194" farm="8" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="320" id="13986079375" isfamily="0" isfriend="0" ispublic="1" latitude="41.828482" license="0" longitude="-87.624506" owner="100231432@N02" pathalias="perspectivesschools" place_id="cF8n.mJTWrhYf0uBEw" secret="f46eef0b1d" server="7308" title="Sean Gallagher, Pulitzer Photojournalist visits MSA" url_n="https://farm8.staticflickr.com/7308/13986079375_f46eef0b1d_n.jpg" width_n="213" woeid="28297331" />
<photo accuracy="12" context="0" dateupload="1394558054" farm="4" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="213" id="13086071753" isfamily="0" isfriend="0" ispublic="1" latitude="51.451914" license="2" longitude="-0.122882" owner="96189004@N04" pathalias="" place_id="JYdWRftQUbMvFA" secret="265103ac38" server="3040" title="" url_n="https://farm4.staticflickr.com/3040/13086071753_265103ac38_n.jpg" width_n="320" woeid="13978" />
<photo accuracy="12" context="0" dateupload="1394558019" farm="8" geo_is_contact="0" geo_is_family="0" geo_is_friend="0" geo_is_public="1" height_n="213" id="13086343854" isfamily="0" isfriend="0" ispublic="1" latitude="51.451914" license="2" longitude="-0.122882" owner="96189004@N04" pathalias="" place_id="JYdWRftQUbMvFA" secret="a6858f84d2" server="7451" title="" url_n="https://farm8.staticflickr.com/7451/13086343854_a6858f84d2_n.jpg" width_n="320" woeid="13978" />
"""

for line in xmllines.strip().splitlines():
    doc = etree.fromstring(line)
    urls = doc.xpath("/photo/@url_n")
    if urls:
        url = urls[0]
        print url
    else:
        print "---no attribute url_n was found---"

会输出：

$ python flickr.py
https://farm8.staticflickr.com/7308/13986079375_f46eef0b1d_n.jpg
https://farm4.staticflickr.com/3040/13086071753_265103ac38_n.jpg
https://farm8.staticflickr.com/7451/13086343854_a6858f84d2_n.jpg

Answer 2

使用正则表达式解析 XML 是 not a good idea。试试BeautifulSoup - 它不仅解析 XML，而且还具有轻松获取与所选对象及其属性相关的下一个/父级/等元素的功能。

使用示例：

from bs4 import BeautifulSoup
(...)
soup = BeautifulSoup(flickr_xml)
for photo in soup.find_all('photo'):
    print(photo.get('url_n'))