ElementTree findall（）返回空的lis

Question

我正在尝试编写一个小脚本来与last.fm API交互。

我有一些使用ElementTree的经验，但是我以前使用它的方式似乎不起作用，而是返回一个空列表。
我删除了API密钥，因为我不知道它到底应该有多私有，并给出了一个在它的位置接收XML的示例。

用于与API交互的类：

from xml.etree import ElementTree
import urllib
import urllib2

class Last_fmWrapper(object):
    def __init__(self):
        self.last_fm_api_key = '*****************************'
        self.api_url = 'http://ws.audioscrobbler.com/2.0/'
    def get_now_playing(self, last_fm_user, method):
        self.last_fm_user = last_fm_user
        self.method = method
        parameters = {'user':self.last_fm_user, 'api_key':self.last_fm_api_key, 'method':self.method}
        encoded_parameters = urllib.urlencode(parameters)
        request = urllib2.Request(self.api_url, encoded_parameters)
        response = urllib2.urlopen(request)
        api_results = ElementTree.parse(response).findall('track')
        # why does api_results == []?

    def compare_tasteometer(self):
        pass

    def register_user(self):
        pass

---

调用get_now_playing method of Last_fmWrapper()：

from last_fm_wrapper import Last_fmWrapper
last_fm = Last_fmWrapper()
now_playing = last_fm.get_now_playing('BiriBiriRG', 'user.getRecentTracks')
if now_playing == None:
    print 'You not currently playing anything.'
else:
    print 'You are now playing {}'.format(now_playing)

---

我收到的xml示例：

<?xml version="1.0" encoding="utf-8"?>
<lfm status="ok">
<recenttracks user="BiriBiriRG" page="1" perPage="10" totalPages="18406" total="184058" >
<track> 
                <artist mbid="01809552-4f87-45b0-afff-2c6f0730a3be">Elvis Presley</artist>
<name>Thrill Of Your Love</name>
<streamable>1</streamable>
<mbid></mbid>
        <album mbid="c445fa3a-3b24-41d4-b955-b6ca560c6f7a">Love Songs</album>
    <url>http://www.last.fm/music/Elvis+Presley/_/Thrill+Of+Your+Love</url>
    <image size="small">http://userserve-ak.last.fm/serve/34s/69037914.png</image>
    <image size="medium">http://userserve-ak.last.fm/serve/64s/69037914.png</image>
    <image size="large">http://userserve-ak.last.fm/serve/126/69037914.png</image>
    <image size="extralarge">http://userserve-ak.last.fm/serve/300x300/69037914.png</image>
        <date uts="1328153196">2 Feb 2012, 03:26</date>
</track>
<track> 
                <artist mbid="efc8a006-d0c6-4a9b-8cb1-91ca770fa2b9">Colbie Caillat</artist>
<name>Oxygen</name>
<streamable>1</streamable>
<mbid></mbid>
        <album mbid="2d297b29-a215-42fe-8a8c-dc8b502903b1">Coco</album>
    <url>http://www.last.fm/music/Colbie+Caillat/_/Oxygen</url>
    <image size="small">http://userserve-ak.last.fm/serve/34s/69229764.png</image>
    <image size="medium">http://userserve-ak.last.fm/serve/64s/69229764.png</image>
    <image size="large">http://userserve-ak.last.fm/serve/126/69229764.png</image>
    <image size="extralarge">http://userserve-ak.last.fm/serve/300x300/69229764.png</image>
        <date uts="1328152962">2 Feb 2012, 03:22</date>
</track>
<track> 

Answer 1

问题是findallonly searches the immediate descendants of an element if it is given a tag name。您需要给它一个XPath表达式，它可以在树下的任何位置找到track。因此，以下方法应该有效，例如：

api_results = ElementTree.parse(response).findall('.//track')