获取xml响应的“tags”元素的每个标记名不起作用

Question

我从一个特定的国家获得顶级艺术家，然后我想储存每个艺术家的名字和标签。名字很好用，但标签不起作用。标签是摇滚、爵士乐等流派

我得到标签的部分：

for child in tree:
    for artist in child:
        print(artist)
        for tag in artist.findall('tags'):
            print(tag)
            bands[i]['Tags'] = tag.text

但打印（标记）返回时不起作用：

<Element 'name' at 0x00000211BBEB0F98>
<Element 'tags' at 0x00000211BBEBD638>

你知道如何在乐队{}中得到和撕碎标签吗？例如，在上面的例子中，标签是摇滚、经典摇滚、爱尔兰、流行和另类

响应的格式如下：

<lfm status="ok">
<artist>
<name>U2</name>
<tags>
<tag>
<name>rock</name>
<url>https://www.last.fm/tag/rock</url>
</tag>
<tag>
<name>classic rock</name>
<url>https://www.last.fm/tag/classic+rock</url>
</tag>
<tag>
<name>irish</name>
<url>https://www.last.fm/tag/irish</url>
</tag>
<tag>
<name>pop</name>
<url>https://www.last.fm/tag/pop</url>
</tag>
<tag>
<name>alternative</name>
<url>https://www.last.fm/tag/alternative</url>
</tag>
</tags>

</artist>
</lfm>

最小可验证示例：

import xml.etree.ElementTree as ET
import requests

ID = 1

api_key = "b088cbedecd40b35dd89e90f55227ac2"  # generated for the example

bands = {}

# GET TOP ARTISTS

artistslist = requests.get(
    'http://ws.audioscrobbler.com/2.0/?method=geo.gettopartists&country=spain&page=1&limit=5&api_key=' + api_key)
tree = ET.fromstring(artistslist.content)
for child in tree:
    for artist in child.findall('artist'):
        name = artist.find('name').text
        bands[ID] = {}
        bands[ID]['ID'] = ID
        bands[ID]['Name'] = name

        ID += 1

# GET ARTIST INFO
for i, v in bands.items():

    chosen = bands[i]['Name'].replace(" ", "+")
    artist = requests.get(
        'http://ws.audioscrobbler.com/2.0/?method=artist.getinfo&artist=' + chosen + '&api_key=' + api_key)
    tree = ET.fromstring(artist.content)
    for child in tree:
        for artist in child:
            print(artist)
            for tag in artist.findall('tags'):
                print(tag['name'])
                bands[i]['Tags'] = tag.text
            if (artist.get('size') == "large"):
                if (artist.text is not None):
                    bands[i]['Image'] = artist.text

    print(bands[i]['Name'] + " RETRIEVED")

Answer 1

在循环中，artist.findall('tags')返回一个包含单个元素的列表<tags>元素。您正试图迭代<tags>元素中的每个<tag>。请改用以下方法：

for tag in artist.find('tags').findall('tag')

另外，请注意tag.text将是None。相反，您可能需要tag.find('name').text和tag.find('url').text