2 回答
TA贡献1876条经验 获得超7个赞
您没有正确解析数据。
该行将for image_sub_dict in sub_dict['type']:返回字典的键'type',而不是值。相反,您可能只想查询'type'dict 的值"emote"。以下是获取所有表情的图标 URL 的示例:
import requests
response = requests.get(
'https://fortnite-api.com/v2/cosmetics/br',
headers={'Accept': 'application/json'},
).json()
for entry in response['data']:
kind = entry['type']['value']
if kind == 'emote':
name = entry['name']
icon = entry['images']['icon']
print(f"{name}: {icon}")
输出:
...
Scorecard: https://fortnite-api.com/images/cosmetics/br/eid_scorecard/icon.png
Team Burger: https://fortnite-api.com/images/cosmetics/br/eid_scorecardburger/icon.png
Team Tomato: https://fortnite-api.com/images/cosmetics/br/eid_scorecardtomato/icon.png
Members Only: https://fortnite-api.com/images/cosmetics/br/eid_secrethandshake/icon.png
...
TA贡献1827条经验 获得超8个赞
您需要添加一些日志记录来查看哪个键是错误的。尝试:
rr = r.json()
print(rr, type(rr))
for sub_dict in rr['data']:
print(sub_dict, type(sub_dict))
for image_sub_dict in sub_dict['type']:
print(image_sub_dict, type(image_sub_dict))
for j in image_sub_dict['value']:
print(j)
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