如何从该字符串格式中提取该字段？
如何从该字符串格式中提取该字段？
提问于 2018-08-22 02:04:38
[{"answerInfo":{"extraData":{"am_answer_type":"NN"}},"content":"MP3：not support.","messageId":"c4d6a2f4649d483a811fcce4b26ae9a1"}]如何使用正则表达式或python代码从该字符串中提取"MP3: not support“？
但根据建议生成了一个错误：
Traceback (most recent call last):
File "/Users/congminmin/PycharmProjects/Misc/csv/csvLoader.py", line 16, in <module>
print(question+ " " + json.loads(answer)[0]['content'])
File "/Users/congminmin/anaconda3/lib/python3.6/json/__init__.py", line 354, in loads
return _default_decoder.decode(s)
File "/Users/congminmin/anaconda3/lib/python3.6/json/decoder.py", line 339, in decode
obj, end = self.raw_decode(s, idx=_w(s, 0).end())
File "/Users/congminmin/anaconda3/lib/python3.6/json/decoder.py", line 357, in raw_decode
raise JSONDecodeError("Expecting value", s, err.value) from None
json.decoder.JSONDecodeError: Expecting value: line 1 column 1 (char 0)回答 2
Stack Overflow用户
发布于 2018-08-22 02:09:34
Python 3.6.5 (default, Jun 17 2018, 12:13:06)
>>> text = '[{"answerInfo":{"extraData":{"am_answer_type":"NN"}},"content":"MP3：not support.","messageId":"c4d6a2f4649d483a811fcce4b26ae9a1"}]'
>>> import json
>>> json.loads(text)[0]['content']
'MP3：not support.'Stack Overflow用户
发布于 2018-08-22 02:11:24
根据您的示例，我建议使用json而不是regex。
Check it running here
import json
json_data = '[{"answerInfo":{"extraData":{"am_answer_type":"NN"}},"content":"MP3：not support.","messageId":"c4d6a2f4649d483a811fcce4b26ae9a1"}]'
python_obj = json.loads(json_data)
print(python_obj[0]["content"])页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51954369
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