json.decoder.JSONDecodeError: 解决方法

通过ast模块处理
Source Code:

Python

import ast str_info = '{"name": "nock", "age": 18}' dict_info = ast.literal_eval(str_info) print("string info type is -->: %s" % (type(str_info))) print("dict info type is -->: %s" % (type(dict_info))) s_info = "{'name': 'nock', 'age': 18}" d_info = ast.literal_eval(s_info) print("s info type is -->: %s" % (type(s_info))) print("d info type is -->: %s" % (type(d_info)))

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< br / > import ast str_info = '{"name": "nock", "age": 18}' dict_info = ast . literal_eval ( str_info )   print ( "string info type is -->: %s" % ( type ( str_info ) ) ) print ( "dict info type is -->: %s" % ( type ( dict_info ) ) )   s_info = "{'name': 'nock', 'age': 18}" d_info = ast . literal_eval ( s_info )   print ( "s info type is -->: %s" % ( type ( s_info ) ) ) print ( "d info type is -->: %s" % ( type ( d_info ) ) )

Result:

Python

string info type is -->: dict info type is -->: s info type is -->: d info type is -->:

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string info type is -- > : < class 'str' > dict info type is -- > : < class 'dict' > s info type is -- > : < class 'str' > d info type is -- > : < class 'dict' >

使用ast.literal_eval进行转换既不存在使用json 模块进行转换的问题，也不存在使用eval模块进行转换的安全性问题，因此推荐大家使用ast.literal_eval的方法。

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