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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