python json dumps与loads有可能犯的错误

Posted 2020-09-16

json.dumps() 是将一个Python数据结构转换为一个JSON编码的字符串

json.loads() 是将一个JSON编码的字符串转换为一个Python数据结构

 

 

如下：

>>> import json

>>> a={‘name‘:‘xiaoming‘}

>>> json.dumps(a)

‘{"name": "xiaoming"}‘

>>> b=‘{"name": "xiaoming"}‘

>>> json.loads(b)

{u‘name‘: u‘xiaoming‘}

 

 

注意：我们原先定义a的元素都是用单引号，而dumps转化后都变成了双引号，而在loads后变成python变量时，元素都变成了单引号，并且字符串前加多了个u。

 

那么当我定义如下的字符串类型时，元素内容用单引号，而整个字符串用双引号括起来：

>>> c="{‘name‘:‘xiaoming‘}"

>>> print json.loads(c)

Traceback (most recent call last):

File "<stdin>", line 1, in <module>

File "/usr/local/python27/lib/python2.7/json/__init__.py", line 339, in loads

return _default_decoder.decode(s)

File "/usr/local/python27/lib/python2.7/json/decoder.py", line 364, in decode

obj, end = self.raw_decode(s, idx=_w(s, 0).end())

File "/usr/local/python27/lib/python2.7/json/decoder.py", line 380, in raw_decode

obj, end = self.scan_once(s, idx)

ValueError: Expecting property name: line 1 column 2 (char 1)

 

程序报错了，所以一般要求当要字符串通过loads转为python数据类型时，得外层用单引号，里面元素key和value用双引号

 

还有，即使满足上述条件，当有u前缀的unicode出现修饰时，也会报错：

>>> c=‘{u"name": u"xiaoming"}‘

>>> json.loads(c)

Traceback (most recent call last):

File "<stdin>", line 1, in <module>

File "/usr/local/python27/lib/python2.7/json/__init__.py", line 339, in loads

return _default_decoder.decode(s)

File "/usr/local/python27/lib/python2.7/json/decoder.py", line 364, in decode

obj, end = self.raw_decode(s, idx=_w(s, 0).end())

File "/usr/local/python27/lib/python2.7/json/decoder.py", line 380, in raw_decode

obj, end = self.scan_once(s, idx)

ValueError: Expecting property name: line 1 column 2 (char 1)

 

所以loads在处理中，还需要把里面的u前缀修饰符去掉

 

 

 

 

 

 

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