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我正在尝试加载我的 JSON 文件，该文件是通过从另一个 JSON 文件复制内容而创建的。ValueError: Expecting property name: line 1 column 1 (char 1)但是当我尝试从复制所有数据的文件中读取 JSON 数据时，我不断收到错误消息，我的 JSON 数据的格式为

{
    "server": {
        "ipaddress": "IP_Sample",
        "name": "name_Sample",
        "type": "type_Sample",
        "label": "label_Sample",
        "keyword": "kwd_Sample",
        "uid": "uid_Sample",
        "start_time": "start_Sample",
        "stop_time": "stop_Sample"
    }
}

我的加载和写入方法是

def load(self, filename):
    inputfile = open(filename,'r')
    self.data = json.loads(inputfile.read())
    print (self.data)
    inputfile.close()
    return

def write(self, filename):
    file = open(filename, "w")
    tempObject = self.data
    print type(tempObject)
    #json.dump(filename, self.data)
    print self.data["server"]
    print >> file, self.data
    file.close()
    return

我无法弄清楚我要去哪里错了，任何人都可以帮我解决这个问题..

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1 回答 1

3

要在文件中保存和加载 JSON，请使用打开的文件对象。您的代码表明您尝试将文件名保存到self.data，这不是文件对象...

以下代码有效：

def write(self, filename):
    with open(filename, 'w') as output:
        json.dump(self.data, output)

def load(self, filename):
    with open(filename, 'r') as input:
        self.data = json.load(input)

我将打开的文件用作上下文管理器，以确保它们在完成读取或写入时关闭。

您的其他尝试只是将python 表示形式print >> file, self.data打印到文件中，而不是 JSON：

>>> print example
{u'server': {u'uid': u'uid_Sample', u'keyword': u'kwd_Sample', u'ipaddress': u'IP_Sample', u'start_time': u'start_Sample', u'label': u'label_Sample', u'stop_time': u'stop_Sample', u'type': u'type_Sample', u'name': u'name_Sample'}}

其中，当从文件中读回时，会给出您指示的错误消息：

>>> json.loads("{u'server': {u'uid': u'uid_Sample', u'keyword': u'kwd_Sample', u'ipaddress': u'IP_Sample', u'start_time': u'start_Sample', u'label': u'label_Sample', u'stop_time': u'stop_Sample', u'type': u'type_Sample', u'name': u'name_Sample'}}")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/json/__init__.py", line 307, in loads
    return _default_decoder.decode(s)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/json/decoder.py", line 319, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/json/decoder.py", line 336, in raw_decode
    obj, end = self._scanner.iterscan(s, **kw).next()
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/json/scanner.py", line 55, in iterscan
    rval, next_pos = action(m, context)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/json/decoder.py", line 171, in JSONObject
    raise ValueError(errmsg("Expecting property name", s, end))
ValueError: Expecting property name: line 1 column 1 (char 1)

您必须改为打印json.dumps()输出：

>>> print json.dumps(example)
'{"server": {"uid": "uid_Sample", "keyword": "kwd_Sample", "ipaddress": "IP_Sample", "start_time": "start_Sample", "label": "label_Sample", "stop_time": "stop_Sample", "type": "type_Sample", "name": "name_Sample"}}'
于 2012-08-25T20:00:04.130 回答