Question

前段时间，我在json文件中编写了python词典。今天，我在这个文件中添加了一些其他信息：

with open('path.json', 'a') as fp:
    json.dump(DataDict, fp)

尝试使用两种方法打开它：

with open('path.json', 'r') as content_file:
    content = content_file.read()
records = json.loads(content)

和

with open('path.json', 'r') as fp:
     File = json.load(fp)

但两人都以错误结束：

JSONDecodeError                           Traceback (most recent call last)
<ipython-input-149-d027cfbf5e86> in <module>()
      3 with open('/home/vladislav/Документы/Diploma/data/News/ВТБ.json', 'r') as content_file:
      4     content = content_file.read()
----> 5 records = json.loads(content)

/home/vladislav/anaconda3/lib/python3.5/json/__init__.py in loads(s, encoding, cls, object_hook, parse_float, parse_int, parse_constant, object_pairs_hook, **kw)
    317             parse_int is None and parse_float is None and
    318             parse_constant is None and object_pairs_hook is None and not kw):
--> 319         return _default_decoder.decode(s)
    320     if cls is None:
    321         cls = JSONDecoder

/home/vladislav/anaconda3/lib/python3.5/json/decoder.py in decode(self, s, _w)
    340         end = _w(s, end).end()
    341         if end != len(s):
--> 342             raise JSONDecodeError("Extra data", s, end)
    343         return obj
    344 

JSONDecodeError: Extra data: line 1 column 462852 (char 462851)

我该如何解决？互联网似乎没有第一眼的答案。

Answer 1

您通过将新的JSON对象连接到它来破坏JSON。也许更好的想法是加载现有数据，扩展它并重新保存它：

import json

DataDict = ...

with open('path.json', 'rb') as fp:
    new_data = json.load(fp)
    new_data.update(DataDict)

with open('path.json', 'w') as fp:
    json.dump(new_data, fp)

根据评论进行修改：

这只是一个想法。如果你一直在连接JSON对象，并假设对象的顶层确实是一个对象（如在{...}中），那么这个hack可能会起作用：

（注意：请在尝试之前保存并备份此文件！）

import json

with open('path.json', 'rb') as fp:
    corrupted = fp.read()
    fixed_raw = b", ".join(corrupted.split(b"}{"))
    fixed = json.loads(str(fixed_raw, "UTF-8"))

with open('path-fixed.json', 'w') as fp:
    json.dump(fixed, fp)

错误，打开json文件

1 个答案: