Reading JSON from a file?

I am getting a bit of headache just because a simple looking, easy statement is throwing some errors in my face.

I have a json file called strings.json like this:

"strings": [{"-name": "city", "#text": "City"}, {"-name": "phone", "#text": "Phone"}, ...,
            {"-name": "address", "#text": "Address"}]

I want to read the json file, just that for now. I have these statements which I found out, but it’s not working:

import json
from pprint import pprint

with open('strings.json') as json_data:
    d = json.loads(json_data)
    json_data.close()
    pprint(d)

The error displayed on the console was this:

Traceback (most recent call last):
  File "/home/.../android/values/manipulate_json.py", line 5, in <module>
    d = json.loads(json_data)
  File "/usr/lib/python2.7/json/__init__.py", line 326, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python2.7/json/decoder.py", line 366, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
TypeError: expected string or buffer
[Finished in 0.1s with exit code 1]

If I use json.load instead of json.loads, I get this error:

Traceback (most recent call last):
  File "/home/.../android/values/manipulate_json.py", line 5, in <module>
    d = json.load(json_data)
  File "/usr/lib/python2.7/json/__init__.py", line 278, in load
    **kw)
  File "/usr/lib/python2.7/json/__init__.py", line 326, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python2.7/json/decoder.py", line 369, in decode
    raise ValueError(errmsg("Extra data", s, end, len(s)))
ValueError: Extra data: line 829 column 1 - line 829 column 2 (char 18476 - 18477)
[Finished in 0.1s with exit code 1]

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

The json.load() method (without “s” in “load”) can read a file directly:

import json

with open('strings.json') as f:
    d = json.load(f)
    print(d)

You were using the json.loads() method, which is used for string arguments only.


The error you get with json.loads is a totally different problem. In that case, there is some invalid json in that file. For that, I would recommend running the file through a json validator.

There are also solutions for fixing json like for example How do I automatically fix an invalid JSON string?

Method 2

Here is a copy of code which works fine for me

import json

with open("test.json") as json_file:
    json_data = json.load(json_file)
    print(json_data)

with the data

{
    "a": [1,3,"asdf",true],
    "b": {
        "Hello": "world"
    }
}

you may want to wrap your json.load line with a try catch because invalid JSON will cause a stacktrace error message.

Method 3

The problem is using with statement:

with open('strings.json') as json_data:
    d = json.load(json_data)
    pprint(d)

The file is going to be implicitly closed already. There is no need to call json_data.close() again.

Method 4

In python 3, we can use below method.

Read from file and convert to JSON

import json
from pprint import pprint

# Considering "json_list.json" is a json file

with open('json_list.json') as fd:
     json_data = json.load(fd)
     pprint(json_data)

with statement automatically close the opened file descriptor.


String to JSON

import json
from pprint import pprint

json_data = json.loads('{"name" : "myName", "age":24}')
pprint(json_data)

Method 5

To add on this, today you are able to use pandas to import json:
https://pandas.pydata.org/pandas-docs/stable/generated/pandas.read_json.html
You may want to do a careful use of the orient parameter.

Method 6

You can use pandas library to read the JSON file.

import pandas as pd
df = pd.read_json('strings.json',lines=True)
print(df)

Method 7

def read_JSON():
    with open("FILE PATH", "r") as i:
        JSON_data = i.read()
    print(JSON_data)


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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