Python multi-line JSON and variables

Question

I'm trying to encode a somewhat large JSON in Python (v2.7) and I'm having trouble putting in my variables!

As the JSON is multi-line and to keep my code neat I've decided to use the triple double quotation mark to make it look as follows:

my_json = """{
        "settings": {
                "serial": "1",
                "status": "2",
                "ersion": "3"
        },
        "config": {
                "active": "4",
                "version": "5"
        }
}"""

To encode this, and output it works well for me, but I'm not sure how I can change the numbers I have there and replace them by variable strings. I've tried:

    "settings": {
            "serial": 'json_serial',

but to no avail. Any help would be appreciated!

Bluntly -- why would you hardcode a JSON string, rather than hardcoding a Python data structure and converting it to JSON when it's time for output? — Charles Duffy
– Charles Duffy, Commented Feb 24, 2017 at 16:56

davidejones · Accepted Answer · 2017-02-24 16:57:41Z

37

Why don't you make it a dictionary and set variables then use the json library to make it into json

import json
json_serial = "123"
my_json = {
    'settings': {
        "serial": json_serial,
        "status": '2',
        "ersion": '3',
    },
    'config': {
        'active': '4',
        'version': '5'
    }
}
print(json.dumps(my_json))

answered Feb 24, 2017 at 16:57

davidejones

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3 Comments

user5740843 Over a year ago

This works great! But, I'm trying to have it show up all tiny and organized, so I have an extra: "json.dumps(json.loads(my_json), indent=4, sort_keys=True)" which is giving me the error: TypeError: expected string or buffer. Any idea's?

Charles Duffy Over a year ago

@user5740843, get rid of the json.loads call -- the input object is just a native Python data type, not JSON at all, so it's already ready to be passed as the first argument to json.dumps() exactly as-is. So: json.dumps(my_json, indent=4, sort_keys=True)

user5740843 Over a year ago

Briliant. Thank you!

Community · Accepted Answer · 2017-05-23 11:53:52Z

3

If you absolutely insist on generating JSON with string concatenation -- and, to be clear, you absolutely shouldn't -- the only way to be entirely certain that your output is valid JSON is to generate the substrings being substituted with a JSON generator. That is:

'''"settings" : {
  "serial"   : {serial},
  "version"  : {version}
}'''.format(serial=json.dumps("5"), version=json.dumps(1))

But don't. Really, really don't. The answer by @davidejones is the Right Thing for this scenario.

edited May 23, 2017 at 11:53

CommunityBot

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answered Feb 24, 2017 at 17:04

Charles Duffy

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