Creating a json parsing function with Python

Question

I have been trying to create a function to use as an importable tool for parsing json. here is my code I have so far

#json parser
def jsonimporter(file_path , env_to_return):
    import json
    a = open(file_path,"r") #opens and sets the file to read

    data = a.read()         #sets variable to read function
    print(data)

    json_data = json.loads(data) #loads the json data and stores it in json_data variable


    print(json_data)
    json_data.get(env_to_return)

    return json_data

What my issue is, is that when I call the function in another file it does not display the parsed json it just displays the json as a a dict in this form;

{u'Environments': [{u'Dev': [u'111', u'222']}, {u'Qa': [u'333', u'444']}, {u'prod': [u'555', u'666']}]}

The print statements you see are just me trying to double check my answers. Also I have passed the call the correct parameters as it prints out the parameters if I ask it too. Thank you for your help!

dev = jsonimporter("test2.json","dev")
# dosomething with value
print(value)

This is how I called the function

And maybe you need to loop through the `Environments` list, and return the one where the dictionary key matches `env_to_return`. — Barmar, Nov 19 '19 at 21:51
BTW, there's no need to call `a.read()` and `json.loads()`, you can just use `json.load(a)` — Barmar, Nov 19 '19 at 21:52
And you should close the file after you're done with it. Best is to use `with` so this happens automatically. — Barmar, Nov 19 '19 at 21:52
Why do you have a list of dictionaries that each have a different key? It would make more sense to have a single dictionary with each environment as a different key. — Barmar, Nov 19 '19 at 21:58
Can you show how you're calling the function and the result you're trying to get? — Barmar, Nov 19 '19 at 21:59
```dev = jsonimporter("test2.json","dev") # dosomething with value print(value)``` This is how I am calling the function — aroe, Nov 19 '19 at 22:10
When you put code blocks in a question, the code has to start on the line *after* the triple backticks, not the same line. Does the little diagram below the question confuse you? See https://meta.stackoverflow.com/questions/391467/users-frequently-start-code-blocks-on-the-same-line-as-the-triple-backticks — Barmar, Nov 19 '19 at 22:18
@Barmar is on top of this, good advice so far, what I will add is that the imports should go at the top of your file. Following proper style, especially this kind of rule, is important. Also, calling `my_dict.get(one_arg)` is pointless, just use `my_dict[one_arg]` instead. — AMC, Nov 19 '19 at 23:15
@AlexanderCécile The latter depends on what you want to happen when the key isn't found. `get()` will return a default value, `my_dict[one_arg]` will throw an error. — Barmar, Nov 20 '19 at 17:21
@Barmar For some reason I thought `.get()` without the default value would behave exactly like `[]`, forgetting that the uh, default default value, is None. — AMC, Nov 20 '19 at 18:35
The `import json` should be at the top of the file, not in the function. — AMC, Nov 20 '19 at 19:44

Barmar · Answer 1 · 2019-11-19T22:25:24.347

0

You need to loop through the Environments property, searching for the dictionary with the key you want.

def jsonimporter(file_path, key, env_to_return):
    import json
    with open(file_path,"r") as a:
        json_data = json.load(a)

    for d in json_data[key]:
        if env_to_return in d:
            return d[env_to_return]

dev = jsonimporter("test2.json", "Environments", "dev")
print(dev)

Also see How to get string objects instead of Unicode from JSON? for how to get JSON with ordinary strings instead of Unicode strings in Python 2.

edited Nov 19 '19 at 22:25

answered Nov 19 '19 at 22:17

Barmar

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I like this loop for the answer however, could I used file_path instead of 'environments' because this is for a utility function and the 'Environments' is specific to this particular json – aroe Nov 19 '19 at 22:22
`file_path` is the filename, not the name of the key in the dictionary. – Barmar Nov 19 '19 at 22:23
2

Why is the parameter called `env_to_return` if this function isn't specifically about environments? – Barmar Nov 19 '19 at 22:24
2

You could make that key another parameter to the function. I've updated the answer to show how to do this. – Barmar Nov 19 '19 at 22:25

score 0 · Answer 2 · answered Nov 20 '19 at 23:00

Here is my final answer I got.

def get_multiple_envs_using_jsonimporter(file_path , env_list):
    var1 = jsonimporter("test2.json","Dev")
    var2 = jsonimporter("test2.json","Qa")

    full_list=[]
    full_list.append(var1)
    full_list.append(var2)
    print(full_list)
    # for each env
    # var = jsonimporter(file_path, env)
    # add environment+accounts to list_to_return
    # dev = jsonimporter("test2.json","Dev")









def jsonimporter(file_path , env_to_return):
    a = open(file_path,"r") #opens and sets the file to read

    data = a.read()         #sets variable to read function
    # print(data)

    json_data = json.loads(data) #loads the json data and stores it in json_data variable


    for env in json_data["Environments"]:
        if env_to_return in env:
            return env[env_to_return]
        else:
             print('Not Found')

I just appended the variables to the list that called the first function from.

Creating a json parsing function with Python

2 Answers2