How can I call variables and strings together from a json file in Python?

Question

I am trying to print variables and strings in a line. like [Hello! name How are you?] -> [Hello! John How are you?]

json file

#json01.json
{
"greet" : ["Hello! *name* How are you?"] # I don't know how to write this part..
}

and this is where I am..

import json

a = 'greet'
name = 'John'

with open('json01.json') as json_file:
json_dict = json.load(json_file)

if a in json_dict:
    print(json_dict[a]) # I don't know how to write this part as well..

How should I do to get the result I want?

Sorry for the poor explanation and Thank you!

@HarshaBiyani it's not a request for debugging, the question is how to make the output string have the name inserted in the right place. The existing code ignores that completely. — Karl Knechtel
– Karl Knechtel, Commented Apr 16, 2020 at 5:35

shx2 · Accepted Answer · 2020-04-16 05:35:14Z

5

There are various ways to format strings in python.

Here are 3 examples:

"Hello! %s How are you?" % name
"Hello! %(name)s How are you?" % locals()
"Hello! {name} How are you?".format(name=name)

Search the web for "python string formatting", and you'll find plenty of useful stuff.

edited Apr 16, 2020 at 5:35

answered Apr 16, 2020 at 5:28

shx2

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5 Comments

Oslo Over a year ago

does it work when it's decoded from json to dictionary?

shx2 Over a year ago

Sure, json_dict[a] is an expression returning a string

Karl Knechtel Over a year ago

How you get the dictionary doesn't matter. What matters is that you pull a string out of the dictionary (like with json_dict[a], like you already have) that has the right "template" stuff in it (you make this happen by editing the .json file to match), and then you use the matching code to replace the place-holder with the desired name value.

kajakIYD Over a year ago

@shx2 - I don't catch the difference between suggested approaches and f-strings. Could give me a clue? Isn't "Hello! {name} How are you?".format(name=name) equivalent of f"Hello! {name} How are you?"?

shx2 Over a year ago

@kajakIYD, as the answer is phrased, there is no difference. However, the 3 examples apply to the question because you can replace ""Hello! ..." with json_dict[a]. I don't think you can use such an expression in conjunction with f-strings (because the template is not fixed and known at runtime).

kajakIYD · Accepted Answer · 2020-04-16 09:40:02Z

In case you want to "substitute" more than only one variable into string, with pointing out explicitly what do you want to instantiate without typing many arguments in .format():

Create a dictionary. Keys should be strings to be replaced in acquired string and values should be variables that will appear in place of keys.
Use double asterisk ** to extract keys from namespace dictionary to "unpack" dictionary into .format() arguments list to substitute keys by values in acquired string

According to your example:

import json

a = 'greet'
name = 'John'
weight = 80
height = 190
age = 23
# 1.
namespace = {'name': name, 'age': age, 'weight': weight, 'height': height}

json_dict_all_keys = json.loads('{"greet" : "Hello! {name}, {age} years old, {height}cm, {weight}kg. How are you?"}')
json_dict_some_keys = json.loads('{"greet" : "Hello! {name}, {weight}kg. How are you?"}') 
json_dict_mixed_keys = json.loads('{"greet" : "Hello! {name}, {other} years old, {key}cm, {weight}kg. How are you?"}')
json_dict_none_of_keys = json.loads('{"greet" : "Hello! {some}, {other} years old, {key}cm, {here}kg. How are you?"}')

if a in json_dict:
    # 2.
    print(json_dict_all_keys[a].format(**namespace))  # Hello! John, 23 years old, 190cm, 80kg. How are you?
    print(json_dict_some_keys[a].format(**namespace))  # Hello! John, 80kg. How are you?
    print(json_dict_mixed_keys[a].format(**namespace))  # Raising KeyError!
    print(json_dict_none_of_keys[a].format(**namespace))  # Raising KeyError too!

As you can see, there is no requirement to use all of "keys" from namespace. But be careful - when in string you want to format appears "key" that is not included in namespace KeyError will occur.

To simplify explaination I used json.loads instead of loading json from file with json.load.

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