Convert nested json to csv to sheets json api - json

Question

I'm want to make my json to csv so that i can upload it on google sheets and make it as json api. Whenever i have change data i will just change it on google sheets. But I'm having problems on converting my json file to csv because it changes the variables whenever i convert it. I'm using https://toolslick.com/csv-to-json-converter to convert my json file to csv.
What is the best way to convert json nested to csv ?
JSON
{
"options": [
{
"id": "1",
"value": "Jumbo",
"shortcut": "J",
"textColor": "#FFFFFF",
"backgroundColor": "#00000"
},
{
"id": "2",
"value": "Hot",
"shortcut": "D",
"textColor": "#FFFFFF",
"backgroundColor": "#FFFFFF"
}
],
"categories": [
{
"id": "1",
"order": 1,
"name": "First Category",
"active": true
},
{
"id": "2",
"order": 2,
"name": "Second Category",
"shortcut": "MT",
"active": true
}
],
"products": [
{
"id": "03c6787c-fc2a-4aa8-93a3-5e0f0f98cfb2",
"categoryId": "1",
"name": "First Product",
"shortcut": "First",
"options": [
{
"optionId": "1",
"price": 23
},
{
"optionId": "2",
"price": 45
}
],
"active": true
},
{
"id": "e8669cea-4c9c-431c-84ba-0b014f0f9bc2",
"categoryId": "2",
"name": "Second Product",
"shortcut": "Second",
"options": [
{
"optionId": "1",
"price": 11
},
{
"optionId": "2",
"price": 20
}
],
"active": true
}
],
"discounts": [
{
"id": "1",
"name": "S",
"type": 1,
"amount": 20,
"active": true
},
{
"id": "2",
"name": "P",
"type": 1,
"amount": 20,
"active": true
},
{
"id": "3",
"name": "G",
"type": 2,
"amount": 5,
"active": true
}
]
}

Answer 1

Using python, this can be easily done or almost done. Maybe this code will help you in some way to understand that.
import json,csv
data = []
with open('your_json_file_here.json') as file:
for line in file:
data.append(json.loads(line))
length = len(data)
with open('create_new_file.csv','w') as f:
writer = csv.writer(f)
writers = csv.DictWriter(f, fieldnames=['header1','header2'])
writers.writeheader()
for iter in range(length):
writer.writerow((data[iter]['specific_col_name1'],data[iter]['specific_col_name2']))
f.close()