Python读取JSON文件并修改

Question

Hi I am trying to take the data from a json file and insert and id then perform POST REST. my file data.json has:

{
    'name':'myname'
}

and I would like to add an id so that the json data looks like:

 {
     'id': 134,
     'name': 'myname'
 }

So I tried:

import json
f = open("data.json","r")
data = f.read()
jsonObj = json.loads(data)

I can't get to load the json format file. What should I do so that I can convert the json file into json object and add another id value.

Answer 1

Set item using data['id'] = ... .

import json

with open('data.json', 'r+') as f:
    data = json.load(f)
    data['id'] = 134 # <--- add `id` value.
    f.seek(0)        # <--- should reset file position to the beginning.
    json.dump(data, f, indent=4)
    f.truncate()     # remove remaining part

Answer 2

falsetru's solution is nice, but has a little bug:

Suppose original 'id' length was larger than 5 characters. When we then dump with the new 'id' (134 with only 3 characters) the length of the string being written from position 0 in file is shorter than the original length. Extra chars (such as '}') left in file from the original content.

I solved that by replacing the original file.

import json
import os

filename = 'data.json'
with open(filename, 'r') as f:
    data = json.load(f)
    data['id'] = 134 # <--- add `id` value.

os.remove(filename)
with open(filename, 'w') as f:
    json.dump(data, f, indent=4)

Answer 3

I would like to present a modified version of Vadim's solution. It helps to deal with asynchronous requests to write/modify json file. I know it wasn't a part of the original question but might be helpful for others.

In case of asynchronous file modification os.remove(filename) will raise FileNotFoundError if requests emerge frequently. To overcome this problem you can create temporary file with modified content and then rename it simultaneously replacing old version. This solution works fine both for synchronous and asynchronous cases.

import os, json, uuid

filename = 'data.json'
with open(filename, 'r') as f:
    data = json.load(f)
    data['id'] = 134 # <--- add `id` value.
    # add, remove, modify content

# create randomly named temporary file to avoid 
# interference with other thread/asynchronous request
tempfile = os.path.join(os.path.dirname(filename), str(uuid.uuid4()))
with open(tempfile, 'w') as f:
    json.dump(data, f, indent=4)

# rename temporary file replacing old file
os.rename(tempfile, filename)

Answer 4

There is really quite a number of ways to do this and all of the above are in one way or another valid approaches... Let me add a straightforward proposition. So assuming your current existing json file looks is this....

{
     "name":"myname"
}

And you want to bring in this new json content (adding key "id")

{
     "id": "134",
     "name": "myname"
 }

My approach has always been to keep the code extremely readable with easily traceable logic. So first, we read the entire existing json file into memory, assuming you are very well aware of your json's existing key(s).

import json 

# first, get the absolute path to json file
PATH_TO_JSON = 'data.json' #  assuming same directory (but you can work your magic here with os.)

# read existing json to memory. you do this to preserve whatever existing data. 
with open(PATH_TO_JSON,'r') as jsonfile:
    json_content = json.load(jsonfile) # this is now in memory! you can use it outside 'open'

Next, we use the 'with open()' syntax again, with the 'w' option. 'w' is a write mode which lets us edit and write new information to the file. Here s the catch that works for us ::: any existing json with the same target write name will be erased automatically.

So what we can do now, is simply write to the same filename with the new data

# add the id key-value pair (rmbr that it already has the "name" key value)
json_content["id"] = "134"

with open(PATH_TO_JSON,'w') as jsonfile:
    json.dump(json_content, jsonfile, indent=4) # you decide the indentation level

And there you go! data.json should be good to go for an good old POST request

Answer 5

try this script:

with open("data.json") as f:
    data = json.load(f)
    data["id"] = 134
    json.dump(data, open("data.json", "w"), indent = 4)

the result is:

{
    "name":"mynamme",
    "id":134
}

Just the arrangement is different, You can solve the problem by converting the "data" type to a list, then arranging it as you wish, then returning it and saving the file, like that:

index_add = 0
with open("data.json") as f:
    data = json.load(f)
    data_li = [[k, v] for k, v in data.items()]
    data_li.insert(index_add, ["id", 134])
    data = {data_li[i][0]:data_li[i][1] for i in range(0, len(data_li))}
    json.dump(data, open("data.json", "w"), indent = 4)

the result is:

{
    "id":134,
    "name":"myname"
}

you can add if condition in order not to repeat the key, just change it, like that:

index_add = 0
n_k = "id"
n_v = 134
with open("data.json") as f:
    data = json.load(f)
    if n_k in data:
        data[n_k] = n_v
    else:
       data_li = [[k, v] for k, v in data.items()]
       data_li.insert(index_add, [n_k, n_v])
       data = {data_li[i][0]:data_li[i][1] for i in range(0, len(data_li))}
    json.dump(data, open("data.json", "w"), indent = 4)

Answer 6

This implementation should suffice:

with open(jsonfile, 'r') as file:
    data = json.load(file)
    data[id] = value

with open(jsonfile, 'w') as file:
    json.dump(data, file)

using context manager for the opening of the jsonfile. data holds the updated object and dumped into the overwritten jsonfile in 'w' mode.