How can I get data from external API JSON file in ODOO - json

[{"Id":"605a321e-7c10-49e4-9d34-ba03c4b34f69","Url":"","Type":"INBOUND_OUTBOUND",
"ClearCurrentData":true,"FillOutFormFields":true,"RequestProtocol":"HTTP_REQUEST",
"FileToSend":"NONE","SendDefaultData":false,"SendDetectionData":false,"ShowDialogMessage":false,"IsActive":false,"SendingTrigger":"MANUALLY","TCPSocketMethod":"","TriggerButtonName":"Get data"}]
This is an External API Call JSON file how can i get the data in ODOO Any Solutions please ?
Mentioned Above code API JSON file
i had only External JSON File,don't have a table name & Database name is it possible to get data in ODOO

create a model for this data for example
class apiCall(models.Model):
_name = 'apiCall.api'
id = fields.Char(string='Id')
url = fields.Char(string='url')
#api.model
def call_api(self):
### result of calling external api is "result"
for line of self:
id = result[0].Id
url = result[0].url

You can call a method on button click and access the json file. Below is the code for accessing json file:-
import json
def api_call(self):
# Opening JSON file
f = open('file_name.json',)
# returns JSON object as
# a dictionary
data = json.load(f)
# Iterating through the json
# list
for i in self:
print(i)
# Closing file
f.close()
Below is the button code which is applied in xml file:-
<button name="api_call" type="object" string="Call API" class="oe_highlight" />
by this way you can call the json file with external api in odoo.

In your custom module (my_custom_module), move your json file to its subdirectory static/src:
import json
from odoo.modules.module import get_module_resource
def call_api(self)
json_filepath = get_module_resource('my_custom_module', 'static/src/', 'my_api_cred.json')
with open(json_filepath) as f:
data = json.load(f)

Related

Reading json file from remote url in groovy

Suppose Some JSON file is # www.github.com/xyz/Hello.json, I want to read this content of this JSON in JSON object in Groovy
You then need:
import groovy.json.JsonSlurper
def slurped = new JsonSlurper().parse('www.github.com/xyz/Hello.json'.toURL())
Here you can find more info.

Flask read uploaded json file

I'm uploading a json file via flask, but I'm having trouble actually reading what is in the file.
# named fJson b/c of other json imports
from flask import json as fJson
#app.route('/upload', methods=['GET', 'POST'])
def upload():
if request.method == 'POST':
file = request.files['file']
# data = fJson.load(file)
# myfile = file.read()
I'm trying to deal with this by using the 'file' variable. I looked at http://flask.pocoo.org/docs/0.10/api/#flask.json.load, but I get the error "No JSON object could be decoded". I also looked at Read file data without saving it in Flask which recommended using file.read(), but that didn't work, returns either "None" or "".
Request.files
A MultiDict with files uploaded as part of a POST or PUT request. Each file is stored as FileStorage object. It basically behaves like a standard file object you know from Python, with the difference that it also has a save() function that can store the file on the filesystem.
http://flask.pocoo.org/docs/0.10/api/#flask.Request.files
You don't need use json, just use read(), like this:
if request.method == 'POST':
file = request.files['file']
myfile = file.read()
For some reason the position in the file was at the end. Doing
file.seek(0)
before doing a read or load fixes the problem.

Value Error while trying to read json file

In Django, I am trying to read the countries to cities json file that's available here: https://raw.githubusercontent.com/David-Haim/CountriesToCitiesJSON/master/countriesToCities.json
I have downloaded the file locally into my static assets folder and I am doing the following to open, read and push all cities into another array
obj = []
filename = 'static/json/countriesToCities.json'
with open(filename, "r") as f:
data = json.loads(f.read())
for key, values in data:
obj.append(key[0])
However, this gives me the following error:
ValueError at /citiesUrl/
No JSON object could be decoded
How do I push all the values of each key into a new array?
use load instead loads (first for files, second for strings)
I've tested your JSON and it works:
json_data = open('/Users/madzohan/Downloads/data.json', 'r')
data = json.load(json_data)

Using Django Rest Framework, how can I upload a file AND send a JSON payload?

I am trying to write a Django Rest Framework API handler that can receive a file as well as a JSON payload. I've set the MultiPartParser as the handler parser.
However, it seems I cannot do both. If I send the payload with the file as a multi part request, the JSON payload is available in a mangled manner in the request.data (first text part until the first colon as the key, the rest is the data). I can send the parameters in standard form parameters just fine - but the rest of my API accepts JSON payloads and I wanted to be consistent. The request.body cannot be read as it raises *** RawPostDataException: You cannot access body after reading from request's data stream
For example, a file and this payload in the request body:
{"title":"Document Title", "description":"Doc Description"}
Becomes:
<QueryDict: {u'fileUpload': [<InMemoryUploadedFile: 20150504_115355.jpg (image/jpeg)>, <InMemoryUploadedFile: Front end lead.doc (application/msword)>], u'{%22title%22': [u'"Document Title", "description":"Doc Description"}']}>
Is there a way to do this? Can I eat my cake, keep it and not gain any weight?
Edit:
It was suggested that this might be a copy of Django REST Framework upload image: "The submitted data was not a file". It is not. The upload and request is done in multipart, and keep in mind the file and upload of it is fine. I can even complete the request with standard form variables. But I want to see if I can get a JSON payload in there instead.
For someone who needs to upload a file and send some data, there is no straight fwd way you can get it to work. There is an open issue in json api specs for this. One possibility i have seen is to use multipart/related as shown here, but i think its very hard to implement it in drf.
Finally what i had implemented was to send the request as formdata. You would send each file as file and all other data as text.
Now for sending the data as text you can have a single key called data and send the whole json as string in value.
Models.py
class Posts(models.Model):
id = models.UUIDField(default=uuid.uuid4, primary_key=True, editable=False)
caption = models.TextField(max_length=1000)
media = models.ImageField(blank=True, default="", upload_to="posts/")
tags = models.ManyToManyField('Tags', related_name='posts')
serializers.py -> no special changes needed, not showing my serializer here as its too lengthy because of the writable ManyToMany Field implimentation.
views.py
class PostsViewset(viewsets.ModelViewSet):
serializer_class = PostsSerializer
parser_classes = (MultipartJsonParser, parsers.JSONParser)
queryset = Posts.objects.all()
lookup_field = 'id'
You will need custom parser as shown below for parsing json.
utils.py
from django.http import QueryDict
import json
from rest_framework import parsers
class MultipartJsonParser(parsers.MultiPartParser):
def parse(self, stream, media_type=None, parser_context=None):
result = super().parse(
stream,
media_type=media_type,
parser_context=parser_context
)
data = {}
# find the data field and parse it
data = json.loads(result.data["data"])
qdict = QueryDict('', mutable=True)
qdict.update(data)
return parsers.DataAndFiles(qdict, result.files)
The request example in postman
EDIT:
see this extended answer if you want to send each data as key value pair
I know this is an old thread, but I just came across this. I had to use MultiPartParser in order to get my file and extra data to come across together. Here's what my code looks like:
# views.py
class FileUploadView(views.APIView):
parser_classes = (MultiPartParser,)
def put(self, request, filename, format=None):
file_obj = request.data['file']
ftype = request.data['ftype']
caption = request.data['caption']
# ...
# do some stuff with uploaded file
# ...
return Response(status=204)
My AngularJS code using ng-file-upload is:
file.upload = Upload.upload({
url: "/api/picture/upload/" + file.name,
data: {
file: file,
ftype: 'final',
caption: 'This is an image caption'
}
});
I send JSON and an image to create/update a product object. Below is a create APIView that works for me.
Serializer
class ProductCreateSerializer(serializers.ModelSerializer):
class Meta:
model = Product
fields = [
"id",
"product_name",
"product_description",
"product_price",
]
def create(self,validated_data):
return Product.objects.create(**validated_data)
View
from rest_framework import generics,status
from rest_framework.parsers import FormParser,MultiPartParser
class ProductCreateAPIView(generics.CreateAPIView):
queryset = Product.objects.all()
serializer_class = ProductCreateSerializer
permission_classes = [IsAdminOrIsSelf,]
parser_classes = (MultiPartParser,FormParser,)
def perform_create(self,serializer,format=None):
owner = self.request.user
if self.request.data.get('image') is not None:
product_image = self.request.data.get('image')
serializer.save(owner=owner,product_image=product_image)
else:
serializer.save(owner=owner)
Example test:
def test_product_creation_with_image(self):
url = reverse('products_create_api')
self.client.login(username='testaccount',password='testaccount')
data = {
"product_name" : "Potatoes",
"product_description" : "Amazing Potatoes",
"image" : open("local-filename.jpg","rb")
}
response = self.client.post(url,data)
self.assertEqual(response.status_code,status.HTTP_201_CREATED)
#Nithin solution works but essentially it means you are sending JSON as strings and hence not using the actual application/json inside the multipart segments.
What we want is to make the backend accept data in the below format
------WebKitFormBoundaryrga771iuUYap8BB2
Content-Disposition: form-data; name="file"; filename="1x1_noexif.jpeg"
Content-Type: image/jpeg
------WebKitFormBoundaryrga771iuUYap8BB2
Content-Disposition: form-data; name="myjson"; filename="blob"
Content-Type: application/json
{"hello":"world"}
------WebKitFormBoundaryrga771iuUYap8BB2
Content-Disposition: form-data; name="isDownscaled"; filename="blob"
Content-Type: application/json
false
------WebKitFormBoundaryrga771iuUYap8BB2--
MultiPartParser works with the above format but will treat those jsons as files. So we simply unmarshal those jsons by putting them to data.
parsers.py
from rest_framework import parsers
class MultiPartJSONParser(parsers.MultiPartParser):
def parse(self, stream, *args, **kwargs):
data = super().parse(stream, *args, **kwargs)
# Any 'File' found having application/json as type will be moved to data
mutable_data = data.data.copy()
unmarshaled_blob_names = []
json_parser = parsers.JSONParser()
for name, blob in data.files.items():
if blob.content_type == 'application/json' and name not in data.data:
mutable_data[name] = json_parser.parse(blob)
unmarshaled_blob_names.append(name)
for name in unmarshaled_blob_names:
del data.files[name]
data.data = mutable_data
return data
settings.py
REST_FRAMEWORK = {
..
'DEFAULT_PARSER_CLASSES': [
..
'myproject.parsers.MultiPartJSONParser',
],
}
This should work now.
The final bit is testing. Since the test client that ships with Django and REST doesn't support multipart JSON, we work around that by wrapping any JSON data.
import io
import json
def JsonBlob(obj):
stringified = json.dumps(obj)
blob = io.StringIO(stringified)
blob.content_type = 'application/json'
return blob
def test_simple(client, png_3x3):
response = client.post(f'http://localhost/files/', {
'file': png_3x3,
'metadata': JsonBlob({'lens': 'Sigma 35mm'}),
}, format='multipart')
assert response.status_code == 200
If you're getting an error along the lines of Incorrect type. Expected pk value, received list., with #nithin's solution, it's because Django's QueryDict is getting in the way - it's specifically structured to use a list for each entry in the dictionary, and thus:
{ "list": [1, 2] }
when parsed by MultipartJsonParser yields
{ 'list': [[1, 2]] }
which trips up your serializer.
Here is an alternative which handles this case, specifically expecting the _data key for your JSON:
from rest_framework import parsers
import json
class MultiPartJSONParser(parsers.MultiPartParser):
def parse(self, stream, *args, **kwargs):
data = super().parse(stream, *args, **kwargs)
json_data_field = data.data.get('_data')
if json_data_field is not None:
parsed = json.loads(json_data_field)
mutable_data = {}
for key, value in parsed.items():
mutable_data[key] = value
mutable_files = {}
for key, value in data.files.items():
if key != '_data':
mutable_files[key] = value
return parsers.DataAndFiles(mutable_data, mutable_files)
json_data_file = data.files.get('_data')
if json_data_file:
parsed = parsers.JSONParser().parse(json_data_file)
mutable_data = {}
for key, value in parsed.items():
mutable_data[key] = value
mutable_files = {}
for key, value in data.files.items():
mutable_files[key] = value
return parsers.DataAndFiles(mutable_data, mutable_files)
return data
It is very simple to use a multipart post and a regular view, if this is an option.
You send the json as a field and files as files, then process in one view.
Here is a simple python client and a Django server:
Client - sending multiple files and an arbitrary json-encoded object:
import json
import requests
payload = {
"field1": 1,
"manifest": "special cakes",
"nested": {"arbitrary":1, "object":[1,2,3]},
"hello": "word" }
filenames = ["file1","file2"]
request_files = {}
url="example.com/upload"
for filename in filenames:
request_files[filename] = open(filename, 'rb')
r = requests.post(url, data={'json':json.dumps(payload)}, files=request_files)
Server - consuming the json and saving the files:
#csrf_exempt
def upload(request):
if request.method == 'POST':
data = json.loads(request.POST['json'])
try:
manifest = data['manifest']
#process the json data
except KeyError:
HttpResponseServerError("Malformed data!")
dir = os.path.join(settings.MEDIA_ROOT, "uploads")
os.makedirs(dir, exist_ok=True)
for file in request.FILES:
path = os.path.join(dir,file)
if not os.path.exists(path):
save_uploaded_file(path, request.FILES[file])
else:
return HttpResponseNotFound()
return HttpResponse("Got json data")
def save_uploaded_file(path,f):
with open(path, 'wb+') as destination:
for chunk in f.chunks():
destination.write(chunk)
I'd just like to add to #Pithikos's answer by modifying the parser to accept lists as well, in line with how DRF parses lists in serializers in utils/html#parse_html_list
class MultiPartJSONParser(parsers.MultiPartParser):
def parse(self, stream, *args, **kwargs):
data = super().parse(stream, *args, **kwargs)
# Any 'File' found having application/json as type will be moved to data
mutable_data = data.data.copy()
unmarshaled_blob_names = []
json_parser = parsers.JSONParser()
for name, blob in data.files.items():
if blob.content_type == 'application/json' and name not in data.data:
parsed = json_parser.parse(blob)
if isinstance(parsed, list):
# need to break it out into [0], [1] etc
for idx, item in enumerate(parsed):
mutable_data[name+f"[{str(idx)}]"] = item
else:
mutable_data[name] = parsed
unmarshaled_blob_names.append(name)
for name in unmarshaled_blob_names:
del data.files[name]
data.data = mutable_data
return data
The following code worked for me.
from django.core.files.uploadedfile import SimpleUploadedFile
import requests
from typing import Dict
with open(file_path, 'rb') as f:
file = SimpleUploadedFile('Your-Name', f.read())
data: Dict[str,str]
files: Dict[str,SimpleUploadedFile] = {'model_field_name': file}
requests.put(url, headers=headers, data=data, files=files)
requests.post(url, headers=headers, data=data, files=files)
'model_field_name' is the name of the FileField or ImageField in your Django model. You can pass other data as name or location as usual by using data parameter.
Hope this helps.
This work for me:
class FileUpload(APIView):
parser_classes = [MultiPartParser]
authentication_classes = [JWTAuthentication]
def post(self, request, filename, format=None):
file = request.data['file']
data = json.loads(request.POST['form'])
#.... just do.....
.
.
.
frontend part: example with fetch (vue frontend)
let data = await new FormData(); // creates a new FormData object
data.append("file", this.files); // add your file to form data
data.append('form',JSON.stringify(body)) //add your json
fetch(`https://endpoint/FileUpload/${body.nombre}`, {
method: 'POST',
body: data,
headers: {Authorization: `Bearer ${accessToken}`}
})
I hope this helps.

django upload file using JSON

i have JSON request like this:
object: { "fields":{ "src" : "http://dss.com/a.jpg", "data" : " //file is here" } }
i have the model like this:
class FileMy(models.Model):
f = models.FileField(upload_to='file_path/',)
How to save the file ?
You may use urllib to read the file and then you can add it to your model.
Take a look at this post:
Django: add image in an ImageField from image url
You may be able to wrap the data in a ContentFile which inherits from File and then save the file to the model directly.
from __future__ import unicode_literals
from django.core.files.base import ContentFile
from .models import FileMy
f1 = ContentFile("esta sentencia está en español")
f2 = ContentFile(b"these are bytes")
m1 = FileMy()
m2 = FileMy()
m1.f.save("filename", f1, save=True)
m2.f.save("filename", f2, save=True)
First of all, encode the raw data in the json request body.
from tempfile import NamedTemporaryFile
from django.core.files import File
def save_file_to_field(field, file_name, raw_content):
# field: reference to the model object instance field
img_temp = NamedTemporaryFile(delete=True)
img_temp.write(raw_content)
field.save(
file_name,
File(img_temp)
)
img_temp.flush()
What does this do:
creates a temporary file on your system that holds the data
uses the file field save method to trigger the usual file handling
deletes the temporary file