I'm using a NAO robot with naoqi version 2.1 and Choregraphe on Windows. I want to parse json from an attached file to the behavior. I attached the file like in that link.
Code:
def onLoad(self):
self.filepath = os.path.join(os.path.dirname(ALFrameManager.getBehaviorPath(self.behaviorId)), "fileName.json")
def onInput_onStart(self):
with open(self.filepath, "r") as f:
self.data = self.json.load(f.get_Response())
self.dataFromFile = self.data['value']
self.log("Data from file: " + str(self.dataFromFile))
But when I run this code on the robot (connected with a router) I'll get an error:
[ERROR] behavior.box :_safeCallOfUserMethod:281 _Behavior__lastUploadedChoregrapheBehaviorbehavior_1136151280__root__AbfrageKontostand_3__AuslesenJSONDatei_1: Traceback (most recent call last):
File "/usr/lib/python2.7/site-packages/naoqi.py", line 271, in _safeCallOfUserMethod
func()
File "<string>", line 20, in onInput_onStart
File "/usr/lib/python2.7/site-packages/inaoqi.py", line 265, in <lambda>
__getattr__ = lambda self, name: _swig_getattr(self, behavior, name)
File "/usr/lib/python2.7/site-packages/inaoqi.py", line 55, in _swig_getattr
raise AttributeError(name)
AttributeError: json
I already tried to understand the code from the correspondending lines but I couldn't fixed the error. But I know that the type of my object f is 'file'. How can I open the json file as a json file?
Your problem comes from this:
self.json.load(f.get_Response())
... there is no such thing as "self.json" on a Choregraphe box, import json and then do json.load. And what is get_Response ? That method doesn't exist on anything in Python that I know of.
You might want to first try making a standalone python script (that doesn't use the robot) that can read your json file before you try it with choregraphe. It will be easier.
Related
I am new to Python (and JSON) so apologies of this is obvious to you.
I pull some data from an API using the following code
import requests
import json
headers = {'Content-Type': 'application/json', 'accept-encoding':'identity'}
api_url = api_url_base+api_token+api_request #variables removed for security
response = requests.get(api_url, headers=headers)
data=response.json()
keys=data.keys
if response.status_code == 200:
print(data["message"], "saving to file...")
print("Found the following keys:")
print(keys)
with open('vulns.json', 'w') as outfile:
json.dump(response.content.decode('utf-8'),outfile)
print("File Saved.")
else:
print('The site returned a', response.status_code, 'error')
this works, I get some data returned and I am able to write the file.
I am trying to change what's returned form a short format to a long format and to check its working I need to see the keys, I was trying to do this offline using the written file (as practice for reading JSON from files).
I wrote these few lines (taken from this site https://www.kite.com/python/answers/how-to-print-the-keys-of-a-dictionary-in-python)
import json
with open('vulns.json') as json_file:
data=json.load(json_file)
print(data)
keys=list(data.keys())
print(keys)
Unfortunately, whenever I run this it returns this error
Python 3.9.1 (tags/v3.9.1:1e5d33e, Dec 7 2020, 17:08:21) [MSC v.1927 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> print(keys)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'keys' is not defined
>>> & C:/Users/xxxx/AppData/Local/Microsoft/WindowsApps/python.exe c:/Temp/read-vulnfile.py
File "<stdin>", line 1
& C:/Users/xxxx/AppData/Local/Microsoft/WindowsApps/python.exe c:/Temp/read-vulnfile.py
^
SyntaxError: invalid syntax
>>> exit()
PS C:\Users\xxxx\Documents\scripts\Python> & C:/Users/xxx/AppData/Local/Microsoft/WindowsApps/python.exe c:/Temp/read-vulnfile.py
Traceback (most recent call last):
File "c:\Temp\read-vulnfile.py", line 6, in <module>
keys=list(data.keys)
AttributeError: 'str' object has no attribute 'keys'
The Print(data) command returns what looks like JSON, this is the opening line:
{"count": 1000, "message": "Vulnerabilities found: 1000", "data":
[{"...
I cant show the content it's sensitive.
why is this looking at a str object rather than a dictionary?
how do I read JSON back into a dictionary please?
You just have that content stored in file as a string. Just open the vulns.json in some editor and there most likely is something like "{'count': 1000, ... instead of {"count": 1000, ....
It's opened by json.load, but translated to string (see this table).
So you should take one step back and take a look what happens during saving to file. You take some content from your response, but dump the string decoded value into a file. Take instead a try with
json.dump(response.json(), outfile)
(or just use data variable you already have provided).
This should allow you to succesfully dump and load data as a dict.
I'm starting using python to validate some json information, i'm using a json schema with reference but i'm having trouble to reference those files. This is the code :
from os.path import join, dirname
from jsonschema import validate
import jsonref
def assert_valid_schema(data, schema_file):
""" Checks whether the given data matches the schema """
schema = _load_json_schema(schema_file)
return validate(data, schema)
def _load_json_schema(filename):
""" Loads the given schema file """
relative_path = join('schemas', filename).replace("\\", "/")
absolute_path = join(dirname(__file__), relative_path).replace("\\", "/")
base_path = dirname(absolute_path)
base_uri = 'file://{}/'.format(base_path)
with open(absolute_path) as schema_file:
return jsonref.loads(schema_file.read(), base_uri=base_uri, jsonschema=True, )
assert_valid_schema(data, 'grandpa.json')
The json data is :
data = {"id":1,"work":{"id":10,"name":"Miroirs","composer":{"id":100,"name":"Maurice Ravel","functions":["Composer"]}},"recording_artists":[{"id":101,"name":"Alexandre Tharaud","functions":["Piano"]},{"id":102,"name":"Jean-Martial Golaz","functions":["Engineer","Producer"]}]}
And i'm saving the schema and reference file, into a schemas folder :
recording.json :
{"$schema":"http://json-schema.org/draft-04/schema#","title":"Schema for a recording","type":"object","properties":{"id":{"type":"number"},"work":{"type":"object","properties":{"id":{"type":"number"},"name":{"type":"string"},"composer":{"$ref":"artist.json"}}},"recording_artists":{"type":"array","items":{"$ref":"artist.json"}}},"required":["id","work","recording_artists"]}
artist.json :
{"$schema":"http://json-schema.org/draft-04/schema#","title":"Schema for an artist","type":"object","properties":{"id":{"type":"number"},"name":{"type":"string"},"functions":{"type":"array","items":{"type":"string"}}},"required":["id","name","functions"]}
And this is my error :
Connected to pydev debugger (build 181.5281.24)
Traceback (most recent call last):
File "C:\Python\lib\site-packages\proxytypes.py", line 207, in __subject__
return self.cache
File "C:\Python\lib\site-packages\proxytypes.py", line 131, in __getattribute__
return _oga(self, attr)
AttributeError: 'JsonRef' object has no attribute 'cache'
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "C:\Python\lib\site-packages\jsonref.py", line 163, in callback
base_doc = self.loader(uri)
<MORE>
python version : 3.6.5
windows 7
Ide : intellijIdea
Can somebody help me?
Thank you
I am not sure why, but on Windows, the file:// needs an extra /. So the following change should do the trick
base_uri = 'file:///{}/'.format(base_path)
Arrived at this answer from a solution posted for a related issue in json schema
i am using Flask as framework for my server, and while returning a response i get the following error:
> Traceback (most recent call last):
File "C:\Python27\lib\site-packages\flask\app.py", line 1612, in full_dispatch_request
rv = self.dispatch_request()
File "C:\Python27\lib\site-packages\flask\app.py", line 1598, in dispatch_request
return self.view_functions[rule.endpoint](**req.view_args)
File "C:\Python27\lib\site-packages\flask_restful\__init__.py", line 480, in wrapper
resp = resource(*args, **kwargs)
File "C:\Python27\lib\site-packages\flask\views.py", line 84, in view
return self.dispatch_request(*args, **kwargs)
File "C:\Python27\lib\site-packages\flask_restful\__init__.py", line 595, in dispatch_request
resp = meth(*args, **kwargs)
File "rest.py", line 27, in get
return jsonify(**solution)
File "C:\Python27\lib\site-packages\flask\json.py", line 263, in jsonify
(dumps(data, indent=indent, separators=separators), '\n'),
File "C:\Python27\lib\site-packages\flask\json.py", line 123, in dumps
rv = _json.dumps(obj, **kwargs)
File "C:\Python27\lib\json\__init__.py", line 251, in dumps
sort_keys=sort_keys, **kw).encode(obj)
File "C:\Python27\lib\json\encoder.py", line 209, in encode
chunks = list(chunks)
File "C:\Python27\lib\json\encoder.py", line 434, in _iterencode
for chunk in _iterencode_dict(o, _current_indent_level):
File "C:\Python27\lib\json\encoder.py", line 408, in _iterencode_dict
for chunk in chunks:
File "C:\Python27\lib\json\encoder.py", line 332, in _iterencode_list
for chunk in chunks:
File "C:\Python27\lib\json\encoder.py", line 332, in _iterencode_list
for chunk in chunks:
File "C:\Python27\lib\json\encoder.py", line 442, in _iterencode
o = _default(o)
File "C:\Python27\lib\site-packages\flask\json.py", line 80, in default
return _json.JSONEncoder.default(self, o)
File "C:\Python27\lib\json\encoder.py", line 184, in default
raise TypeError(repr(o) + " is not JSON serializable")
TypeError: {'origin': u'porto', 'dest': u'lisboa', 'price': '31', 'date': '2017-12-23', 'url': u'https://www.google.pt/flights/#search;f=opo;t=lis;d=2017-12-23;r=2017-12-24'} is not JSON serializable
i have the following function:
from flask import Flask, request, jsonify
from flask_restful import Resource, Api
from flask_cors import CORS, cross_origin
from json import dumps
import flights
import solveProblem
app = Flask(__name__)
api = Api(app)
CORS(app)
class Flights(Resource):
def get(self, data):
print 'received data from client: ' + data
solution = solveProblem.solve(data)
print 'got the solution from the script! \nSOLUTION: \n'
print solution
return jsonify(solution)
api.add_resource(Flights, '/flights/<string:data>')
if __name__ == '__main__':
app.run()
while debugging the problem, i found the following solutions which did not work:
1) return solution instead of {'solution': solution}
2) do jsonify(solution)
3) do jsonify(**solution)
none of the above worked for me;
i wonder why this happens, when i am trying to return a valid dictionary:
{'flights': [[{'origin': u'porto', 'dest': u'lisboa', 'price': '31', 'date': '2017-12-23', 'url': u'https://www.google.pt/flights/#search;f=opo;t=lis;d=2017-12-23;r=2017-12-24'}]], 'cost': '31'}
any help is appreciated.
Thanks
My guess is when you were creating 'solution', the data that got assigned to it was an incorrectly formatted dictionary
{'item', 'value'}
Instead of:
{'item': 'value'}
Thus creating a set instead of a dict
we cannot directly use the jsonify when your trying to converting list of data into json.
there is two approaches are there you can convert list into dictionary for that we need to write function that convert your list data into dictionary which is complicated task .
there is one smart work you can use Marshmallow library . it serialized you list data after that you can use jsonify.
In flask-restful, Resource class get method will just need to return python data structure. So just remove jsonify. For User Defined Object, you can use marshal_with() decorator.
See more: https://flask-restful.readthedocs.io/en/latest/quickstart.html#a-minimal-api
Since most of your functions are declared elsewhere, I worked a toy Flask program just to pass the dictionary you got stuck with.
[Edit] Before I was using the standard python json module. I edited it to use flask's own jsonify, and it works with the direct dictionary still. So the error is not where the OP is looking for.
{'flights': [[{'origin': u'porto', 'dest': u'lisboa', 'price': '31', 'date': '2017-12-23', 'url': u'https://www.google.pt/flights/#search;f=opo;t=lis;d=2017-12-23;r=2017-12-24'}]], 'cost': '31'}
The following program runs and returns the dictionary as a JSON object:
import flask
app = flask.Flask(__name__)
#app.route('/')
def hello():
jdic = flask.jsonify( {'origin': u'porto', 'dest': u'lisboa', 'price': '31', 'date': '2017-12-23', 'url': u'https://www.google.pt/flights/#search;f=opo;t=lis;d=2017-12-23;r=2017-12-24'} )
return jdic
if __name__ == '__main__':
app.run()
As I found out, this error generally occurs when the response is not a pure python dictionary. This happened to me because I was trying to pass a class object. So, to solve the problem, i created a class method which returns a dictionary describing the object, and use this to create the json response.
Conclusion: Use Pure python objects, which are easily translated to JSON.
I had the same problem with a 3 level Nested Dictionary; it was valid, json serializable and via command line json.dumps had no issue. However, Flask did not want to output it: "TypeError", not json serializable. The only difference is that I am using Python 3.5.
So I made a copy of it as a string (that on command line was json serializable!) and passed to Flask output, it worked.
Try to pass the nested json as
eval(str(solution))
and see the error. It's not a definitive solution but more a workaround.
Hope it helps.
I have a view that sends back json data from events as described in Arsawh's fullcalendar jquery plugin. Using this worked fine. But after installing Django 1.6.2 I get the follwoing error
Internal Server Error: /customer/todaysfeed/
Traceback (most recent call last):
File "/usr/local/lib/python2.7/dist-packages/django/core/handlers/base.py", line 114, in get_response
response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "/home/avlahop/development/django/rhombus2/rhombus/utils/decorators.py", line 18, in wrapper
return view(request, *args, **kw)
File "/home/avlahop/development/django/rhombus2/rhombus/customer/views.py", line 103, in todaysfeed
json_list = json.dumps(json_list, cls=DjangoJSONEncoder)
File "/usr/lib/python2.7/dist-packages/simplejson/__init__.py", line 369, in dumps
**kw).encode(obj)
TypeError: __init__() got an unexpected keyword argument 'namedtuple_as_object
What could be wrong? This is the code again
if request.is_ajax():
doctor = Doctor.objects.get(user=request.user)
try:
start = datetime.fromtimestamp(int(request.GET.get('start',''))).replace(tzinfo=utc)
end = datetime.fromtimestamp(int(request.GET.get('end',''))).replace(tzinfo=utc)
except ValueError:
start = datetime.now().replace(tzinfo=utc)
end = datetime.now().replace(tzinfo=utc)
entries = Entry.objects.filter(start__gte=start, end__lte=end, creator=doctor)
json_list = []
for entry in entries:
id=entry.id
title=str(entry.patient)
start=entry.start
end=entry.end
color= doctor.color if doctor.color else "blue"
allDay=False
item = dict(id=id, title=title, start=start, end=end, allDay=allDay, color=color)
json_list.append(item)
json_list = json.dumps(json_list, cls=DjangoJSONEncoder)
return HttpResponse(json_list, content_type='application/json')
EDIT: Found it!! Clamsy of me, i was importing an older version of json. Now everything works great just used import json. Thank you!
This answer should help.
In short, you've got a system simplejson vs Django-packaged json clash as mentioned in the release notes
There's a bit of code giving me trouble. It was working great in another script I had but I must have messed it up somehow.
The if csv: is primarily because I was relying on a -csv option in an argparser. But even if I were to run this with proper indents outside the if statement, it still returns the same error.
import csv
if csv:
with open('output.csv', 'wb') as csvfile:
csvout = csv.writer(csvfile, delimiter=',',
quotechar=',', quoting=csv.QUOTE_MINIMAL)
csvout.writerow(['A', 'B', 'C'])
csvfile.close()
Gives me:
Traceback (most recent call last):
File "import csv.py", line 34, in <module>
csvout = csv.writer(csvfile, delimiter=',',
AttributeError: 'str' object has no attribute 'writer'
If I remove the if statement, I get:
Traceback (most recent call last):
File "C:\import csv.py", line 34, in <module>
csvout = csv.writer(csvfile, delimiter=',',
AttributeError: 'NoneType' object has no attribute 'writer'
What silly thing am I doing wrong? I did try changing the file name to things like test.py as I saw that in another SO post, didn't work.
For me I had named my file csv.py. So when I import csv from that file I was essentially trying to import the same file itself.
If you've set something that assigns to csv (looks like a string) then you're shadowing the module import. So, the simplest thing is to just change whatever's assigning to csv that isn't the module and call it something else...
In effect what's happening is:
import csv
csv = 'bob'
csvout = csv.writer(somefile)
Remove the further assignment to csv and go from there...
For my case, my function name happened to be csv(). Once I renamed my function, the error disappeared.