A bit new to JSON... Does anyone know how to properly iterate through and grab the symbol and change for example? I've tried wrapping everything in json.loads and using strings, but I keep getting errors regarding tuples. FYI, I'm using ticker inside the string, but I changed it to be YHOO for this question for convenience of anyone trying to run the same code.
from flask import Flask
from flask.ext.compress import Compress
from flask import render_template
from httplib2 import Http
import json
http = Http()
app = Flask(__name__)
Compress(app)
app.config['DEBUG'] = True
app.config['TESTING'] = True
#app.route('/<ticker>', methods=['GET'])
def check(ticker):
yahoo_api = http.request("http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20where%20symbol%20IN%20(%22YHOO%22)&format=json&env=http://datatables.org/alltables.env")
return yahoo_api[1]
if __name__ == '__main__':
app.run()
yahoo_api[1] is a string, use json.loads to get the json.
import json
from httplib2 import Http
yahoo_api = Http().request('http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20where%20symbol%20IN%20(%22YHOO%22)&format=json&env=http://datatables.org/alltables.env')
yahoo_json = json.loads(yahoo_api[1])
change = yahoo_json['query']['results']['quote']['Change']
symbol = yahoo_json['query']['results']['quote']['symbol']
Anthoer way is using requests, no worry about the json, it is esay to use.
import requests
r = requests.get('http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20where%20symbol%20IN%20%28%22YHOO%22%29&format=json&env=http://datatables.org/alltables.env')
change = r.json()['query']['results']['quote']['Change']
symbol = r.json()['query']['results']['quote']['symbol']
I'd think you might have forgotten to take the second part of the tuple (the content), although that seemed unlikely as you do do this for the return statement. Or maybe you forgot the UTF-8 decode?
import json
import pprint
from httplib2 import Http
http = Http()
url = "http://query.yahooapis.com/v1/public/yql?q=select%20*%20from%20yahoo.finance.quotes%20where%20symbol%20IN%20(%22YHOO%22)&format=json&env=http://datatables.org/alltables.env"
yahoo_api = http.request(url)
result = json.loads(yahoo_api[1].decode('utf-8'))
pprint.pprint(result)
Related
I'm trying to get the url of the minecraft skin through the api using python programming but I can't get the url, let's see if someone could
This is the code I'm currently using...
import json
import requests
import base64
response = requests.get(f"https://sessionserver.mojang.com/session/minecraft/profile/11f1cc006cc84499a174bc9b7fa1982a")
id = response.json()["properties"][0]["value"]
####
msg = f"{id}"
msg_bytes = msg.encode('ascii')
base64_bytes = base64.b64decode(msg_bytes)
base64_msg = base64_bytes.decode('ascii')
print(base64_msg)
Thank you very much in advance!
You will have to convert the string to json first.
import requests
import base64
import json
response = requests.get(
f"https://sessionserver.mojang.com/session/minecraft/profile/11f1cc006cc84499a174bc9b7fa1982a"
)
msg = response.json()["properties"][0]["value"]
base64_bytes = base64.b64decode(msg)
print(json.loads(base64_bytes)["textures"]["SKIN"]["url"])
You should always check that your HTTP request succeeds.
This is all you need:
from requests import get as GET
from base64 import b64decode as DECODE
from json import loads as LOADS
URL = 'https://sessionserver.mojang.com/session/minecraft/profile/11f1cc006cc84499a174bc9b7fa1982a'
(response := GET(URL)).raise_for_status()
data = response.json()['properties'][0]['value']
sd = LOADS(DECODE(data))
print(sd['textures']['SKIN']['url'])
Output:
http://textures.minecraft.net/texture/516ca747cee2cf895c02e0b4da4f1fe23495a140326538f960debaaa6fd67045
Note:
This code assumes that the JSON structures are as expected and that the keys always exist
I am trying to extract a table using an API but I am unable to do so. I am pretty sure that I am not using it correctly, and any help would be appreciated.
Actually I am trying to extract a table from this API but unable to figure out the right way on how to do it. This is what is mentioned in the website. I want to extract Latest_full_data table.
This is my code to get the table but I am getting error:
import urllib
import requests
import urllib.request
locu_api = 'api_Key'
def locu_search(query):
api_key = locu_api
url = 'https://www.quandl.com/api/v3/databases/WIKI/metadata?api_key=' + api_key
response = urllib.request.urlopen(url).read()
json_obj = str(response, 'utf-8')
datanew = json.loads(json_obj)
return datanew
When I do print(datanew). Update: Even if I change it to return data new, error is still the same.
I am getting this below error:
name 'datanew' is not defined
I had the same issues with urrlib before. If possible, try to use requests it's a better designed and working library in my opinion. Also, it is capable of reading JSON with a single function so no need to run it through multiple lines Sample code here:
import requests
locu_api = 'api_Key'
def locu_search():
url = 'https://www.quandl.com/api/v3/databases/WIKI/metadata?api_key=' + api_key
return requests.get(url).json()
locu_search()
Edit:
The endpoint that you are calling might not be the correct one. I think you are looking for the following one:
import requests
api_key = 'your_api_key_here'
def locu_search(dataset_code):
url = f'https://www.quandl.com/api/v3/datasets/WIKI/{dataset_code}/metadata.json?api_key={api_key}'
req = requests.get(url)
return req.json()
data = locu_search("FB")
This will return with all the metadata regarding a company. In this case Facebook.
Maybe it doesn't apply to your specific problem, but what I normally do is the following:
import requests
def get_values(url):
response = requests.get(url).text
values = json.loads(response)
return values
I'm working on some code that pulls course info from Canvas. As pure python, it works fine. If I try to incorporate it with Flask, I get the following error
requests.exceptions.MissingSchema: Invalid URL 'run/api/v1/courses/1234567': No schema supplied. Perhaps you meant http://run/api/v1/courses/1234567?
This is the code in question:
Canvas file
import sys
from canvasapi import Canvas
def getinfo():
canvasurl = "https://canvas.instructure.com/";
canvastoken = #Redacted for this example
try:
canvastoken = sys.argv[1];
canvasurl = sys.argv[2];
except:
print()
#Create a new canvas object passing in the newly aquired url and token
canvas = Canvas(canvasurl, canvastoken);
#print(canv)
# Create a new course oject -- passing in course number as a parameter
# Course number is currently hard coded
print(canvas.get_course(1234567))
Flask file code (the file that I'm trying to run):
from flask import Flask
import canvas
canvas.getinfo()
app = Flask(__name__)
#app.route('/')
def hello_world():
return 'Hello World!'
if __name__ == '__main__':
app.run()
No schema provided usually means you haven't specified the http:// or https:// in the URL.
In the code you provided, I don't see any reference to a run/api/v1/courses/1234567. One possibility is if you are using the url_for method from requests anywhere in your code, try setting _external=True:
url = url_for('relativeURL', _external=True)
This allows Flask to construct an absolute URL (i.e., a URL with domain included).
If you aren't using url_for, check other places in your code where you might be omitting the http or https from the URL.
If you update your question to include the part that refers to the offending URL, we might be able to provide more specific help.
I am trying to GET a URL using Python and the response is JSON. However, when I run
import urllib2
response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
html=response.read()
print html
The html is of type str and I am expecting a JSON. Is there any way I can capture the response as JSON or a python dictionary instead of a str.
If the URL is returning valid JSON-encoded data, use the json library to decode that:
import urllib2
import json
response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
data = json.load(response)
print data
import json
import urllib
url = 'http://example.com/file.json'
r = urllib.request.urlopen(url)
data = json.loads(r.read().decode(r.info().get_param('charset') or 'utf-8'))
print(data)
urllib, for Python 3.4
HTTPMessage, returned by r.info()
"""
Return JSON to webpage
Adding to wonderful answer by #Sanal
For Django 3.4
Adding a working url that returns a json (Source: http://www.jsontest.com/#echo)
"""
import json
import urllib
url = 'http://echo.jsontest.com/insert-key-here/insert-value-here/key/value'
respons = urllib.request.urlopen(url)
data = json.loads(respons.read().decode(respons.info().get_param('charset') or 'utf-8'))
return HttpResponse(json.dumps(data), content_type="application/json")
Be careful about the validation and etc, but the straight solution is this:
import json
the_dict = json.load(response)
resource_url = 'http://localhost:8080/service/'
response = json.loads(urllib2.urlopen(resource_url).read())
Python 3 standard library one-liner:
load(urlopen(url))
# imports (place these above the code before running it)
from json import load
from urllib.request import urlopen
url = 'https://jsonplaceholder.typicode.com/todos/1'
you can also get json by using requests as below:
import requests
r = requests.get('http://yoursite.com/your-json-pfile.json')
json_response = r.json()
Though I guess it has already answered I would like to add my little bit in this
import json
import urllib2
class Website(object):
def __init__(self,name):
self.name = name
def dump(self):
self.data= urllib2.urlopen(self.name)
return self.data
def convJSON(self):
data= json.load(self.dump())
print data
domain = Website("https://example.com")
domain.convJSON()
Note : object passed to json.load() should support .read() , therefore urllib2.urlopen(self.name).read() would not work .
Doamin passed should be provided with protocol in this case http
This is another simpler solution to your question
pd.read_json(data)
where data is the str output from the following code
response = urlopen("https://data.nasa.gov/resource/y77d-th95.json")
json_data = response.read().decode('utf-8', 'replace')
None of the provided examples on here worked for me. They were either for Python 2 (uurllib2) or those for Python 3 return the error "ImportError: No module named request". I google the error message and it apparently requires me to install a the module - which is obviously unacceptable for such a simple task.
This code worked for me:
import json,urllib
data = urllib.urlopen("https://api.github.com/users?since=0").read()
d = json.loads(data)
print (d)
I am trying to GET a URL using Python and the response is JSON. However, when I run
import urllib2
response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
html=response.read()
print html
The html is of type str and I am expecting a JSON. Is there any way I can capture the response as JSON or a python dictionary instead of a str.
If the URL is returning valid JSON-encoded data, use the json library to decode that:
import urllib2
import json
response = urllib2.urlopen('https://api.instagram.com/v1/tags/pizza/media/XXXXXX')
data = json.load(response)
print data
import json
import urllib
url = 'http://example.com/file.json'
r = urllib.request.urlopen(url)
data = json.loads(r.read().decode(r.info().get_param('charset') or 'utf-8'))
print(data)
urllib, for Python 3.4
HTTPMessage, returned by r.info()
"""
Return JSON to webpage
Adding to wonderful answer by #Sanal
For Django 3.4
Adding a working url that returns a json (Source: http://www.jsontest.com/#echo)
"""
import json
import urllib
url = 'http://echo.jsontest.com/insert-key-here/insert-value-here/key/value'
respons = urllib.request.urlopen(url)
data = json.loads(respons.read().decode(respons.info().get_param('charset') or 'utf-8'))
return HttpResponse(json.dumps(data), content_type="application/json")
Be careful about the validation and etc, but the straight solution is this:
import json
the_dict = json.load(response)
resource_url = 'http://localhost:8080/service/'
response = json.loads(urllib2.urlopen(resource_url).read())
Python 3 standard library one-liner:
load(urlopen(url))
# imports (place these above the code before running it)
from json import load
from urllib.request import urlopen
url = 'https://jsonplaceholder.typicode.com/todos/1'
you can also get json by using requests as below:
import requests
r = requests.get('http://yoursite.com/your-json-pfile.json')
json_response = r.json()
Though I guess it has already answered I would like to add my little bit in this
import json
import urllib2
class Website(object):
def __init__(self,name):
self.name = name
def dump(self):
self.data= urllib2.urlopen(self.name)
return self.data
def convJSON(self):
data= json.load(self.dump())
print data
domain = Website("https://example.com")
domain.convJSON()
Note : object passed to json.load() should support .read() , therefore urllib2.urlopen(self.name).read() would not work .
Doamin passed should be provided with protocol in this case http
This is another simpler solution to your question
pd.read_json(data)
where data is the str output from the following code
response = urlopen("https://data.nasa.gov/resource/y77d-th95.json")
json_data = response.read().decode('utf-8', 'replace')
None of the provided examples on here worked for me. They were either for Python 2 (uurllib2) or those for Python 3 return the error "ImportError: No module named request". I google the error message and it apparently requires me to install a the module - which is obviously unacceptable for such a simple task.
This code worked for me:
import json,urllib
data = urllib.urlopen("https://api.github.com/users?since=0").read()
d = json.loads(data)
print (d)