I'm trying to get url from object data, but it isn't right. This program has stopped on line 4. Code is under.
My code:
import requests
gifs = str(requests.get("https://api.giphy.com/v1/gifs/random?
api_key=APIKEY"))
dump = json.dumps(gifs)
json.loads(dump['data']['url'])
Your description is not clear enough. You expect to read a json and select a field that brings you something?
I recommend you check this section of requests quickstart guide this i suspect you want to read the data to json and extract from some fields.
Maybe something like this might help:
r = requests.get('http://whatever.com')
url = r.json()['url']
Related
When I try to parse a rest api data,
it raises TypeError.
This is my code:
def get_contracts():
response_object = requests.get(
"https://testnet-api.phemex.com/md/orderbook?symbol=BTCUSD"
)
print(response_object.status_code)
for contract in response_object.json()["result"]["book"]:
print(contract["asks"])
get_contracts()
Any tip or solution will be very welcomed. Thanks in advance.
Edit/Update:
For some reason I am not able to select a specific key in the format above, its only possible if I do it like this:
data = response_object.json()['result']['book']['asks']
print(data)
I will try to work my code around that. Thanks for everyone who helped.
This code review may help you:
import requests
url = "https://testnet-api.phemex.com/md/orderbook?symbol=BTCUSD"
response_object = requests.get(url)
data = response_object.json()
# Printing your data helps to inspect the structure
# print(data)
# This is the list you are looking for:
asks = data['result']['book']['asks']
for ask in asks:
print(ask)
You need to iterate through asks, not book.
You have a nested dictionary where asks is a nested list.
If you simply click on the link you get getting, or print out your response_object.json() you would see the structure.
for foo in response_object.json()['result']['book']['asks']:
print(foo)
Although generally it's better to assign your response_object to a variable.
data = response_object.json()
for foo in data['result']['book']['asks']:
print(foo)
It looks like you are trying to access something that is not there, hence the KeyError.
I would debug, a simple print, the JSON object you are getting as answer and make sure that the keys you are trying to access are there.
Long story short, i get the query from spotify api which is JSON that has data about newest albums. How do i get the specific info from that like let's say every band name or every album title. I've tried a lot of ways to get that info that i found on the internet and nothing seems to work for me and after couple of hours im kinda frustrated
JSON data is on jsfiddle
here is the request
endpoint = "https://api.spotify.com/v1/browse/new-releases"
lookup_url = f"{endpoint}"
r = requests.get(lookup_url, headers=headers)
print(r.json())
you can find the
When you make this request like the comments have mentioned you get a dictionary which you can then access the keys and values. For example if you want to get the album_type you could do the following:
print(data["albums"]["items"][0]["album_type"])
Since items contains a list you would need to get the first values 0 and then access the album_type.
Output:
single
Here is a link to the code I used with your json.
I suggest you look into how to deal with json data in python, this is a good place to start.
I copied the data from the jsfiddle link.
Now try the following code:
import ast
pyobj=ast.literal_eval(str_cop_from_src)
later you can try with keys
pyobj["albums"]["items"][0]["album_type"]
pyobj will be a python dictionary will all data.
When sending data through python-requests a GET request, I have a need to specifically add something at the beginning of the query string. I have tried passing the data in through dicts and json strings with no luck.
The request as it appears when produced by requests:
/apply/.../explain?%7B%22......
The request as it appears when produced by their interactive API documentation (Swagger):
/apply/.../explain?record=%7B%22....
Where the key-value pairs of my data follow the excerpt above.
Ultimately, I think the missing piece is the record= that gets produced by their documentation. It is the only piece that is different from what is produced by Requests.
At the moment I've got it set up something like this:
import requests
s = requests.Session()
s.auth = requests.auth.HTTPBasicAuth(username,password)
s.verify = certificate_path
# with data below being a dictionary of the values I need to pass.
r = s.get(url,data=data)
I am trying to include an image of the documentation below, but don't yet have enough reputation to do so:
apply/model/explain documentation
'GET' requests don't have data, that's for 'POST' and friends.
You can send the query string arguments using params kwarg instead:
>>> params = {'record': '{"'}
>>> response = requests.get('http://www.example.com/explain', params=params)
>>> response.request.url
'http://www.example.com/explain?record=%7B%22'
From the comments i felt the need to explain this.
http://example.com/sth?key=value&anotherkey=anothervalue
Let's assume you have a url like the above in order to call with python requests you only have to write
response = requests.get('http://example.com/sth', params={
'key':'value',
'anotherkey':'anothervalue'
})
Have in mind that if your value or your keys have any special character in them they will be escaped thats the reason for the %7B%2 part of url in your question.
I am trying to create a Telegram bot from scratch using python. I've done all the initial steps and got the bot token, and now what I want to do is, for easy manipulation of the data it sends me (Like getting the first_name of the person from getupdates method) I want the data neatly arranged into a python dictionary.
When I try /getme, I get this:
b'{"ok":true,"result":{"id":999999999,"first_name":"telebotsrock","username":"sample_bot"}}'
Since the b' at the beginning and ' at the end causes an error when I do json.loads(data) (Where data is the thing given above converted to a string).
So I do data[2:-1] to remove the b' and ' and json.loads() works just fine, but when I change the /getme to /getupdates, a bunch of new errors pop up.
All in all, it's a mess. Can someone give me a clean way to get data from the bot and sort it into a python dictionary? Please do not tell me to use a different language or just copy an existing bot framework.
My current code:
from urllib.request import urlopen
import json
token="999999999:xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
site="https://api.telegram.org/bot"+token
content=str(urlopen(site+"/getme").read())
#content=str(urlopen(site+"/getupdates").read())
data=content[2:-1]
print(data)
info=json.loads(data)
print(info)
This code correctly coverts the output of /getme to a python dictionary, but gives errors when I try /getupdates instead.
Output of /getupdates before I slice it is:
b'{"ok":true,"result":[{"update_id":66666666,\n"message":{"message_id":1,"from":{"id":777777777,"first_name":"Aswin","last_name":"G","username":"MatrixHunter"},"chat":{"id":777777777,"first_name":"Aswin","last_name":"G","username":"MatrixHunter","type":"private"},"date":1459932293,"text":"\\/start"}},{"update_id":88888888,\n"message":{"message_id":2,"from":{"id":777777777,"first_name":"Aswin","last_name":"G","username":"MatrixHunter"},"chat":{"id":777777777,"first_name":"Aswin","last_name":"G","username":"MatrixHunter","type":"private"},"date":1459932298,"text":"Oy"}}]}'
This should work for you. You can use the .decode('utf-8') to get rid of the byte prefix.
token = "999999999:xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
url="https://api.telegram.org/bot" +token + "/getme"
req = Request(url)
response = urlopen(req)
data = response.read().decode('utf-8')
json_data = json.loads(data)
print(str(data['ok'])) #should print True
According to the feedparser documentation, I can turn an RSS feed into a parsed object like this:
import feedparser
d = feedparser.parse('http://feedparser.org/docs/examples/atom10.xml')
but I can't find anything showing how to go the other way; I'd like to be able do manipulate 'd' and then output the result as XML:
print d.toXML()
but there doesn't seem to be anything in feedparser for going in that direction. Am I going to have to loop through d's various elements, or is there a quicker way?
Appended is a not hugely-elegant, but working solution - it uses feedparser to parse the feed, you can then modify the entries, and it passes the data to PyRSS2Gen. It preserves most of the feed info (the important bits anyway, there are somethings that will need extra conversion, the parsed_feed['feed']['image'] element for example).
I put this together as part of a little feed-processing framework I'm fiddling about with.. It may be of some use (it's pretty short - should be less than 100 lines of code in total when done..)
#!/usr/bin/env python
import datetime
# http://www.feedparser.org/
import feedparser
# http://www.dalkescientific.com/Python/PyRSS2Gen.html
import PyRSS2Gen
# Get the data
parsed_feed = feedparser.parse('http://reddit.com/.rss')
# Modify the parsed_feed data here
items = [
PyRSS2Gen.RSSItem(
title = x.title,
link = x.link,
description = x.summary,
guid = x.link,
pubDate = datetime.datetime(
x.modified_parsed[0],
x.modified_parsed[1],
x.modified_parsed[2],
x.modified_parsed[3],
x.modified_parsed[4],
x.modified_parsed[5])
)
for x in parsed_feed.entries
]
# make the RSS2 object
# Try to grab the title, link, language etc from the orig feed
rss = PyRSS2Gen.RSS2(
title = parsed_feed['feed'].get("title"),
link = parsed_feed['feed'].get("link"),
description = parsed_feed['feed'].get("description"),
language = parsed_feed['feed'].get("language"),
copyright = parsed_feed['feed'].get("copyright"),
managingEditor = parsed_feed['feed'].get("managingEditor"),
webMaster = parsed_feed['feed'].get("webMaster"),
pubDate = parsed_feed['feed'].get("pubDate"),
lastBuildDate = parsed_feed['feed'].get("lastBuildDate"),
categories = parsed_feed['feed'].get("categories"),
generator = parsed_feed['feed'].get("generator"),
docs = parsed_feed['feed'].get("docs"),
items = items
)
print rss.to_xml()
If you're looking to read in an XML feed, modify it and then output it again, there's a page on the main python wiki indicating that the RSS.py library might support what you're after (it reads most RSS and is able to output RSS 1.0). I've not looked at it in much detail though..
from xml.dom import minidom
doc= minidom.parse('./your/file.xml')
print doc.toxml()
The only problem is that it do not download feeds from the internet.
As a method of making a feed, how about PyRSS2Gen? :)
I've not played with FeedParser, but have you tried just doing str(yourFeedParserObject)? I've often been suprised by various modules that have str methods to just output the object as text.
[Edit] Just tried the str() method and it doesn't work on this one. Worth a shot though ;-)