Develop Reference

Handling final page in Python paginated API request

I'm requesting Microsoft's Graph API, where I'm using the following function to request multiple pages. I'm trying to request all pages, merge the json files and finally write them to a pandas dataframe.
v = "v1.0"
r = "/users?$filter=userType eq 'Member'&$select=displayName,givenName,jobTitle,mail,department&$top=200"
def query(v, r):
all_records = []
url = uri.format(v=v, r=r)
while True:
if not url:
break
result = requests.get(url, headers=headers)
if result.status_code == 200:
json_data = json.loads(result.text)
all_records = all_records + json_data["value"]
url = json_data["#odata.nextLink"]
return all_records
The while-loop goes through all the pages, but when I run the function I'm getting a error:
KeyError: '#odata.nextLink'
I assume this is because the loop reaches the final page, and thus the '#odata.nextLink' cannot be found. But how can I handle this?

You are doing
url = json_data["#odata.nextLink"]
which suggest json_data is dict, so you should be able to use .get method which returns default value when key not found (None by default), please try doing following and write if it does work as excepted:
url = json_data.get("#odata.nextLink")
if url is None:
print("nextLink not found")
else:
print("nextLink found")

data scraping on discord using python

I'm currently trying to learn web scraping and decided to scrape some discord data. Code follows:
import requests
import json
def retrieve_messages(channelid):
num=0
headers = {
'authorization': 'here we enter the authorization code'
}
r = requests.get(
f'https://discord.com/api/v9/channels/{channelid}/messages?limit=100',headers=headers
)
jsonn = json.loads(r.text)
for value in jsonn:
print(value['content'], '\n')
num=num+1
print('number of messages we collected is',num)
retrieve_messages('server id goes here')
The problem: when I tried changing the limit here messages?limit=100 apparently it only accepts numbers between 0 and 100, meaning that the maximum number of messages I can get is 100. I tried changing this number to 900, for example, to scrape more messages. But then I get the error TypeError: string indices must be integers.
Any ideas on how I could get, possibly, all the messages in a channel?
Thank you very much for reading!

APIs that return a bunch of records are almost always limited to some number of items.
Otherwise, if a large quantity of items is requested, the API may fail due to being out of memory.
For that purpose, most APIs implement pagination using limit, before and after parameters where:
limit: tells you how many messages to fetch
before: get messages before this message ID
after: get messages after this message ID
Discord API is no exception as the documentation tells us.
Here's how you do it:
First, you will need to query the data multiple times.
For that, you can use a while loop.
Make sure to add an if the condition that will prevent the loop from running indefinitely - I added a check whether there are any messages left.
while True:
# ... requests code
jsonn = json.loads(r.text)
if len(jsonn) == 0:
break
for value in jsonn:
print(value['content'], '\n')
num=num+1
Define a variable that has the last message that you fetched and save the last message id that you already printed
def retrieve_messages(channelid):
last_message_id = None
while True:
# ...
for value in jsonn:
print(value['content'], '\n')
last_message_id = value['id']
num=num+1
Now on the first run the last_message_id is None, and on subsequent requests it has the last message you printed.
Use that to build your query
while True:
query_parameters = f'limit={limit}'
if last_message_id is not None:
query_parameters += f'&before={last_message_id}'
r = requests.get(
f'https://discord.com/api/v9/channels/{channelid}/messages?{query_parameters}',headers=headers
)
# ...
Note: discord servers give you the latest message first, so you have to use the before parameter
Here's a fully working example of your code
import requests
import json
def retrieve_messages(channelid):
num = 0
limit = 10
headers = {
'authorization': 'auth header here'
}
last_message_id = None
while True:
query_parameters = f'limit={limit}'
if last_message_id is not None:
query_parameters += f'&before={last_message_id}'
r = requests.get(
f'https://discord.com/api/v9/channels/{channelid}/messages?{query_parameters}',headers=headers
)
jsonn = json.loads(r.text)
if len(jsonn) == 0:
break
for value in jsonn:
print(value['content'], '\n')
last_message_id = value['id']
num=num+1
print('number of messages we collected is',num)
retrieve_messages('server id here')

To answer this question, we must look at the discord API. Googling "discord api get messages" gets us the developer reference for the discord API. The particular endpoint you are using is documented here:
https://discord.com/developers/docs/resources/channel#get-channel-messages
The limit is documented here, along with the around, before, and after parameters. Using one of these parameters (most likely after) we can paginate the results.
In pseudocode, it would look something like this:
offset = 0
limit = 100
all_messages=[]
while True:
r = requests.get(
f'https://discord.com/api/v9/channels/{channelid}/messages?limit={limit}&after={offset}',headers=headers
)
all_messages.append(extract messages from response)
if (number of responses < limit):
break # We have reached the end of all the messages, exit the loop
else:
offset += limit
By the way, you will probably want to print(r.text) right after the response comes in so you can see what the response looks like. It will save a lot of confusion.

Here is my solution. Feedback is welcome as I'm newish to Python. Kindly provide me w/ credit/good-luck if using this. Thank you =)
import requests
CHANNELID = 'REPLACE_ME'
HEADERS = {'authorization': 'REPLACE_ME'}
LIMIT=100
all_messages = []
r = requests.get(f'https://discord.com/api/v9/channels/{CHANNELID}/messages?limit={LIMIT}',headers=HEADERS)
all_messages.extend(r.json())
print(f'len(r.json()) is {len(r.json())}','\n')
while len(r.json()) == LIMIT:
last_message_id = r.json()[-1].get('id')
r = requests.get(f'https://discord.com/api/v9/channels/{CHANNELID}/messages?limit={LIMIT}&before={last_message_id}',headers=HEADERS)
all_messages.extend(r.json())
print(f'len(r.json()) is {len(r.json())} and last_message_id is {last_message_id} and len(all_messages) is {len(all_messages)}')

Read column/list containing urls by skipping blank cells using python

I have a column all_sheet_url in google sheet which has some url links in it.
I need to read them one by one and fetch data from them.
As there are some blanks/NA also in the column which I want to skip.
List item
I have tried with following code, but its not working to read only url and skip blanks if any in the column.
sheet_url = df['Links']
for line in sheet_url:
#if line in sheet_url:
try:
url = line
req = requests.get(url, stream=True)
r = requests.get(url, timeout=10)
r.raise_for_status()
# checking if it is an html page
content_type = req.headers.get('content-type')
if 'html' in content_type or 'application/xhtml+xml' in content_type:
# reading the contents
html = req.content
# req.close()
output = html
print(output)
continue
else:
print("\t{} is not an HTML file".format(url))
#req.close()
except Exception:
return
Kindly please suggest a method/corrections for the same, Thanks.
I need to read range of data from url and write in main google sheet.
I am able to read and write for single url at once.But, I want it to be dynamic so that it should read and write range of data from urls continously automatically by taking index of that url placed in the column.
This is the code I have tried with,
wks = gc.open_by_url(url)
wks1 = gc.open_by_url(url1)
# Defining Range of cells to read
range1 = wks1.range('A5:A9')
range2 = wks1.range('A5:B9')
range3 = wks1.range('G11:L19')
range4 = wks1.range('B12:C13')
# Defining Range of cells to write
range1n = wks.range('CO43:CS43')
range2n = wks.range('CT43:DR43')
range3n = wks.range('DS43:FK43')
range4n = wks.range('FL43:FM43')
range_names = [['range1', 'range2', 'range3', 'range4']]
for range1_cell, range1n_cell in zip(range1, range1n):
range1n_cell.value = range1_cell.value
wks.update_cells(range1n)

pandas has a dropna method to do exactly what you need.
Documentation: pandas.Series.dropna
In your cas that could just be the same processing code with the first line being:
all_sheet_url = df['Workbook Link'].dropna()

Index out of range when sending requests in a loop

I encounter an index out of range error when I try to get the number of contributors of a GitHub project in a loop. After some iterations (which are working perfectly) it just throws that exception. I have no clue why ...
for x in range(100):
r = requests.get('https://github.com/tipsy/profile-summary-for-github')
xpath = '//span[contains(#class, "num") and following-sibling::text()[normalize-space()="contributors"]]/text()'
contributors_number = int(html.fromstring(r.text).xpath(xpath)[0].strip().replace(',', ''))
print(contributors_number) # prints the correct number until the exception
Here's the exception.
----> 4 contributors_number = int(html.fromstring(r.text).xpath(xpath)[0].strip().replace(',', ''))
IndexError: list index out of range

It seems likely that you're getting a 429 - Too many requests since you're firing requests of one after the other.
You might want to modify your code as such:
import time
for index in range(100):
r = requests.get('https://github.com/tipsy/profile-summary-for-github')
xpath = '//span[contains(#class, "num") and following-sibling::text()[normalize-space()="contributors"]]/text()'
contributors_number = int(html.fromstring(r.text).xpath(xpath)[0].strip().replace(',', ''))
print(contributors_number)
time.sleep(3) # Wait a bit before firing of another request
Better yet would be:
import time
for index in range(100):
r = requests.get('https://github.com/tipsy/profile-summary-for-github')
if r.status_code in [200]: # Check if the request was successful
xpath = '//span[contains(#class, "num") and following-sibling::text()[normalize-space()="contributors"]]/text()'
contributors_number = int(html.fromstring(r.text).xpath(xpath)[0].strip().replace(',', ''))
print(contributors_number)
else:
print("Failed fetching page, status code: " + str(r.status_code))
time.sleep(3) # Wait a bit before firing of another request

Now this works perfectly for me while using the API. Probably the cleanest way of doing it.
import requests
import json
url = 'https://api.github.com/repos/valentinxxx/nginxconfig.io/commits?&per_page=100'
response = requests.get(url)
commits = json.loads(response.text)
commits_total = len(commits)
page_number = 1
while(len(commits) == 100):
page_number += 1
url = 'https://api.github.com/repos/valentinxxx/nginxconfig.io/commits?&per_page=100'+'&page='+str(page_number)
response = requests.get(url)
commits = json.loads(response.text)
commits_total += len(commits)

GitHub is blocking your repeated requests. Do not scrape sites in quick succession, many website operators actively block too many requests. As a result, the content that is returned no longer matches your XPath query.
You should be using the REST API that GitHub provides to retrieve project stats like the number of contributors, and you should implement some kind of rate limiting. There is no need to retrieve the same number 100 times, contributor counts do not change that rapidly.
API responses include information on how many requests you can make in a time window, and you can use conditional requests to only incur rate limit costs when the data actually has changed:
import requests
import time
from urllib.parse import parse_qsl, urlparse
owner, repo = 'tipsy', 'profile-summary-for-github'
github_username = '....'
# token = '....' # optional Github basic auth token
stats = 'https://api.github.com/repos/{}/{}/contributors'
with requests.session() as sess:
# GitHub requests you use your username or appname in the header
sess.headers['User-Agent'] += ' - {}'.format(github_username)
# Consider logging in! You'll get more quota
# sess.auth = (github_username, token)
# start with the first, move to the last when available, include anonymous
last_page = stats.format(owner, repo) + '?per_page=100&page=1&anon=true'
while True:
r = sess.get(last_page)
if r.status_code == requests.codes.not_found:
print("No such repo")
break
if r.status_code == requests.codes.no_content:
print("No contributors, repository is empty")
break
if r.status_code == requests.codes.accepted:
print("Stats not yet ready, retrying")
elif r.status_code == requests.codes.not_modified:
print("Stats not changed")
elif r.ok:
# success! Check for a last page, get that instead of current
# to get accurate count
link_last = r.links.get('last', {}).get('url')
if link_last and r.url != link_last:
last_page = link_last
else:
# this is the last page, report on count
params = dict(parse_qsl(urlparse(r.url).query))
page_num = int(params.get('page', '1'))
per_page = int(params.get('per_page', '100'))
contributor_count = len(r.json()) + (per_page * (page_num - 1))
print("Contributor count:", contributor_count)
# only get us a fresh response next time
sess.headers['If-None-Match'] = r.headers['ETag']
# pace ourselves following the rate limit
window_remaining = int(r.headers['X-RateLimit-Reset']) - time.time()
rate_remaining = int(r.headers['X-RateLimit-Remaining'])
# sleep long enough to honour the rate limit or at least 100 milliseconds
time.sleep(max(window_remaining / rate_remaining, 0.1))
The above uses a requests session object to handle repeated headers and ensure that you get to reuse connections where possible.
A good library such as github3.py (incidentally written by a requests core contributor) will take care of most of those details for you.
If you do want to persist on scraping the site directly, you do take a risk that the site operators block you altogether. Try to take some responsibility by not hammering the site continually.
That means that at the very least, you should honour the Retry-After header that GitHub gives you on 429:
if not r.ok:
print("Received a response other that 200 OK:", r.status_code, r.reason)
retry_after = r.headers.get('Retry-After')
if retry_after is not None:
print("Response included a Retry-After:", retry_after)
time.sleep(int(retry_after))
else:
# parse OK response

How to add all the pages of an api into a pandas dataframe

I do know that python has the read_json function to effectively get data from an api into a pandas dataframe. But is there any way to actually read through all the pages of the api and input it into the same dataframe.
import requests
import pandas as pd
import config
api_key = config.api_key
url = " http://api.themoviedb.org/3/discover/movie?release_date.gte=2017-12-
01&release_date.lte=2017-12-31&api_key=" + api_key
payload = "{}"
response = requests.request("GET", url, data=payload)
print(response.text.encode("utf-8"))
I tried with the requests method but this only gives me the 1st page of the api. But I wanted to see if there is any way I can do it with the df method as below. I am unable to understand how to write a loop to effectively loop over all the pages and then input it all into 1 dataframe for further analysis.
df = pd.read_json('http://api.themoviedb.org/3/discover/movie?
release_date.gte=2017-12-01&release_date.lte=2017-12-
31&api_key=''&page=%s' %page)

You can read each page into a dataframe and concatenate them:
page = 0
df = []
while True:
try:
next_page = pd.read_json('http://api.themoviedb.org/3/discover/movie?
release_date.gte=2017-12-01&release_date.lte=2017-12-
31&api_key=''&page=%s' %page)
# doesn't get any content, stop
if len(next_page) == 0:
break
else:
# move on to the next page
df.append(next_page)
page += 1
except:
# if we got error from the API call, maybe the URL for that page doesn't exist
# the stop
break
df = pd.concat(df, axis=0)
Documentation for pd.concat here. Hope it helps :)