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Home/ Questions/Q 7663879
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T14:09:40+00:00 2026-05-31T14:09:40+00:00

I’m trying to create a program that will go through a bunch of tumblr

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I’m trying to create a program that will go through a bunch of tumblr photos and extract the username of the person who uploaded them.

http://www.tumblr.com/tagged/food

If you look here, you can see multiple pictures of food with multiple different uploaders. If you scroll down you will begin to see even more pictures with even more uploaders. If you right click in your browser to view the source, and search “username”, however, it will only yield 10 results. Every time, no matter how far down you scroll.

Is there any way to counter this and have instead have it display the entire source for all images, or for X amount of images, or for however far you scrolled?

Here is my code to show what I’m doing:

#Imports
import requests
from bs4 import BeautifulSoup
import re

#Start of code
r = requests.get('http://www.tumblr.com/tagged/skateboard')
page = r.content

soup = BeautifulSoup(page)
soup.prettify()
arrayDiv = []

for anchor in soup.findAll("div", { "class" : "post_info" }):
    anchor = str(anchor)
    tempString = anchor.replace('</a>:', '')
    tempString = tempString.replace('<div class="post_info">', '') 
    tempString = tempString.replace('</div>', '')
    tempString = tempString.split('>')
    newString = tempString[1]
    newString = newString.strip()

    arrayDiv.append(newString)

print arrayDiv
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T14:09:41+00:00Added an answer on May 31, 2026 at 2:09 pm

    I had solved a similiar problem using beautifulsoup. what I did is looping through the paged pages. check with beautifulsoup is there is a continue element – here(in the tumbler page) for example this is an element with an id “next_page_link”
    if there is one I would loop the photo scraping code while changing the url fetched by requests. you would need to encapsulate all the code in a function ofcourse

    good luck.

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