🏯 Instagram Scraper (Pay Per Result)
Pay $0.50 for 1,000 posts
🏯 Instagram Scraper (Pay Per Result)
Pay $0.50 for 1,000 posts
Famous for its blazing-fast speed and completeness, it offers rapid and efficient scraping of many URLs such as; location, audio, tag, and profile. It's incredibly cost-effective at only $0.50 for every 1000 posts, capable of scraping 100-200 posts each second!
When I run a test run with a location URL this is the only result:
[{'noResults': True, 'error': True, 'code': 'C006', 'message': 'Could not get posts for location ID: locations'}]
This is true on the console and in python:
location_id = "102624392520735" limit = 10 actor_id = "culc72xb7MP3EbaeX"
run_input = { "startUrls": ["https://www.instagram.com/explore/locations/327485353771492"], # "startUrls": ["https://www.instagram.com/explore/tags/burgerlove"], "timeout": 3600, # default is 60s which is likely too short "maxItems": limit, "until": "2023-12-31", "customMapFunction": "(object) => { return {...object} }", }
run = client.actor(actor_id).call(run_input=run_input)
results = [] for item in client.dataset(run["defaultDatasetId"]).iterate_items(): results.append(item) results
Hello,
Unfortunately we cannot give you code level support. However you need to use other actor if you want to fetch locations https://apify.com/apidojo/instagram-location-scraper
Cheers
Actor Metrics
164 monthly users
-
31 stars
>99% runs succeeded
6.5 hours response time
Created in Dec 2023
Modified 9 hours ago