Web Applications Stack Exchange is a question and answer site for power users of web applications. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can I automatically retrieve the exact URL of a Flickr photo when I pass the ID to the API?

I have the IDs of all pictures and I'm currently using the flickr.photos.getInfo method that takes the ID as parameter and it gives me an XML file with the picture page URL as an answer.

However the picture page URL is not the URL of the picture (which is displayed in this page).

How can I automatically retrieve the exact URL of the picture?

share|improve this question
up vote 2 down vote accepted

From the Official Flickr documentation:

Photo Source URLs

You can construct the source URL to a photo once you know its ID, server ID, farm ID and secret, as returned by many API methods.

The URL takes the following format:

http://farm{farm-id}.static.flickr.com/{server-id}/{id}_{secret}.jpg
or
http://farm{farm-id}.static.flickr.com/{server-id}/{id}_{secret}_[mstzb].jpg
or
http://farm{farm-id}.static.flickr.com/{server-id}/{id}_{o-secret}_o.(jpg|gif|png)

share|improve this answer
    
The answer says "once you know its ID, server ID, farm ID and secret". So my question is how do you find those? – user10433 Apr 25 '11 at 7:07

If you have your pictures in a Photoset then you can use this method: flickr.photosets.getPhotos

What you'll need is API key, Secret key and the ID of the photoset which can be found in its URL.

share|improve this answer
1  
Thanks for your suggestion, but please don't rely on the link being around forever. Without the link there isn't any useful information here, and linkrot is a fact of life. Please at least summarize the information to be found there. – Al E. Nov 6 '15 at 16:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.