I have a twitter feed and I can find the instagram url in a tweet.
{
indices: [
90,
110
],
url: "http://t.co/kF9EXifn",
expanded_url: "http://instagr.am/p/QC8hWKL_4K/",
display_url: "instagr.am/p/QC8hWKL_4K/"
}
What I need is to take the expanded url from that object "http://instagr.am/p/QC8hWKL_4K/" and get the link to just the image itself, not the image on the Instagram site. Anyone know how to do that? Can I even derive the that link from the url?
You can get just the image by appending /media/
to the URL. Using your example: http://instagr.am/p/QC8hWKL_4K/media/.
You can even specify a size,
One of t (thumbnail), m (medium), l (large). Defaults to m.
So for a thumbnail: http://instagr.am/p/QC8hWKL_4K/media/?size=t
Documentation here.