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Probably just missing something in the documentation, but I have been pecking at this for a while.

if I use the command

$a=$f->images_upload( "AlbumID=$albumid", "File=$filepath", "Caption='$caption'", "Heavy=1");

I seem not to be able to extract any kind of response. Can someone put me out of my misery and tell me what I need to do to $a or $f to get the response to see if my upload was successful? Tried a variety of print_r and print to analyze this, but no luck

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1 Answer

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Answering: And the answer is that the response is indeed in the variable $a above except if the upload fails, in which case the variable has no value.

Easy enough to test, might be a good thing to have in the docs. Or at least, if it's there, would have been nice to be easier to find. Silent fails can be hard. Ask any mime.

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Correct. The images_upload method should behave like all the other methods and return the response from the API. I'll check this this week and update the phpSmug docs to reflect this. – lildude Dec 30 '12 at 16:44

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