Unable to post image to twitter

Givinv following error

Error: it was not possible to retrieve the API call: 3 it was not possible to access the file php-oauth.png

The file must exist with that path in the same machine you are running PHP. Does the file exist?

Sorry ! File path was not correct.

Thanks!

Is there a way to post an image from a remote url instead of a directory on the same server?

Do you want to attach a file with contents from a remote URL, or do you want to insert the URL of that file in the Twitter post message?

$media = "image.png"; //works perfect!
$media = "http://domain/image.png"; //This is what I need!

"https://api.twitter.com/1.1/statuses/update_with_media.json",
'POST', array('status'=>$text,
'media[]'=>$media),array(
'FailOnAccessError'=>true,
'Files'=>array('media[]'=>array()
), $osid,$user);

$media = "http://domain/image.png";

You can try the latest class version using a call like this:

$success = $client->CallAPI(
"https://api.twitter.com/1.1/statuses/update_with_media.json",
'POST', array(
'status'=>'This is a test tweet to evaluate the PHP OAuth API support to upload image files sent at '.strftime("%Y-%m-%d %H:%M:%S"),
'media[]'=>file_get_contents($media)
),
array(
'FailOnAccessError'=>true,
'Files'=>array(
'media[]'=>array(
'FileName'=>'image.png'
)
)
), $upload);

Sorry you also need to set the type of parameter to 'Data'

$media = "http://domain/image.png";
$success = $client->CallAPI(
"https://api.twitter.com/1.1/statuses/update_with_media.json",
'POST', array(
'status'=>'This is a test tweet to evaluate the PHP OAuth API support to upload image files sent at '.strftime("%Y-%m-%d %H:%M:%S"),
'media[]'=>file_get_contents($media)
),
array(
'FailOnAccessError'=>true,
'Files'=>array(
'media[]'=>array(
'Type'=>'Data',
'FileName'=>'image.png'
)
)
), $upload);

I am still not able to upload the image. Can you explain what should be included in the $upload parameter? in all other examples you use the $user parameter. I do not find any reference to $upload in your API code.

As explained in the documentation, that last parameter is a variable that is passed by reference to return the response of the request.

You can name it the way you want. In that case, I called it $upload because you are uploading some data, but you may choose some other name.