$file = isset($_FILES['fileupload']) ? $_FILES['fileupload']['tmp_name'] : false;
$image = chunk_split(base64_encode(file_get_contents($_FILES['fileupload']['tmp_name']))); //store image in $image
//need help here nothing prints
$exif_ifd0 = read_exif_data($image,'IFD0' ,0);
$notFound = "Unavailable";
if (@array_key_exists('Make', $image))
{
$camMake = $image['Make'];
}
else
{
$camMake = $notFound;
}
echo"$camMake";
hwoarang69
11
Newbie Poster
Recommended Answers
Jump to PostThis works for me:
<?php $img = "img2.jpg"; $exif = exif_read_data($img,'IFD0'); echo $exif['Make']; ?>
I think you don't need to do all this:
$image = chunk_split(base64_encode(file_get_contents($_FILES['fileupload']['tmp_name']))); //store image in $image
at least not for the exif_read_data() function, just provide the name and …
All 3 Replies
cereal
1,524
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hwoarang69
11
Newbie Poster
cereal
1,524
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