An image created by PHP has no extension and is not treated as an image
I am loading an image from a remote server (google for example), but PHP never treats it as an image. This is my code:
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
$filename = $photo["name"];
$ext = strtolower(pathinfo($filename, PATHINFO_EXTENSION)); //returns ""
$filepath = $photo["tmp_name"];
is_array(getimagesize($filepath); //returns false
What am I doing wrong?
php
asked Nov 24 '18 at 13:28
ruler23ruler23
3914
add a comment |
I am loading an image from a remote server (google for example), but PHP never treats it as an image. This is my code:
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
$filename = $photo["name"];
$ext = strtolower(pathinfo($filename, PATHINFO_EXTENSION)); //returns ""
$filepath = $photo["tmp_name"];
is_array(getimagesize($filepath); //returns false
What am I doing wrong?
php
asked Nov 24 '18 at 13:28
ruler23ruler23
3914
1
Maybe addvar_dump()after each line to see in which line we have a flaw.
– Colin Cline
Nov 24 '18 at 13:31
@ColinCline var_dump($photo) returns "resource(9) of type (gd)", $photo['name'] and ['tmp_name'] are NULL
– ruler23
Nov 24 '18 at 13:38
add a comment |
I am loading an image from a remote server (google for example), but PHP never treats it as an image. This is my code:
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
$filename = $photo["name"];
$ext = strtolower(pathinfo($filename, PATHINFO_EXTENSION)); //returns ""
$filepath = $photo["tmp_name"];
is_array(getimagesize($filepath); //returns false
What am I doing wrong?
php
asked Nov 24 '18 at 13:28
ruler23ruler23
3914
I am loading an image from a remote server (google for example), but PHP never treats it as an image. This is my code:
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
$filename = $photo["name"];
$ext = strtolower(pathinfo($filename, PATHINFO_EXTENSION)); //returns ""
$filepath = $photo["tmp_name"];
is_array(getimagesize($filepath); //returns false
What am I doing wrong?
php
php
asked Nov 24 '18 at 13:28
ruler23ruler23
3914
asked Nov 24 '18 at 13:28
ruler23ruler23
3914
asked Nov 24 '18 at 13:28
ruler23ruler23
3914
asked Nov 24 '18 at 13:28
ruler23ruler23
3914
asked Nov 24 '18 at 13:28
ruler23ruler23
3914
3914
1
Maybe addvar_dump()after each line to see in which line we have a flaw.
– Colin Cline
Nov 24 '18 at 13:31
@ColinCline var_dump($photo) returns "resource(9) of type (gd)", $photo['name'] and ['tmp_name'] are NULL
– ruler23
Nov 24 '18 at 13:38
add a comment |
1
Maybe addvar_dump()after each line to see in which line we have a flaw.
– Colin Cline
Nov 24 '18 at 13:31
@ColinCline var_dump($photo) returns "resource(9) of type (gd)", $photo['name'] and ['tmp_name'] are NULL
– ruler23
Nov 24 '18 at 13:38
1
1
Maybe add
var_dump() after each line to see in which line we have a flaw.– Colin Cline
Nov 24 '18 at 13:31
Maybe add
var_dump() after each line to see in which line we have a flaw.– Colin Cline
Nov 24 '18 at 13:31
@ColinCline var_dump($photo) returns "resource(9) of type (gd)", $photo['name'] and ['tmp_name'] are NULL
– ruler23
Nov 24 '18 at 13:38
@ColinCline var_dump($photo) returns "resource(9) of type (gd)", $photo['name'] and ['tmp_name'] are NULL
– ruler23
Nov 24 '18 at 13:38
add a comment |
1 Answer
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imagecreatefromstring returns an image not array.
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
ob_end_clean();
header('Content-type: image/jpeg');
imagejpeg($photo, null, 80);
If this code doesn't produce an image then start debugging from finding out the return value of file_get_contents().
Also - after creating an image (using any of imagecreatefrom... functions) you are dealing with image resource, not file system or any other object. Image resource does not have information you are trying to extract (name, pathinfo etc.)
answered Nov 24 '18 at 13:58
Sven LiivakSven Liivak
99229
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
imagecreatefromstring returns an image not array.
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
ob_end_clean();
header('Content-type: image/jpeg');
imagejpeg($photo, null, 80);
If this code doesn't produce an image then start debugging from finding out the return value of file_get_contents().
Also - after creating an image (using any of imagecreatefrom... functions) you are dealing with image resource, not file system or any other object. Image resource does not have information you are trying to extract (name, pathinfo etc.)
answered Nov 24 '18 at 13:58
Sven LiivakSven Liivak
99229
add a comment |
imagecreatefromstring returns an image not array.
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
ob_end_clean();
header('Content-type: image/jpeg');
imagejpeg($photo, null, 80);
If this code doesn't produce an image then start debugging from finding out the return value of file_get_contents().
Also - after creating an image (using any of imagecreatefrom... functions) you are dealing with image resource, not file system or any other object. Image resource does not have information you are trying to extract (name, pathinfo etc.)
answered Nov 24 '18 at 13:58
Sven LiivakSven Liivak
99229
add a comment |
imagecreatefromstring returns an image not array.
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
ob_end_clean();
header('Content-type: image/jpeg');
imagejpeg($photo, null, 80);
If this code doesn't produce an image then start debugging from finding out the return value of file_get_contents().
Also - after creating an image (using any of imagecreatefrom... functions) you are dealing with image resource, not file system or any other object. Image resource does not have information you are trying to extract (name, pathinfo etc.)
answered Nov 24 '18 at 13:58
Sven LiivakSven Liivak
99229
imagecreatefromstring returns an image not array.
$photo = imagecreatefromstring(file_get_contents("https://some.com/image.jpg"));
ob_end_clean();
header('Content-type: image/jpeg');
imagejpeg($photo, null, 80);
If this code doesn't produce an image then start debugging from finding out the return value of file_get_contents().
Also - after creating an image (using any of imagecreatefrom... functions) you are dealing with image resource, not file system or any other object. Image resource does not have information you are trying to extract (name, pathinfo etc.)
answered Nov 24 '18 at 13:58
Sven LiivakSven Liivak
99229
edited Nov 24 '18 at 14:19
answered Nov 24 '18 at 13:58
Sven LiivakSven Liivak
99229
answered Nov 24 '18 at 13:58
Sven LiivakSven Liivak
99229
answered Nov 24 '18 at 13:58
Sven LiivakSven Liivak
99229
99229
add a comment |
add a comment |
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1
Maybe add
var_dump()after each line to see in which line we have a flaw.– Colin Cline
Nov 24 '18 at 13:31
@ColinCline var_dump($photo) returns "resource(9) of type (gd)", $photo['name'] and ['tmp_name'] are NULL
– ruler23
Nov 24 '18 at 13:38