PHP - Read an image with URL

Question

Here my problem, I do a query to MySQL (PDO) for give me the last 5 URLs of a table nammed avatar who contains the ID and the URL :

$response = $dbh->query("SELECT url FROM avatar ORDER BY id_URL DESC LIMIT 0,5 ");

And I did :

  while ($donnees = $response->fetch())
{

$urlImage = $donnees['url']; //'url' contains the URL
$result = file_get_contents($urlImage);        
header('Content-Type: image/png');
echo $result;   
?>

But the header just return a small empty white square. However, the "$result = file_get_contents($urlImage);" takes properly the URL because when I do :

   $urlImage = $donnees['url']; //'url' contains the URL
$result = file_get_contents($urlImage);        
echo $result;   
?>

It just shows the "encodage of the image" (a ton of special characters) but not displays the image.

I also try with "imagick" but it says to me that the class doesn't exist and I don't think that the imagecreatefrompng can be use with URL.

Thanks !

@Hexadect Yes. The first integer is simply the offset. So LIMIT 10,5 would fetch 5 records, with an offset of 10. LIMIT 5 OFFSET 0 == LIMIT 0,5 == LIMIT 5. — BenM
– BenM, Commented Feb 14, 2017 at 19:04
@Hexadect Yes, the "LIMIT 0, 5" is here to indicate that the limit is between 0 and 5 but I think it's need to be reversed with the comma. There is an example in W3Schools. — Lafdoma
– Lafdoma, Commented Feb 14, 2017 at 19:12
@Mofallo999 Yes I know. I'm saying that LIMIT 0,5 is the same as LIMIT 5. — BenM
– BenM, Commented Feb 14, 2017 at 19:14

Raj Nandan Sharma · Accepted Answer · 2017-02-15 06:07:39Z

3

Can you try this and see if it works?

 $image = file_get_contents($donnees['url']);
 $finfo = new finfo(FILEINFO_MIME_TYPE);
 header('content-type: ' . $finfo->buffer($image));
 echo $image;

This is suppose to handle one Image. One php script can return one image. If you want to combine the images and render a big long image then probably you should look at http://image.intervention.io/

EDIT

What I understood after trying out the above code is that if you put file_get_contents before header then the raw characters are shown. However you put it after header then everything seems to be working

$image="http://www.hillspet.com/HillsPetUS/v1/portal/en/us/cat-care/images/HP_PCC_md_0130_cat53.jpg";
$filename = basename($image);
$file_extension = strtolower(substr(strrchr($filename,"."),1));
switch( $file_extension ) {
    case "gif": $ctype="image/gif"; break;
    case "png": $ctype="image/png"; break;
    case "jpeg":
    case "jpg": $ctype="image/jpeg"; break;
    default:
}

header('Content-type: ' . $ctype);
$image = file_get_contents($image);
echo $image;

Working fiddle

edited Feb 15, 2017 at 6:07

answered Feb 14, 2017 at 19:11

Raj Nandan Sharma

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6 Comments

Lafdoma Over a year ago

No, I want to return one image, this is a loop so the variable will always render one image ? But that's the same that's my code, it's return meta data of the image.

Raj Nandan Sharma Over a year ago

You have written the header and echo inside the loop. That means it will echo it 5 times. So even though the client side thinks it is an image but it cannot render it as it is some combination for 5 image urls. Try doing limit 0,1

Lafdoma Over a year ago

Ah, yes I understand what you say, but I make the Limit 0,1 and remove the loop, that's give the same result.

Lafdoma Over a year ago

Hi, I just copy paste your code but this just returns a small empty white square.

Raj Nandan Sharma Over a year ago

@Mofallo999 please try out the demo.. the picture is coming fine. Also note that your url cannot have https, so replace that with http (if working with http) using $image = str_replace("https", "http", $image); after the first line

|

Altimus Prime · Accepted Answer · 2017-02-14 19:26:39Z

1

If you're trying to use a dynamic image source where your url is the image source, and it isn't working, then the problem might be that there's a space or extra character somewhere on the page, which will make the browser treat it like a document instead of an image in some cases.

Your problem is that the browser isn't understanding that it's supposed to be an image.

You could always do:

<img src="<?=$urlImage?>">

edited Feb 14, 2017 at 19:26

answered Feb 14, 2017 at 19:01

Altimus Prime

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4 Comments

BenM Over a year ago

1. <img /> is a self-closing tag. 2. That's the purpose of header('Content-Type: image/png');.

Altimus Prime Over a year ago

@BenM. I come on and help someone where you couldn't and you give me a downvote?

BenM Over a year ago

@AuntJamaima I downvoted because this doesn't actually solve the problem. Hotlinking to an image inside of <img /> is not the same objective as the OP posted. They wish to output the image directly from a PHP script; those are two distinct problem sets.

Altimus Prime Over a year ago

@BenM. Where does he say he wants to output directly from the script? He's running a while loop and said he tried imagick and couldn't make it display. This is a case of server client confusion. Case in point he already said that this works in his comment above.

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