How to insert an image by parameter in the url and display it on the screen

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I need the code to check if this parameter with the image exists, if it exists, I have to display it on the screen. How can I do?

Note: The parameter with the image should be passed by the user in the url. That is, with GET.

  • @Jorgeb.Actually this does not answer my question, my question is different.

  • Oh I misunderstood the question. When you talk about passing this parameter with the image you mean with the right image link?

2 answers

6


if (isset($_REQUEST['url_da_imagem']))
{
    echo '<img src="'.$_REQUEST['url_da_imagem'].'">';
}

In place of $_REQUESTyou can use $_GET, is that $_REQUEST works for both GET and POST, so if you need it you can receive the URL by POST as well, without having to change your code.

Edit:

Ah, the URL would look like this: http://seudomain.com?url_da_imagem=http://dominiodaimagem.com/imagem.jpg

  • And to see if $_GET exists? I’m trying to use : <?php if(isset($_GET['image'])) { echo 'existe'; }Else ː echo 'does not exist; } ? > But the site breaks...

  • Define me better as "break". And whenever you post code, use markups to make it easier to read. Place the piece of code in the middle of the text.

  • The screen turns white, and taking the if back to work

  • I edited the answer with a verification code ready, but if you prefer to change to $_GET, do not forget to change as much in condition as in block.

  • Thanks brother, thank you very much!

3

I imagine Curl can help.

$foto = $_GET['imagem_url'];

$ch = curl_init($foto);
curl_exec($ch);
$HTTPCode = curl_getinfo($ch, CURLINFO_HTTP_CODE);
curl_close($ch);

if($HTTPCode == '200'){
    echo '<img src=".'$foto.'">';
}
else 
    echo 'Imagem não encontrada';

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