PHP: Check image exists on URL

Question

I am getting a image source from a URL dynamically, but sometimes the URL returns empty or the image does not exist so i created a PHP function to check if the image source do exist or not so i can display a default image to my users when the image is not accessible.

My concern, is my function enough to catch the data?

My current php function which works when its sure that a image do exist.

function get_image_by_id($ID) {

  $url = 'http://proweb/process/img/'.$ID.'.jpg'; 

  if(empty($url)){

   return 'default.jpg';

   }else{

    return $url;

  }

}

My index.php

<img src="<?php echo get_image_by_id($ID);?>" />

Your empty check seems redundant

Rotimi
– Rotimi

2017-11-03 05:13:41 +00:00
Commented Nov 3, 2017 at 5:13 — Rotimi
– Rotimi, Commented Nov 3, 2017 at 5:13

Sahil Gulati · Accepted Answer · 2017-11-03 05:31:56Z

2

Here CURL can also be helpful, tested it and working fine. Here we are using curl to get HTTP response code for the URL. If status code is 200 it means it exists.

function get_image_by_id($ID)
{
    $url = 'http://proweb/process/img/' . $ID . '.jpg';
    $ch = curl_init($url);
    curl_setopt($ch, CURLOPT_HEADER, true);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    curl_exec($ch);
    $info = curl_getinfo($ch);
    if ($info["http_code"] != 200)
    {
        return 'default.jpg';
    }
    else
    {
        return $url;
    }
}

edited Nov 3, 2017 at 5:31

answered Nov 3, 2017 at 5:18

Sahil Gulati

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Comments

Rotimi · Accepted Answer · 2017-11-03 05:15:44Z

1

You can use php is_file()to determine.

if (is_file($url) {
//its a file 
   }

You can use a ternary operator(one liner)

    return is_file($url) ? $url : 'default.jpg';

answered Nov 3, 2017 at 5:15

Rotimi

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1 Comment

Rotimi Over a year ago

It does work. Try accessing the URL from the browser and see what it returns

Jinesh · Accepted Answer · 2017-11-03 05:26:13Z

0

you can use like this

<?PHP 
function get_image_by_id($ID) {

  $url = 'http://proweb/process/img/'.$ID.'.jpg';
// Checks to see if the reomte server is running 
       if(file_get_contents(url) !== NULL) 
        { 
            return  $url; 
        } 
       else
        { 
             return 'default.jpg';
        } 
}
?>

answered Nov 3, 2017 at 5:26

Jinesh

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Comments

hasnain · Accepted Answer · 2017-11-03 05:15:33Z

-1

function get_image_by_id($ID) {
 if(empty($url)){
$url = 'http://proweb/process/img/default.jpg; 
}else{
$url = 'http://proweb/process/'.$ID.'.jpg'; 
 }
return $url
}

answered Nov 3, 2017 at 5:15

hasnain

1271 silver badge10 bronze badges

1 Comment

mickmackusa Over a year ago

This answer has a syntax error, no explanation, scope issues, and is poorly formatted. It adds no value to the page. Please delete this answer.

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