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Home/ Questions/Q 7965063
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T06:03:20+00:00 2026-06-04T06:03:20+00:00

I would like to add an image in case a URL is present/active. Ive

  • 0

I would like to add an image in case a URL is present/active. Ive tried many things myself to make it work but without success.

What I want is: if url1 exist show image1 else nothing, if url2 exist show image2 else nothing, etc.

foreach ($urlsArray as $url) {
  if(@fopen($url,'r')){
    echo 'Exist';
  }
  else {
      echo 'Doesnt exist';
  }
}

Sorry I have to be more specific!

$urlArray = array(
'http://www.domain.com/page1.php' => 'images/image1.jpg',
'http://www.domain.com/page2.php' => 'images/image2.jpg', 
etc);

foreach($urlsArray as $url){
    if(@fopen($url,'r')){
        echo '<img src="$image" />' /* if url1 exist show image1, etc */
    }else{
        echo '';  
    }
}

How to get this to work?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T06:03:21+00:00Added an answer on June 4, 2026 at 6:03 am

    assuming that $url = path/to/pic/pic.gif – meaning that url actually leads to a pic to be displayed

    foreach ($urlsArray as $url) {
      if(@fopen($url,'r')){
        echo '<img src="'.$url.'" />';
      }
    }
    

    updated (use $key => $value for your foreach):

    foreach ($urlsArray as $url => $img)  {
      if(@fopen($url,'r')){
        echo '<img src="'.$img.'" />';
      }
    }
    
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