当第一个没有加载php时，如何加载辅助img

I try to load a other img when the first one is not loading.

<?php
$id2 = intval($_GET['id']);
$jpg = ".jpg";
$id = $id2 . $jpg;

$imagepath= $id;
$image=imagecreatefromjpeg( $imagepath);


header('Content-Type: image/jpeg');
imagejpeg($image);
?>

this loads my img when i have the id in /img

now i want to load a img with the same id when $imgsize = filesize( $id ); is 0.

from a external link like: http://xxxx.xx/getimg.php?id=28771

it should be easy with if else but i can't get it.

edit:

$id2 = intval($_GET['id']);
$jpg = ".jpg";
$id = $path . $id2 . $jpg;
if(isset($_GET['id']) && $_GET['id']!=0){
    $id2 = intval($_GET['id']);
}
else{
    $id2 = 1;
}

$imagepath= $id;
$image=imagecreatefromjpeg( $imagepath);
header('Content-Type: image/jpeg');
imagejpeg($image);

so this is the code I use now.

Try something like this considering that 1.jpg is a default file in your imagepath folder

<?php
if(isset($_GET['id']) && $_GET['id']!=0){
    $id2 = intval($_GET['id']);
}
else{
    $id2 = 1;
}
?>

Something like this might solve your problem.This is just and idea you can modify the loops as per your need but i hope you got the overall idea.

if(isset($_GET['id']) && $_GET['id']!=0){ 
   $id2 = intval($_GET['id']); 
 } else{ 
   $id2 = 1; 
 } 

$jpg = ".jpg"; 
$id = $path . $id2 . $jpg;
$imagepath= $id; 
$image=imagecreatefromjpeg( $imagepath); 
header('Content-Type: image/jpeg'); 
imagejpeg($image);

Use this code and tell me if it works or not

Another way to do it is in Javascript, there might be some scenarios where $_GET['id'] is set but the actual image is missing. This uses jquery.

$('img').error(function () {
  if ($(this).hasClass("optional_class")) {
    $(this).unbind("error").attr("src", "/no-picture.jpg");
  } else {
    $(this).remove();
  }
});