IMAGE COMPRESSOR Script [ Notice: Only variables should be passed by reference in ]

Question

I tried to take it off this error I do not understand why it happens, because I am uploading the file correctly... Right now is uploading the files, it shows me the error Error Notice: Only variables should be passed by reference in...

Notice: Only variables should be passed by reference in D:\xampp\htdocs\reduce-image\upload.php on line 10 ** I am facing in upload.php code. It's showing this error:

**

```   
**upload.php**
<pre>
<?php 
    $file_type_error = '';
    if($_FILES['upload_images']['name']) {    
        $upload_dir = "uploads/";   
        if (($_FILES["upload_images"]["type"] == "image/gif") ||
           ($_FILES["upload_images"]["type"] == "image/jpeg") ||
           ($_FILES["upload_images"]["type"] == "image/png") ||
           ($_FILES["upload_images"]["type"] == "image/pjpeg")) {
            $file_name = $_FILES["upload_images"]["name"];
            $extension = end((explode(".", $file_name)));
            $upload_file = $upload_dir.$file_name;      
            if(move_uploaded_file($_FILES['upload_images']['tmp_name'],$upload_file)){            
                 $source_image = $upload_file;
                 $image_destination = $upload_dir."min-".$file_name;
                 $compress_images = compressImage($source_image, $image_destination);            
            }        
        } else {
            $file_type_error = "Upload only jpg or gif or png file type";
        }   
    }
    // created compressed JPEG file from source file
    function compressImage($source_image, $compress_image) {
        $image_info = getimagesize($source_image);  
        if ($image_info['mime'] == 'image/jpeg') { 
            $source_image = imagecreatefromjpeg($source_image);
            imagejpeg($source_image, $compress_image, 75);
        } elseif ($image_info['mime'] == 'image/gif') {
            $source_image = imagecreatefromgif($source_image);
            imagegif($source_image, $compress_image, 75);
        } elseif ($image_info['mime'] == 'image/png') {
            $source_image = imagecreatefrompng($source_image);
            imagepng($source_image, $compress_image, 6);
        }       
        return $compress_image;
    }
?>
</pre> ```

  <pre>
<?php 
include('header.php');
?>
<script type="text/javascript" src="scripts/jquery.form.js"></script>
<script type="text/javascript" src="scripts/upload.js"></script>
<link type="text/css" rel="stylesheet" href="style.css" />

<link type="text/css" rel="stylesheet" href="style.css" />
<div class="container">
 <h2>Reduce or Compress Image Size While Uploading in PHP</h2>       
 <form method="post" name="upload_form" id="upload_form" enctype="multipart/form-data" action="upload.php">   
 <label>Choose Images to Upload</label>  
 <input type="file" name="upload_images" id="image_file">
 <div class="file_uploading hidden">
     <label> </label>
     <img src="uploading.gif" alt="Uploading......"/>
 </div>
 </form>
 <div id="uploaded_images_preview">
 </div>  
</div>
<?php include('footer.php');?>
</pre>`





**js.file**
<pre>
$(document).ready(function(){
 $('#image_file').on('change',function(){
     $('#upload_form').ajaxForm({           
         target:'#uploaded_images_preview',
         beforeSubmit:function(e){
             $('.file_uploading').show();
         },
         success:function(e){
             $('.file_uploading').hide();
         },
         error:function(e){
         }
     }).submit();
 });
});
</pre> ```

score 0 · Answer 1 · answered May 07 '21 at 16:55

From the documentation of end

This array is passed by reference because it is modified by the function. This means you must pass it a real variable and not a function returning an array because only actual variables may be passed by reference.

However you are passing it a function that returns an array.

Try

$parts = explode(".", $file_name);
$extension = end($parts);

IMAGE COMPRESSOR Script [ Notice: Only variables should be passed by reference in ]

1 Answers1