How to display multiple images after uploading them?

Question

I looked up for multiple file uploading script, so I stumbled on this: http://techstream.org/Web-Development/PHP/Multiple-File-Upload-with-PHP-and-MySQL I know this is an old post, but it's working very well. I deleted SQL part because I don't need it. So as you look at script, it only uploads multiple files to server, and I tried ti display uploaded images, like this:

if(isset($_FILES['files'])){
    $errors= array();
    foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){
        $file_name = $key.$_FILES['files']['name'][$key];
        $file_size =$_FILES['files']['size'][$key];
        $file_tmp =$_FILES['files']['tmp_name'][$key];
        $file_type=$_FILES['files']['type'][$key];  
        if($file_size > 2097152){
            $errors[]='File size must be less than 2 MB';
        }
        $desired_dir="../img/full-size";
        if(empty($errors)==true){
            if(is_dir($desired_dir)==false){
                mkdir("$desired_dir", 0700);        // Create directory if it does not exist
            }
            if(is_dir("$desired_dir/".$file_name)==false){
                move_uploaded_file($file_tmp,"$desired_dir/".$file_name);
            }else{                                  // rename the file if another one exist
                $new_dir="$desired_dir/".$file_name.time();
                 rename($file_tmp,$new_dir) ;               
            }       
        }else{
                print_r($errors);
        }
    }
    if(empty($error)){
        echo "Success";
        echo "<img src='" . $desired_dir . '/' . $file_name . "'>";
        echo "<img src='" . $desired_dir . '/' . $file_name . "'>";

    }
}

HTML:

<form action="" method="POST" enctype="multipart/form-data">
    <input type="file" name="files[]" multiple/>
    <input type="submit"/>
</form>

This works, but only for one Image, not array. Any help?

Answer 1

In comments you said:

Output is just one image two times

Sure it is so, because you did this:

for each uploaded image
    move image somewhere
end for
if no errors
    show 2 same images (it will be last images shown 2 times)
end if

You do this part:

if(empty($error)){
    echo "Success";
    echo "<img src='" . $desired_dir . '/' . $file_name . "'>";
    echo "<img src='" . $desired_dir . '/' . $file_name . "'>";
}

not in for, but right after it. Your code output last uploaded image 2 times.

You need to put output in for in loop (updated code with array of images links):

$images = array();
foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){
    $file_name = $key.$_FILES['files']['name'][$key];
    $file_size =$_FILES['files']['size'][$key];
    $file_tmp =$_FILES['files']['tmp_name'][$key];
    $file_type=$_FILES['files']['type'][$key];  
    if($file_size > 2097152){
        $errors[]='File size must be less than 2 MB';
    }
    $desired_dir="../img/full-size";
    if(empty($errors)==true){
        if(is_dir($desired_dir)==false){
            mkdir("$desired_dir", 0700);        // Create directory if it does not exist
        }
        if(is_dir("$desired_dir/".$file_name)==false){
            move_uploaded_file($file_tmp,"$desired_dir/".$file_name);
        }else{                                  // rename the file if another one exist
            $new_dir="$desired_dir/".$file_name.time();
             rename($file_tmp,$new_dir) ;               
        }  
        $images[] ="<img src='" . $desired_dir . '/' . $file_name . "'>"; //  <----- there    
    }else{
            print_r($errors);
    }
}

foreach($images as $image) {
     echo $image;
}

Answer 2

I think you have to display images from your folder after uploading, so try this:

Remove this code :

if(empty($error)){
        echo "Success";
        echo "<img src='" . $desired_dir . '/' . $file_name . "'>";
        echo "<img src='" . $desired_dir . '/' . $file_name . "'>";

    }

And add this

if(empty($error)){

    $files = glob("../img/full-size/*.*");
    for ($i=1; $i<count($files); $i++)
    {

      $num = $files[$i];
      echo '<img src="'.$num.'" align="absmiddle" /> ';
      print substr(substr($num,6,100),0,-4);

    }

}

Answer 3

Just try ,

<?php 
if(isset($_FILES['files'])){
    $errors= array();
    foreach($_FILES['files']['tmp_name'] as $key => $tmp_name ){
        $file_name = $key.$_FILES['files']['name'][$key];
        $file_size =$_FILES['files']['size'][$key];
        $file_tmp =$_FILES['files']['tmp_name'][$key];
        $file_type=$_FILES['files']['type'][$key];  
        if($file_size > 2097152){
            $errors[]='File size must be less than 2 MB';
        }
        $desired_dir="ne_img";
        if(empty($errors)==true){
            if(is_dir($desired_dir)==false){
                mkdir("$desired_dir");
                chmod("$desired_dir", 0755);       // Create directory if it does not exist
            }
            if(is_dir("$desired_dir/".$file_name)==false){
                move_uploaded_file($file_tmp,"$desired_dir/".$file_name) ;
                $files_name[] = $file_name;
            }else{                                  // rename the file if another one exist
                $new_dir="$desired_dir/".$file_name.time();
                 rename($file_tmp,$new_dir) ;               
            }       
        }else{
                print_r($errors);
        }
    }
    if(empty($error)){
        if(is_array($files_name))
        {
            foreach($files_name as $file_name)
            {
                echo "<img src='" . $desired_dir . '/' . $file_name . "'>";
            }
        }


    }
}

?>