how to display multiple images in php from database?

Question

I used the following code to upload the images. It works well. But I dont know how to display all the images in a display page.

Please help for the code for display.

    <?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';
    }       
    $query="INSERT into upload_data (`USER_ID`,`FILE_NAME`,`FILE_SIZE`,`FILE_TYPE`) 
            VALUES('$user_id','$file_name','$file_size','$file_type'); ";

    $desired_dir="gallery";
    if(empty($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($file_tmp,"gallery/".$file_name);
        }else{                                  //rename the file if another one exist
            $new_dir="gallery/".$file_name.time();
             rename($file_tmp,$new_dir) ;               
        }
        mysql_query($query) or die(mysql_error());          
    }else{
            print_r($errors);
    }
 }
if(empty($error)){
    echo "Success";
}
}
?>

Before you write any more SQL code, you must read up on proper SQL escaping to avoid serious SQL escaping bugs. You cannot use string interpolation to compose queries. It's also a bad idea to be using mysql_query as it's deprecated and will be removed from future versions of PHP. A modern replacement like PDO is not hard to learn. — tadman
– tadman, Commented Jul 2, 2013 at 13:55

Matheno · Accepted Answer · 2013-07-02 13:48:15Z

5

You can show images something like this:

$sql="select * from tbl_image";
$query=mysql_query($sql);
while($row=mysql_fetch_array($query))
{
$image=$row ['photo'];

echo '<img src="path/'.$image.'" width="360" height="150">';

}

answered Jul 2, 2013 at 13:48

Matheno

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4 Comments

Eric Over a year ago

What's up with all the error-suppressing? Terrible for debugging

Twisted1919 Over a year ago

the fact you have edited your code doesn't mean much. Is there a reason behind the fact you are suppressing even a variable declaration with the @ sign? Are you aware of it's purpose?

Matheno Over a year ago

Yes I know the purpose of it. It temporarily sets the error reporting level to 0 for that line. If that line triggers an error, the error handler will still be called, but it will be called with an error level of 0. I sometimes use it while testing some scripts, but it's better to not show it him that way. You guys are right.

SS Nath Over a year ago

It dos n't shows any error. But image not displayed. Image border displays for all images, but image shows like unavailable or broken image.

Nitesh Mishra · Accepted Answer · 2013-07-02 13:58:09Z

1

Try this:

$result= //select query for retrieving filenames for a particular userid 
         //or all depending upon your requirement.

while($row=mysql_fetch_array($result))
{
 echo "<div><img src=\"gallery/".$row['filename']."\" /></div>"
}

answered Jul 2, 2013 at 13:58

Nitesh Mishra

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