Just add the Div and class, where u want a static image, and in css work, add image by using background:url('#'); The simplest and easiest way.

Tables are the best way to arrange data in a pretty layout and in exact form. Every website use forms to arrange data.

Create subdomain with the directory as, public_html/subdomain this will work for you and will not redirect.

Try this:

.form-inline input[type="text"], 
 .form-inline input[type="email"]{
    margin-left:.8em;
    margin-top:.4em;
    }

its working fine with this code.

Troy is right, you must use href attribute in <a href='#'> like that. now this code will work.

I think no, You have to customize your navbar directly into dreamweaver with javascript or you can use jquery.

If you are a wordpress developer, i recomment you NetBeans IDE. this is working perfect for me. have more than feature of PHP strom and Dreamweaver.

you can use this code:
first line of existing image is that image, which always fetch from database.

$existing_image = $ri['upload'];

if(isset($_FILES['upload']) && ($_FILES['upload']['size']>0)){
        if(trim($existing_image)!='1'){$photo=$existing_image;}
    else { $photo = date('dmY-his');
    $photo .="-".$_FILES['upload']['name'];

    }
    move_uploaded_file($_FILES['upload']['tmp_name'], 'profile/'.$photo);
    }
    else { $photo = $existing_image; }

Please ask your Question again and in some simple words, you want to ask about cookie storage in your local system or in browser cookies? and also just fail to view or also fail to get result back of that cookie?

Thank You for Quering.... Let see below code and reply me

$existing_image = $r['image'];

if(isset($_FILES['uploads']) && ($_FILES['uploads']['size']>0)){
        if(trim($existing_image)!='No'){$photo=$existing_image;}
        else { $photo = date('dmY-his');
        $photo .="-".$_FILES['uploads']['name'];

        }
        move_uploaded_file($_FILES['uploads']['tmp_name'], 'directory/image/'.$photo);
        }
        else { $photo = $existing_image; }

I am using here mysqli not mysqli, after this set query:

$q = "update users set image='$photo'";
$result=mysqli_query($q) or die(mysqli_error());