提交表单时，Post方法的行为类似于get方法

Question

//db connect
    $server     = 'localhost';
    $username   = 'root';
    $password   = '';
    $database   = 'resort';
    $connect    = mysqli_connect($server, $username, $password ) or die(mysqli_error.'error connecting to db');
//select database
    mysqli_select_db ($database, $connect) or die(mysqli_error.'error selecting db');/*Jquery Function*/<script>
                    $(function(){
    //CONTACT FORM AJAX SUBMIT
    $('#contactForm').submit(function(){
          $.ajax({
        url:'mailer.php',
        type : 'POST',
        dataType: 'json',
        data: $(this).serialize(),
           success: function(data){
         if(data.error){
            $('#error').css('display','block');
        }else {
            $('#note').show();
            $('#error').hide();
            $(".contact-form").hide();
        }
    }
   });
   return false;
 });
});
                    </script>

issue is that when i am trying to click on submit the webpage refreshes and the contents are shown in the url. Errors does not occur so no way to find out what the problem is. I am trying to build a contact form that can send mail. So you will also get some code related to it.The main issue is that why the page refreshes every time and why there is not any error if, database is connected or not,query written for insertion of data is correct or not, variables are correctly declared or not.

Answer 1

If your images are already in the right format then use Convert.ToBase64String Method (Byte[]) method to convert them to a base64 string and send this base64 string back.

where you can use the string in the src attribute of the img tag to display the images. Which would be something like

<img src=”data:<MIMETYPE>;base64,<BASE64_ENCODED_IMAGE>”>