View mysql data using PHP/Jquery with confirmation dialog box

I have a question about jQuery UI Dialog box and showing dynamic content from a database. Here I have a table which is generating blog post using php and mysql and in that table, there is a column to view contents which are belong to each blog post.

That link is something like this -

$html .= "  <td align='center'>\n";
$html .= "      <a href='#' id='blog-$blog_id' class='view' >\n";
$html .= "          <span class='icon-small ico-view-blog' title='View This Blog Post'></span>\n";
$html .= "      </a>\n";
$html .= "  </td>\n";

Clicking on above link I need to pop-up a jquery dialog to display all blog content. Eg: blog-title, author, image, blog etc.

I tried it with jqury and using separate php script to fetch blog contets like this. But it is not pop-up the dialog as I expect.

This is jquery I have used for the dialog -

$( "#dialog-view-blog" ).dialog({
        autoOpen: false,
        height: 450,
        width: 650,
        modal: true,
        buttons: {
            Cancel: function() {
            $( this ).dialog( "close" );
            }
        }, 
        position: { 
            my: "center top", 
            at: "center top",
            of: "#content"
        }
    });

This is how I send a ajax request for the data from the php file to update the content in the dialog -

$( "a.view" ).click(function(e) {
    e.preventDefault();     
    var clickblogID = this.id.split('-'); //Split string 
    var DbNumberID = clickedID[1]; //and get number from array
    var blogId = 'blog_id='+ DbNumberID; //build a post data structure  
    $.ajax({
        url: 'update_blog.php',
        type: 'POST',
        data: blogId,
        success: function(data){

            //alert(data);

            //construct the data however, update the HTML of the popup div 
            //$('#dialog-view-blog').html(data);
            $('#dialog-view-blog').dialog('open');
        }
    });             
}); 

My code from update_blog.php page

if (isset($_POST['blog_id'])) { 
    //blog_id 
    $blogId = $_POST['blog_id'];

    // If there is no any blog to this user display a string. 
    $q = "SELECT * FROM userblogs WHERE blog_id = ?";
    // Prepare the statement:
    $stmt = mysqli_prepare($dbc, $q);
    // Bind the variables:
    mysqli_stmt_bind_param($stmt, 'i', $blogId);                            
    // Execute the query:
    mysqli_stmt_execute($stmt); 
    //store result  
    mysqli_stmt_store_result($stmt); 
    // Get the number of rows returned: 
    $rows = mysqli_stmt_num_rows ($stmt);

    if ( $rows == 1 ) { 
        $viewBlog  = "<div id='dialog-view-blog' title='View Blogs'>\n";
        $viewBlog .= "      <h2>$blog_title</h2>\n";
        $viewBlog .= "  <p>$blog_author | $blog_added_date</p>\n";
        $viewBlog .= "  <p>";
        $viewBlog .= "          <img src='".UPLOAD_DIR.$userName."/".$blog_image."' alt='Image for Blog Title' />";
        $viewBlog .= "      $blog</p>";
        $viewBlog .= "</div>\n";        

        echo $viewBlog;
    } 

Can anybody pointed me where I have gone wrong? Any comments are greatly appreciated.

Thank you.