## MySQLi stmt issue

FORM

PHP Code:
``` <?php if(\$_POST['submit']){     \$section = mysqli_prep(\$_POST['table']);     \$linkid = mysqli_prep(\$_POST['linkid']);     \$title = mysqli_prep(\$_POST['title']);     \$des = mysqli_prep(\$_POST['des']);     \$date = mysqli_prep(\$_POST['date']);     if(isset(\$_POST['facebook'])){\$facebook = mysqli_prep(\$_POST['facebook']); } else { \$facebook = NULL; }          \$sql = "INSERT INTO ? (             `linkid`, `title`, `description`, `date`, `facebook`             ) VALUES (             ?, ?, ?, ?, ?             )";     \$stmt = mysqli_stmt_init(\$connect);     mysqli_stmt_prepare(\$stmt, \$sql);     mysqli_stmt_bind_param(\$stmt, 'sissss', \$section, \$linkid, \$title, \$des, \$date, \$facebook);     if(mysqli_stmt_execute(\$stmt)){         // Success!         \$display_message = "<h6 class=\"displaymessage\">Ministry news article created successfully!</h6>\n";     }else{         // Failed!         \$display_message = "<h6 class=\"displaymessage\">Ministry news article creation failed.</h6>\n";         \$display_message = "<h6 class=\"displaymessage\">".mysqli_error(\$connect)."</h6>\n";     }     mysqli_stmt_close(\$stmt); } ?> ```
Returns...

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '? ( `linkid`, `title`, `description`, `date`, `facebook` ) VALUES ( ?, ' at line 1
Is it possible that the selected table cannot be a (?) value? I have done this no problem with basic sql queries in the past...