Първо, трябва да задействате AJAX заявката с помощта на Javascript. Но аз ще ви напътствам с помощта на jQuery (библиотека на Javascript).
Вашият HTML:
<select name="allbooks" id="allbooks">
<option value="none" ></option>
<option value="allbooks" >All Books</option>
</select>
<div id="show">
<!-- ITEMS TO BE DISPLAYED HERE -->
</div>
След това изтеглете jQuery .
След това нека направим скрипта:
<script src="jquery-1.9.1.min.js"></script> <!-- CHANGE THE JQUERY FILE DEPENDING ON THE VERSION YOU HAVE DOWNLOADED -->
<script type="text/javascript">
$(document).ready(function(){ /* PREPARE THE SCRIPT */
$("#allbooks").change(function(){ /* WHEN YOU CHANGE AND SELECT FROM THE SELECT FIELD */
var allbooks = $(this).val(); /* GET THE VALUE OF THE SELECTED DATA */
var dataString = "allbooks="+allbooks; /* STORE THAT TO A DATA STRING */
$.ajax({ /* THEN THE AJAX CALL */
type: "POST", /* TYPE OF METHOD TO USE TO PASS THE DATA */
url: "get-data.php", /* PAGE WHERE WE WILL PASS THE DATA */
data: dataString, /* THE DATA WE WILL BE PASSING */
success: function(result){ /* GET THE TO BE RETURNED DATA */
$("#show").html(result); /* THE RETURNED DATA WILL BE SHOWN IN THIS DIV */
}
});
});
});
</script>
След това нека създадем get-data.php
който ще получи данните, изпратени чрез AJAX.
if(!empty($_POST["allbooks"])){
/* DO YOUR QUERY HERE AND GET THE OUTPUT YOU WANT */
echo $output; /* PRINT THE OUTPUT YOU WANT, IT WILL BE RETURNED TO THE ORIGINAL PAGE */
}
Можете да проверите този пример - JSfiddle .