將jquery翻轉狀態保存到php會話變量中？

Question

我正在使用本教程http://www.w3schools.com/jquery/tryit.asp?filename=tryjquery_slide_toggle

這是沒有風格的代碼。

<script type="text/javascript"> 
$(document).ready(function() 
{ 


$(".flip").click(function() 
    { 
    var panel = "open"; 
    $(".panel").slideToggle("slow"); 
    }); 
}); 
</script> 

我將如何去保存這個狀態，所以如果我刷新頁面，它將保持打開或關閉狀態。我想象一個PHP會話將是正確的方式，但我怎麼寫在JavaScript？

Answer 1

在JS：

var readWirite='write'; //or 'read' 
$.ajax({ 
    type: "POST", 
    url: "myPhpFile.php", 
    data: "panel="+panel+"&readWrite="+readWrite; 
    success: function(msg){ 
     if(msg == '1'){ 
     alert('Horay panel saved!'); 
     } else { 
     $('#panelId').html(msg); //Write saved panel back to html 
     } 
    }      
}); 

在myPhpFile.php：

<?php 
    if(!isset($_SESSION)) session_start(); 
    if(isset($_POST['readWrite']) && isset($_POST['panel'])){ 
     if($_POST['readWrite'] == 'write'){ 
      $result = '0'; 
      if($_SESSION['panel'] = $_POST['panel']) $result = '1'; 
      echo $result; 
     } else if($_POST['readWrite'] == 'read') { 
      echo $_SESSION['panel']; 
     } 
    } 
?>