检查Cookie，如果它不存在运行脚本

Question

我很努力在html页面上使用Cookie。我想有一个脚本检查，看看如果用户有一个cookie，如果不把这个脚本打开一个示范窗口

<script type="text/javascript"> 
jQuery(document).ready(function() { 
    jQuery("#myModal").reveal(); 
}); 
</script 

该窗口包含“是”和“否”按钮，如果用户选择“是”它会保存一个cookie，以便在将来访问该页面时不会打开该模式，如果用户选择“否”，它将简单地引导他们到另一个页面。

任何帮助将不胜感激！

Answer 1

1）使用jQuery cookie库以方便使用。 https://github.com/carhartl/jquery-cookie
2）检查cookie存在并处理相应

if ($j.cookie("NameOfYourCookie") == null) { 
//Do stuff if cookie doesn't exist like set a cookie with a value of 1 
$j.cookie("NameOfYourCookie", 1, {expires: 10, path:'/'}); 
} 

else { 
//They have a cookie so do something 
alert ('You have a cookie with name NameOfYourCookie'); 
} 

*编辑*

想我可以编辑我的答案是更适合您的问题。

 <button class="yes">YES</button> 
     <button class="no">No</button> 

    var $j = jQuery.noConflict(); 

$j(document).ready(function(){ 
//Show modal on page load   
$j("#myModal").show(); 



$j('.yes').click(function(){ 
     $j("#myModal").hide(); 
     $j.cookie("YesButtonCookie", 1, {expires: 10, path:'/'}); 
     alert ('You have clicked yes, a cookie has been dropped'); 

    }); 


     if ($j.cookie("YesButtonCookie") == 1){ 
      // Don't show the div 
      $j("#myModal").hide(); 
      alert ('Cookie found, you cant see myModal'); 
     } 


     if ($j.cookie("YesButtonCookie") == null){ 
        // Cookie Not Found 
        alert ('Cookie Not found, you can see myModal'); 
        $j("#myModal").show(); 
     } 




    $j('.clearcookie').click(function(){ 
     $j.cookie("YesButtonCookie", null); 
     alert('Cookie Cleared'); 
    }); 


}); 

JSFIDDLE TO TEST