JS cookie，用于在第x次网页浏览后显示内容 [英] JS cookie for displaying content after x-th pageview

问题描述

我试图找到如何设置JavaScript cookie，以便在访问者说第4次网页浏览后显示fancybox弹出窗口.

这是我用来显示Fancybox弹出窗口的代码，但是您可能会看到，它仅在第一次浏览时显示该代码，并且一天后就会过期

function setCookie(c_name,value,exdays)
 {
     var exdate=new Date();
     exdate.setDate(exdate.getDate() + exdays);
     var c_value=escape(value) + ((exdays==null) ? "" : "; expires="+exdate.toUTCString());
     document.cookie=c_name + "=" + c_value + ";domain=.mysite.net;path=/";
 }

function getCookie(c_name)
 {
     var i,x,y,ARRcookies=document.cookie.split(";");
     for (i=0;i<ARRcookies.length;i++)
  {
     x=ARRcookies[i].substr(0,ARRcookies[i].indexOf("="));
     y=ARRcookies[i].substr(ARRcookies[i].indexOf("=")+1);
     x=x.replace(/^\s+|\s+$/g,"");
      if (x==c_name)
       {
       return unescape(y);
       }
  }
 }

jQuery(document).ready(function() {
     var show = getCookie("fancyreg");
     if(show==null || show=="") {
       $.fancybox(
 {
        'type' : 'iframe',
        'href' : 'http://www.mysite.net/popup/', //URL to popup page or image
        'autoDimensions' : false,
        'width' : 480,
        'height' : 260,
        'transitionIn' : 'none',
        'transitionOut' : 'none'
 }
);

  setCookie('fancyreg','1',1); 

 }
 });

如果您能帮助我如何在现有代码中添加延迟，以便在3秒钟(3000毫秒)后显示弹出窗口，我也希望如此.

我尝试使用setTimeout(function()如下

    <script type="text/javascript"> 
jQuery(document).ready(function() {
    setTimeout(function() {
        $.fancybox({
            'type' : 'iframe',
        'href' : 'http://www.mysite.net/popup/', //URL to popup page or image
        'autoDimensions' : false,
        'width' : 480,
        'height' : 260,
        'transitionIn' : 'none',
        'transitionOut' : 'none'
        })
    }, 4000); 
});
</script>

但它不起作用.

关于控制浏览量，我不知道如何设置，也找不到任何资源来帮助我.

非常感谢

解决方案

2019年4月2日更新

如@Gregdebrick的评论中所述，正在更新答案以使用JavaScript Cookie插件 https://github.com/js-cookie/js-cookie 而不是已弃用的jQuery Cookie插件.

因此，从页面中的CDN加载JS Cookie插件后:

 <script src="https://cdn.jsdelivr.net/npm/js-cookie@2/src/js.cookie.min.js"></script>
 

使用以下脚本:

 $(document).ready(function () {
    // if the cookie is undefined, create it
    if(typeof Cookies.get('visited') === "undefined"){
        Cookies.set("visited", 0);
    }
    // Cookies.get('visited') returns a string
    if (parseInt(Cookies.get('visited')) >= 3) {
        // open fancybox after 3 secs on 4th visit
        setTimeout(function () {
            $.fancybox.open({
                // your fancybox API options here
            });
        }, 3000);
    } else {
        let increase = parseInt(Cookies.get('visited')) + 1;
        Cookies.set('visited', increase, { expires: 1 });
        return false;
    }
}); // ready
 

请参见 已更新了有效的演示版

---

假设您正在使用 jQuery Cookie插件，则可以使用以下代码3秒后启动fancybox，同一天访问第4页:

$(document).ready(function () {
    // create cookie
    var visited = $.cookie('visited'); // visited = 0
    if (visited >= 3) {
        // open fancybox after 3 secs on 4th visit or further on the same day
        setTimeout(function () {
            $.fancybox.open({
                // your fancybox API options here
            });
        }, 3000);
    } else {
        visited++; // increase counter of visits
        // set new cookie value to match visits
        $.cookie('visited', visited, {
            expires: 1 // expires after one day
        });
        return false;
    }
}); // ready

请参见 工作演示

I am trying to find how may I set a javascript cookie so it displays a fancybox popup after say 4-th pageview of visitor.

Here is the code that I am using to display a Fancybox popup, but as you may see, this displays the code only on first pageview and it expires after a day

function setCookie(c_name,value,exdays)
 {
     var exdate=new Date();
     exdate.setDate(exdate.getDate() + exdays);
     var c_value=escape(value) + ((exdays==null) ? "" : "; expires="+exdate.toUTCString());
     document.cookie=c_name + "=" + c_value + ";domain=.mysite.net;path=/";
 }

function getCookie(c_name)
 {
     var i,x,y,ARRcookies=document.cookie.split(";");
     for (i=0;i<ARRcookies.length;i++)
  {
     x=ARRcookies[i].substr(0,ARRcookies[i].indexOf("="));
     y=ARRcookies[i].substr(ARRcookies[i].indexOf("=")+1);
     x=x.replace(/^\s+|\s+$/g,"");
      if (x==c_name)
       {
       return unescape(y);
       }
  }
 }

jQuery(document).ready(function() {
     var show = getCookie("fancyreg");
     if(show==null || show=="") {
       $.fancybox(
 {
        'type' : 'iframe',
        'href' : 'http://www.mysite.net/popup/', //URL to popup page or image
        'autoDimensions' : false,
        'width' : 480,
        'height' : 260,
        'transitionIn' : 'none',
        'transitionOut' : 'none'
 }
);

  setCookie('fancyreg','1',1); 

 }
 });

I would also like if you may help me how to add a delay in my existing code so teh popup display after 3 seconds (3000 ms).

I tried with setTimeout(function() as below

    <script type="text/javascript"> 
jQuery(document).ready(function() {
    setTimeout(function() {
        $.fancybox({
            'type' : 'iframe',
        'href' : 'http://www.mysite.net/popup/', //URL to popup page or image
        'autoDimensions' : false,
        'width' : 480,
        'height' : 260,
        'transitionIn' : 'none',
        'transitionOut' : 'none'
        })
    }, 4000); 
});
</script>

but it does not work.

As for controlling the pageviews, I have no idea how to set nor I could find any resource to help me through this.

Thanks a lot

解决方案

Update April 02, 2019

As mentioned in the comments by @Gregdebrick the answer is being update to use the JavaScript Cookie plugin https://github.com/js-cookie/js-cookie instead of the deprecated jQuery Cookie plugin.

So, after loading the JS Cookie plugin from CDN in your page:

<script src="https://cdn.jsdelivr.net/npm/js-cookie@2/src/js.cookie.min.js"></script>

Use the following script:

$(document).ready(function () {
    // if the cookie is undefined, create it
    if(typeof Cookies.get('visited') === "undefined"){
        Cookies.set("visited", 0);
    }
    // Cookies.get('visited') returns a string
    if (parseInt(Cookies.get('visited')) >= 3) {
        // open fancybox after 3 secs on 4th visit
        setTimeout(function () {
            $.fancybox.open({
                // your fancybox API options here
            });
        }, 3000);
    } else {
        let increase = parseInt(Cookies.get('visited')) + 1;
        Cookies.set('visited', increase, { expires: 1 });
        return false;
    }
}); // ready

See updated working DEMO

---

Assuming that you are using the jQuery Cookie Plugin, then you could use the following code to launch fancybox after 3 seconds, the 4th page visit on the same day :

$(document).ready(function () {
    // create cookie
    var visited = $.cookie('visited'); // visited = 0
    if (visited >= 3) {
        // open fancybox after 3 secs on 4th visit or further on the same day
        setTimeout(function () {
            $.fancybox.open({
                // your fancybox API options here
            });
        }, 3000);
    } else {
        visited++; // increase counter of visits
        // set new cookie value to match visits
        $.cookie('visited', visited, {
            expires: 1 // expires after one day
        });
        return false;
    }
}); // ready

See working DEMO

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