在使用 JS Cookie 浏览 x 页后显示模式？

Question

我正在尝试显示在我的网站上查看的 Bootstrap modal avec 3 页面，并使用 JS-Cookie 插件找到this 问题的实际解决方案。

我试过这段代码，但它不起作用：

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

上面的代码应该做什么：在 4 页重新加载后提示模式，like here
什么不工作：我重新加载页面但模式没有出现

这是我的 cookie 状态（在 chrome 上）： Image on this link

这是一个实际用于在网站首次访问时显示模式的代码示例：

$(document).ready(function() {
  if (Cookies('pop_welcome') == null) {
         $('#my_modal').modal('show');
     Cookies('pop_welcome', '31', { expires: 31 });
  }
 });

** -- 解决方案 -- **

您可以找到解决方案into @greedchikara code snip

代码：

function createCookie(name,value,days) {
    if (days) {
        var date = new Date();
        date.setTime(date.getTime()+(days*24*60*60*1000));
        var expires = "; expires="+date.toGMTString();
    }
    else var expires = "";
    document.cookie = name+"="+value+expires+"; path=/";
}

function readCookie(name) {
    var nameEQ = name + "=";
    var ca = document.cookie.split(';');
    for(var i=0;i < ca.length;i++) {
        var c = ca[i];
        while (c.charAt(0)==' ') c = c.substring(1,c.length);
        if (c.indexOf(nameEQ) == 0) return c.substring(nameEQ.length,c.length);
    }
    return null;
}

function eraseCookie(name) {
    createCookie(name,"",-1);
}

var visited = readCookie('tester') || 1;

if (visited > 2) {
  $('#my_modal').modal('show');
} else {
  visited++;
  createCookie('tester', visited, 1);
}

---

你能帮帮我吗？ 谢谢

Here is the code of the JS cookie plugin.

Answer 1

您可以使用这三个简单的函数来执行cookie任务，不需要第三方库。

function createCookie(name,value,days) {
    if (days) {
        var date = new Date();
        date.setTime(date.getTime()+(days*24*60*60*1000));
        var expires = "; expires="+date.toGMTString();
    }
    else var expires = "";
    document.cookie = name+"="+value+expires+"; path=/";
}

function readCookie(name) {
    var nameEQ = name + "=";
    var ca = document.cookie.split(';');
    for(var i=0;i < ca.length;i++) {
        var c = ca[i];
        while (c.charAt(0)==' ') c = c.substring(1,c.length);
        if (c.indexOf(nameEQ) == 0) return c.substring(nameEQ.length,c.length);
    }
    return null;
}

function eraseCookie(name) {
    createCookie(name,"",-1);
}

Answer 2

$(document).ready(function() {
  // create cookie
  var visited = $.cookie('visited'); // visited = 0
  if (visited >= 3) {
    setTimeout(function() {
      $('#my_modal').modal('show'); //make sure model is already created with id my_model 
    }, 1000);
  } else {
    visited++; // increase counter of visits
    // set new cookie value to match visits
    $.cookie('visited', visited, {
      expires: 1 // expires after one day
    });
    return false;
  }
});


Make sure all dependent js are included.

jQuery 插件在这里 https://github.com/carhartl/jquery-cookie