【php】php输出jquery的轮询，5秒跳转指定url

1、在php中直接输出jquery的轮询，5秒后跳转指定url

2、代码稍微改动，即可在html中使用

3、代码：

public function alpha(){
            $html = '<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
                    <html xmlns="http://www.w3.org/1999/xhtml">
                    <head>
                    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
                    <script src="http://weixin.jimuc.com/app/resource/js/lib/jquery-1.11.1.min.js"></script>
                    <title>积慕商城午夜订购</title>
                    </head>
                    <body>';
            $html.= "<img src='http://pms.jimuc.com/index.php?m=file&f=read&t=png&fileID=130' style='width: 100%;height: 100%;'/>
                    <input class='loginTime' value='5' type='hidden'>";
            $html.=    "<script type='text/javascript'>
                        var loginTime = parseInt($('.loginTime').val());
                        var time = setInterval(function(){
                            console.log(loginTime);
                            loginTime = loginTime-1;
                            $('.loginTime').val(loginTime);
                            if(loginTime == 0){
                                window.location.href = '$url';
                            }
                              
                        },1000);
                     </script>";
            $html.= '</body>
                     </html>';
            print_r($html);exit;
}

4、效果：

 

 

posted @ 2018-10-10 11:32  PHP急先锋  阅读(808)  评论(0)  编辑  收藏  举报