0

Haxe/OpenFL 代码：

import openfl.net.URLRequest;
import openfl.Lib;

Lib.getURL (new URLRequest (url), "_self");  
// Opens the linked document in the same window or tab as it was clicked

Lib.getURL (new URLRequest (url), "_blank"); 
// Opens the linked document in a new window or tab. (this is default)

但是，第二个选项会生成被 Chrome 阻止的弹出窗口。

如何在不被阻止的情况下在另一个选项卡中打开链接？

使用Javascript这项工作：

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">

<title>OpenNewTab</title>

<meta id="viewport" name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=no" />
<meta name="apple-mobile-web-app-capable" content="yes">

</head>
<body>  

<center>
    <canvas id="myCanvas" width="200" height="200" style="border-style: solid; border-width: 1px"> </canvas>
</center>

<script>   

    var canvas;
    var linkURL = "http://www.google.com";        

    createLink();

    function createLink() {

        canvas = document.getElementById("myCanvas");           
        canvas.addEventListener("click", Link_click, false);

    }

    function Link_click(e) {

         window.open(linkURL,'_blank'); 

    }     
</script>

</body>
</html>

Ps：我使用 Stencyl 和 HTML/JavaScript。

4

2 回答 2

0

虽然我没有找到更好的解决方案，但我会使用这个：

import openfl.net.URLRequest;
import openfl.Lib;

class Web
{

public static function open(s:String, code:Int)
{   
    var type:String = "_self";
    var s:String = s;
    var code:Int = code;

    if(code==1){
        type = "_self";
    }else if(code==2){
        type = "_blank";
    }  

#if js      
         untyped __js__('

                var canvas;
                var linkURL = s;
                var lock = 0;

                if(lock==0){
                   lock =1;
                   createLink();
                }

                function createLink() {

                    canvas = document.getElementById("openfl-content");           
                    canvas.addEventListener("click", Link_click, false);

                }

                function Link_click(e) {

                     window.open(linkURL,type); 

                }       

        ');             

#else
     Lib.getURL (new URLRequest (s), type);   
#end
}

}
于 2016-07-19T20:47:07.733 回答
0

我相信如果弹出窗口是从用户触发的事件（如指针向下，单击）打开的，则任何弹出阻止程序都不应阻止打开它。

注意：我个人觉得开发人员决定如何打开窗口很烦人，为什么不让用户自己决定呢？

于 2016-07-19T18:08:31.587 回答