弹出Java脚本 [英] Java Script pop up

问题描述

如何确定客户端计算机上是否打开了任何弹出窗口?我只需要打开一个弹出窗口，然后所有弹出窗口都被拒绝.

在C#中开发应用程序时可以在Java脚本中声明应用程序级变量吗?

有人可以帮我吗?

How can I determine whether any pop up window is open on client machine or not? I need to open only one pop up and after that all pop up opening is denied.

Can I declare application level variable in java script while developing application in c#?

Can anyone help me?

推荐答案

它不是C#的一个函数，它将是服务器端的，它是客户端的JavaScript的函数. br/>
Window.Open返回对打开的window对象的引用.将其存储为全局变量，并在打开新变量之前进行检查.

It isn''t a function for C#, which would be server side, it''s a function of JavaScript on the client side.

Window.Open returns a reference to the opened window object. Store this as a global variable and check before opening a new one.

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