如何在彈出窗口中打開彈出窗口？

Question

我想在另一個窗口打開彈出窗口。

請檢查代碼 -

myjavascript.js

var win = null; 
function NewWindow(mypage,myname,w,h,scroll){ 
    LeftPosition = (screen.width) ? (screen.width-w)/2 : 0; 
    TopPosition = (screen.height) ? (screen.height-h)/2 : 0; 
    settings = 
    'height='+h+',width='+w+',top='+TopPosition+',left='+LeftPosition+',scrollbars='+scroll+',resizable' 
    win = window.open(mypage,myname,settings) 
} 

的index.html

<script src="myjavascript.js"></script> 
<a href="#" onclick="NewWindow('popup1.html','name','500','500','yes');return false" class="link-2">My Account</a> 

popup1.html

<script src="myjavascript.js"></script> 
<a href="#" onclick="NewWindow('popup2.html','name','500','500','yes');return false" class="link-2">Click to open popup2</a> 

popup2.html

<h1>You are now popup2.html page</h1> 

上面的代碼是爲我工作，但第二彈出無法打開。請檢查並讓我知道。

感謝 @chatfun

Answer 1

的問題是，您使用的是兩個window.open(mypage,myname,settings)相同的標識符。

更改name在第二次調用：

的index.html

<a href="#" onclick="NewWindow('popup1.html','name','500','500','yes');return false" class="link-2">My Account</a>

popup1.html

<a href="#" onclick="NewWindow('popup2.html','NEWNAME','500','500','yes');return false" class="link-2">Click to open popup2</a>

希望這會有所幫助，