如何将弹出窗口定位在屏幕的右上角

时间:2010-07-01 20:48:53

标签: javascript browser popup resize window

我想要做的是更改窗口的大小并在其旁边创建一个窗口,使它们完全相邻。从这个意义上讲,我需要将新窗口定位到屏幕的右上角,我这样做的方式不起作用(代码如下),我需要帮助:))

function () { 
   var viewportwidth = document.documentElement.clientWidth;
   var viewportheight = document.documentElement.clientHeight;
   window.resizeBy(-300,0); 
   window.open("something.htm",
     "mywindow",
     "width=300,
     height=viewportheight,
     left=(viewportwidth - 300),
     top=0,
     screenX=0,
     screenY=0"); 
}

2 个答案:

答案 0 :(得分:7)

var viewportwidth = document.documentElement.clientWidth;
var viewportheight = document.documentElement.clientHeight;
window.resizeBy(-300,0);
window.moveTo(0,0);

window.open("http://google.com",
            "mywindow",
            "width=300,left="+(viewportwidth-300)+",top=0");

答案 1 :(得分:1)

我还没有测试出实际的窗口大小数学;不确定这是否正确。但是你遇到的第一个显而易见的问题是将变量嵌入到window.open的调用中。尝试更改

  window.open("something.htm", "mywindow",
 "width=300, height=viewportheight, left=(viewportwidth - 300), top=0, screenX=0, screenY=0"); 

  window.open("something.htm", "mywindow",
 "width=300, height=" + viewportheight + ", left=" + (viewportwidth - 300) + ", top=0, screenX=0, screenY=0");

基本上,如果你想要解析变量或数学,它们必须在字符串之外。