ExtJS打开窗口,最大高度 [英] ExtJS open window with max height
问题描述
我试图通过点击一个按钮(Ext.Button)打开一个窗口(Ext.Window)。问题是这个窗口必须具有80%的宽度和100%用户屏幕的高度,就是它应该覆盖所有的垂直空间。
I am trying to open a window (Ext.Window) by clicking on a button (Ext.Button). The problem is that this window must have size 80% width and 100% height of the screen of the user, that is it should cover all the vertical space.
我是真的不能创造一个工作的例子。我使用的按钮是隐藏/显示这个窗口,它总是有错误的大小..
I am really not able to create a working example. The button I use is hiding/showing this window, which always has wrong size..
有人可以帮助我吗?
谢谢
推荐答案
您正在寻找 Ext.getBody()。getViewSize() EM> 即可。更具体地说,Ext.getBody()。getViewSize()。height和Ext.getBody()。getViewSize()。width
You are looking for Ext.getBody().getViewSize(). More Specifically, Ext.getBody().getViewSize().height and Ext.getBody().getViewSize().width
这应该在你的按钮的处理程序中工作 - / p>
This should work in your button's handler -
new Ext.Window({
title:'Hello World Window',
html:'Am I the right size?',
height:Ext.getBody().getViewSize().height,
width:Ext.getBody().getViewSize().width*0.8 //80%
}).show();
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