ExtJS 4“总是在顶部”窗口 [英] ExtJS 4 "always on top" Window

问题描述

我需要实现可以永远在上面的Window。我该怎么办？所有我使用WindowManager的尝试都给我没有结果：（

I need to implement Window which can be always on top. How can I to do it? All my tries with WindowManager give me no results :(

推荐答案

在Ext.window.Window中，有一个名为' > modal '：将其设置为true

In Ext.window.Window, there's a property called 'modal': set it to true.

否则，使用 WindowManager 来管理您的窗口：在这种情况下，您有按照以下步骤：

Otherwise, use the WindowManager to manage your windows: in this case you have to follow the following steps:

1. 注册您的窗口到WindowManager（ Ext.WindowManager.register （winId））


2. 使用 bringToFront 方法将窗口设置在顶部（ Ext.WindowManager.bringToFront（winId））


3. 最后，使用 getActive 方法（ Ext.WindowManager.getActive（））

1. register your windows to the WindowManager (Ext.WindowManager.register (winId))
2. use bringToFront method to set your window on top (Ext.WindowManager.bringToFront (winId))
3. finally, check the element on top with the getActive method (Ext.WindowManager.getActive ())

例如：

Ext.create ('Ext.window.Window', {
  title: 'Your window' ,
  width: 300 ,
  height: 300 ,
  html: 'ciao ciao' ,
  modal: true
}).show ();

或者：

var win1 = Ext.create ('Ext.window.Window', {
  title: 'Your window' ,
  id: 'firstWin' ,
  width: 300 ,
  height: 300 ,
  html: 'ciao ciao' ,
});
win1.showAt (50, 50);

var win2 = Ext.create ('Ext.window.Window', {
  title: 'Your window' ,
  id: 'secondWin' ,
  width: 300 ,
  height: 300 ,
  html: 'I love pizza' ,
});
win2.showAt (60, 60);

// Register your floating objects (window in this case) to the WindowManager
Ext.WindowManager.register (win1);
Ext.WindowManager.register (win2);

// Bring 'firstWin' on top
Ext.WindowManager.bringToFront ('firstWin');

// Then, check the zIndexStack
alert (Ext.WindowManager.getActive().getId ()); // this is firstWin, the window with the highest zIndex

希望这可以帮助您。

Cyaz

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