生成一个正常的链接（<a href="...">) in ExtJS

Question

I'm struggling with a really simple problem here.

I want to generate a normal link in frontend but it seems to don't work somehow. Here is the code i use for the generation of the link (the link was a button before which obened a new window with a specified URL on click).

{ 
    xtype: 'button', 
    id: 'PrintTool', 
    tooltip: 'Printer Friendly', 
    iconCls: 'icon-printerFriendly', 
    html: '<a href=\"http:\/\/www.google.at\">x<\/a>', 
    listeners: { 
    'click': function(button,event) { 
     console.log(this); 
     console.log(event); 
     this.restoreClick(); 
     return true; 
    } 
} 

}

When ExtJS renders the button it makes, hence i add an html attribute to the object, a link out of it.

I can see the link with FireBug. When I click the link I get the output

console.log(this); 
    console.log(event); 

in the console

So the event is fired. But the link gets never opened.

I think it has something to do with the CLICK-event getting stopped from ExtJS.

It seems that despite in Firebug there is no button in the HTML the object passed to the click event is still the button.

So me question is, how can a create normal HTML in ExtJS without setting a xtype of button. Or how can I generate a normal link.

I used to open a popup after the button was clicked. The popup got blocked from chrome, IE and various other browser so I have to use a normal link with a URL insted.

Answer 1

Got an answer to my question in ExtJS forum

Use LinkButton an extension written by Animal

http://www.sencha.com/forum/showthread.php?80639-Ext.LinkButton-A-button-class-which-renders-an-lt-a-gt-element./page3

Answer 2

您也可以使用标签组件的HTML财产，并调用控制器功能，例如：

{xtype: 'label', 
html: 'bla bla? <a href="#" onClick="javascript:appName.app.getController(\'myController\').showRegistration();">Register</a>' 
}