ExtJs 手动触发 Click 事件，按钮参数与鼠标点击不同 [英] ExtJs manually firing Click event, button param is different from mouse click

问题描述

所以，我有一个登录控制器，您可以用鼠标单击登录或按 Enter 键，如下所示:

So, I have a login controller, you can click login with mouse or press Enter key, like this:

Ext.define('My.controller.Login', {
    extend: 'Ext.app.Controller',

    init: function(application) {
        this.control({
            "#idLogin button": {click: this.onButton},
            "#idLogin form > *": {specialkey: this.onKey}
        });
    },

    onButton: function(button, e, eOpts) {
        var win = button.up('window'); // the login window
        //do more stuff...
    },

    onKey: function (field, el) {
        if (el.getKey() == Ext.EventObject.ENTER) //ENTER key performs Login
            Ext.getCmp('#idLogin button').fireEvent('click');
    }
});

我发现当我用鼠标点击登录按钮时，onButton 函数正常工作，button.up() 返回我登录window.

I realised when I use the mouse to click the Login button, onButton function works properly, button.up() returns me the Login window.

但是，如果我按下 Enter 键并触发 onKey 函数来执行 fireEvent('click')，在本例中为 onButton启动但参数 button 与您点击鼠标时收到的 button 参数不同！而这一次，button.up() 函数未定义.

However, if I pressed Enter key and fires the onKey function to do fireEvent('click'), in this case the onButton fires up but parameter button NOT the same as the button parameter received when you click by mouse! And this time, button.up() function is undefined.

问题是，为什么 fireEvent('click') 给我一个不同的按钮参数?

Question is, why does fireEvent('click') give me a different button parameter?

推荐答案

你必须像这样使用 fireEvent 函数:

You must use the fireEvent function like that:

var myBtn = Ext.getCmp('#idLogin button');

myBtn.fireEvent('click', myBtn);

试试看.

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