我必须删除同一视图中的当前视图... 如果我在父视图中我可以做到
parentView.remove(childView);
但是在子视图上我没有 parentView 所以如何弹出 childView 以获得 parentView ,因为它发生了按下iOS中的后退按钮? 请帮忙
这是我的childView文件..
function DetailView(){
var self = Ti.UI.createView({
backgroundColor:'#fff'
});
// Create a Button.
var aButton = Ti.UI.createButton({
title : 'aButton',
height : '50',
width : '100',
top : '10',
left : '20'
});
// Listen for click events.
aButton.addEventListener('click', function() {
alert('\'aButton\' was clicked!');
我必须回顾一下按钮,我应该把它放在这里做什么
});
// Add to the parent view.
self.add(aButton);
return(self);
}
module.exports = DetailView;
这是我的父视图:
//FirstView Component Constructor
var self = Ti.UI.createView();
function FirstView() {
//create object instance, a parasitic subclass of Observable
var DetailView = require('ui/common/DetailView');
var data = [{title:"Row 1"},{title:"Row 2"}];
var table = Titanium.UI.createTableView({
data:data
});
table.addEventListener('click', rowSelected);
self.add(table);
return self;
}
function rowSelected()
{
var DetailView = require('ui/common/DetailView');
//construct UI
var detailView = new DetailView();
self.add(detailView);
}
module.exports = FirstView;
答案 0 :(得分:4)
此时您可以将parentView传递给子视图的构造函数:
//construct UI
var detailView = new DetailView(parentView);
self.add(detailView);
并点击事件
aButton.addEventListener('click', function() {
if ( parentView != null ) {
parentView.remove(childView);
}
});