列表视图不同的布局通胀的每一行 [英] Listview with different layout inflation for each row

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本文介绍了列表视图不同的布局通胀的每一行的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我是相当新的在Android的用户界面设计(和相当新的Andr​​oid开发以及)。我目前正在开发一个Android应用程序,它看起来很像谷歌+的各界页面,Facebook的用户家中,在那里你能看到你的朋友共享的内容。

I'm fairly new to designing UIs in Android (And fairly new to Android development as well). I'm currently developing an Android application that looks a lot like Google+'s "All Circles" page, and Facebook's user home, where you get to see contents shared by your friends.

要更清楚,请看看这是采取从谷歌+的Andr​​oid应用程序下面的截图:

To make things clearer, please take a look at the following screenshot that's taken from Google+'s Android application:

正如你所看到的,保罗哈珀的职位是在一个小框,而Android和我的帖子是另一回事。而你向下滚动的越多,共享的东西,你会看到,在其各自的框架。

As you can see, Paul Harper's post is in a little Frame, and "Android and me"'s post is in another. And the more you scroll downwards, the more shared stuff you'll see, each in its own "Frame".

我真的不知道如何做到这一点的结果(我敢肯定,它涉及到一个ListView成分),所以任何人都可以请告诉我,我应该用做UI组件?

I'm really not sure how to achieve this result(I'm sure that it involves a ListView component), so could anyone please tell me about the UI components that I should be using to do it?

由于一吨。

推荐答案

您应该看看在链接的视频。

You should have a look at the video in the link.

http://www.youtube.com/watch?v=wDBM6wVEO70

private static final int TYPE_ITEM1 = 0;
private static final int TYPE_ITEM2 = 1;
private static final int TYPE_ITEM3 = 2;    

整型;

@Override
public int getItemViewType(int position) {

    if (position== 0){
        type = TYPE_ITEM1;
    } else if  (position == 1){
        type = TYPE_ITEM2;
    }
    return type;
}


@Override  
public View getView(int position, View convertView, ViewGroup parent) {
View row = convertView;
LayoutInflater inflater = null;
int type = getItemViewType(position);
   if (row  == null) {
    if (type == FIRST_TYPE) {
                 //infalte layout of type1
      }
    if (type == FIRST_TYPE) {
                 //infalte layout of type2
    }  else {
                 //infalte layout of normaltype
 }
} 

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