创建的TextView的在Android的清单

问题描述

我想以我的名义的和组织的Andr​​oid应用程序创建的TextView的列表。然而，整个列表是相同的，例如：

I am trying to create a list of TextView's in my Android app of name's and organizations. However, the whole list is the same, for example:

Joe's work for Construction Co.
15hrs of 60hrs
Joe's work for Construction Co.
15hrs of 60hrs
Joe's work for Construction Co.
15hrs of 60hrs
...

重复五次在应用程序列表中。下面是相关code：

is repeated five times in the list in the app. Here is the relevant code:

class IconicAdapter extends ArrayAdapter<Long> {
private ArrayList<Long> items;
private FakeIDO ido;

public IconicAdapter(Context context,int textViewResourceId, ArrayList<Long> ids, FakeIDO ido){
    super(WorkList.this, R.layout.feed_list, ids);
    this.items = ids;
    this.ido = ido;
}


public View getView(int position, View convertView, ViewGroup parent) {
    View v = convertView;
    if (v == null) {
        LayoutInflater vi = (LayoutInflater)getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        v = vi.inflate(R.layout.row, null);
    }
    for(long id : items){ 
        TextView tt = (TextView) v.findViewById(R.id.toptext);
        TextView bt = (TextView) v.findViewById(R.id.bottomtext);
        if (tt != null) {
              tt.setText(ido.getName(id) + "'s work for " + ido.getOrganization(id));
        }
        if(bt != null){
              bt.setText( ido.totalWorked(id) + "hrs of " + ido.estimatedHours(id) + "hrs");
        }
    }
    return v;
}

下面是XML针对这一类工作断：

Here is the xml view this class is working off of:

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="fill_parent"
android:layout_height="?android:attr/listPreferredItemHeight"
android:padding="6dip">
<ImageView
    android:id="@+id/icon"
    android:layout_width="wrap_content"
    android:layout_height="fill_parent"
    android:layout_marginRight="6dip"
    android:src="@drawable/icon" />
<LinearLayout
    android:orientation="vertical"
    android:layout_width="0dip"
    android:layout_weight="1"
    android:layout_height="fill_parent">
    <TextView
        android:id="@+id/toptext"
        android:layout_width="fill_parent"
        android:layout_height="0dip"
        android:layout_weight="1"
        android:gravity="center_vertical"
    />
    <TextView
        android:layout_width="fill_parent"
        android:layout_height="0dip"
        android:layout_weight="1" 
        android:id="@+id/bottomtext"
        android:singleLine="true"
        android:ellipsize="marquee"
    />
</LinearLayout>

现在，我明白，当我打电话同一个实例被退回的TextView TT =（TextView中）v.findViewById（R.id.toptext）;，但我不知道是什么改变来获得一个新的例如每次对象。我缺少什么？

Now, I understand that the same instance is being returned when I call "TextView tt = (TextView) v.findViewById(R.id.toptext);", but I don't know what to change to get a new instance object each time. What am I missing?

推荐答案

使用 getView 像这样

   public View getView(int position, View convertView, ViewGroup parent) {
    View v = convertView;
    if (v == null) {
        LayoutInflater vi = (LayoutInflater)getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        v = vi.inflate(R.layout.row, null);
    }

        TextView tt = (TextView) v.findViewById(R.id.toptext);
        TextView bt = (TextView) v.findViewById(R.id.bottomtext);
        // get the 'id' for position.
        int id = items.get(position);
        if (tt != null) {
              tt.setText(ido.getName(id) + "'s work for " + ido.getOrganization(id));
        }
        if(bt != null){
              bt.setText( ido.totalWorked(id) + "hrs of " + ido.estimatedHours(id) + "hrs");
        }

    return v;
}

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