Question

我正在尝试在名称和组织的Android应用中创建TextView列表。但是，整个列表是相同的，例如：

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
...

在应用列表中重复五次。以下是相关代码：

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;
}

以下是此类正在处理的xml视图：

<?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，我明白当我调用“TextView tt =（TextView）v.findViewById（R.id.toptext）;”时会返回相同的实例，但是我不知道要改变什么来获得一个新的每次都是实例对象。我错过了什么？

Answer 1

像这样使用你的getView

   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;
}

Answer 2

我会尝试更改

TextView tt = (TextView) v.findViewById(R.id.toptext);

到

TextView tt = new TextView(this);
CharSequence text = ((TextView)v.findViewById(R.id.toptext)).getText();
tt.setText(text);

这将实例化一个具有相同内容的新TextView，我认为这就是您所说的潜在问题。

希望有所帮助！

在Android中创建TextView列表

2 个答案: