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我在将一些工作 XML 文件转换为代码时遇到了一些麻烦。我有一个 ListView,我需要能够在运行时从高级未知的资源(因此为什么不使用 XML)切换其按下/检查的可绘制对象;

以下配置效果很好:

主.xml:

<?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="fill_parent"
    android:orientation="vertical" >

    <ListView
        android:id="@+id/listView1"
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:choiceMode="singleChoice" >
    </ListView>

</LinearLayout>

选择器.xml:

<?xml version="1.0" encoding="utf-8"?>
<selector xmlns:android="http://schemas.android.com/apk/res/android">
    <item android:drawable="@drawable/blue" android:state_pressed="true"/>
    <item android:drawable="@drawable/green" android:state_checked="true"/>
    <item android:drawable="@drawable/orange"/>
</selector>

list_row.xml:

<?xml version="1.0" encoding="utf-8"?>
<CheckedTextView xmlns:android="http://schemas.android.com/apk/res/android"
    android:layout_width="fill_parent"
    android:layout_height="wrap_content"
    android:background="@drawable/selector"
    android:padding="10dp" />

主.java:

public class Main extends Activity {

    StateListDrawable selector; 

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        ListView listView1 = (ListView) findViewById(R.id.listView1);
        StringAdapter adapter = new StringAdapter(this, R.layout.list_row);
        adapter.add("one");     adapter.add("two");     adapter.add("three");       adapter.add("four");
        adapter.add("five");        adapter.add("six");     adapter.add("seven");       adapter.add("eight");
        adapter.add("nine");        adapter.add("ten");     adapter.add("eleven");      adapter.add("twelve");
        listView1.setAdapter(adapter);
    }

    private class StringAdapter extends ArrayAdapter<String>{

        public StringAdapter(Context context, int textViewResourceId) {
            super(context, textViewResourceId);         
        }

        @Override
        public View getView(int position, View convertView, ViewGroup parent) {
            LayoutInflater inflater = (LayoutInflater) getContext().getSystemService(Context.LAYOUT_INFLATER_SERVICE);
            final CheckedTextView  tv = (CheckedTextView ) inflater.inflate(R.layout.list_row, parent, false);
            tv.setText(getItem(position));
            return tv;
        }
    }
}

结果是这样的,非常棒(按下:蓝色,选中:绿色,否则:橙色):

结果1

但是,如果我删除

android:background="@drawable/selector

从 list_row_xml,并通过代码应用它:

    Drawable blue = getResources().getDrawable(R.drawable.blue);
    Drawable green = getResources().getDrawable(R.drawable.green);
    Drawable orange = getResources().getDrawable(R.drawable.orange);

    selector = new StateListDrawable();
    selector.addState(new int[] { android.R.attr.state_pressed }, blue);
    selector.addState(new int[] { android.R.attr.state_checked }, green);
    selector.addState(new int[] { }, orange);

       tv.setBackgroundDrawable(selector);

我得到以下信息(一切都得到 state_pressed 可绘制,蓝色):

在此处输入图像描述

什么地方出了错?我很确定我已将选择器适当地转换为代码。

4

1 回答 1

4

我复制了你的代码,一切正常。不过,您应该确保创建一个selectorin的新实例getView()。否则,如果您selector对所有项目使用相同的内容,则按一项将影响您的所有项目。

这是你的getView()方法应该是什么样子

    @Override
    public View getView(int position, View convertView, ViewGroup parent) {
        LayoutInflater inflater = (LayoutInflater) getContext().getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        final CheckedTextView tv = (CheckedTextView) inflater.inflate(R.layout.list_row, parent, false);
        tv.setText(getItem(position));

        Drawable blue = getResources().getDrawable(R.drawable.blue);
        Drawable green = getResources().getDrawable(R.drawable.green);
        Drawable orange = getResources().getDrawable(R.drawable.orange);

        selector = new StateListDrawable();
        selector.addState(new int[] { android.R.attr.state_pressed }, blue);
        selector.addState(new int[] { android.R.attr.state_checked }, green);
        selector.addState(new int[] {}, orange);

        tv.setBackgroundDrawable(selector);

        return tv;
    }

当然,您可以进行一些优化,但这会起作用。

于 2012-06-22T16:56:26.780 回答