https://github.com/nadavfima/cardsui-for-android
我正在使用这个库来创建一张卡片,它的标题很简单,内容中的列表视图更复杂一些。我遇到的问题是我无法从文件“card_ex.xml”访问任何 xml 对象,因为“错误”视图被设置为内容视图。
MainActivity.java
package com.cardsui.example;
import android.app.Activity;
import android.content.Context;
import android.os.Bundle;
import android.view.LayoutInflater;
import android.view.Menu;
import android.view.View;
import android.widget.ArrayAdapter;
import android.widget.ListView;
import android.widget.RelativeLayout;
import com.fima.cardsui.views.CardUI;
public class MainActivity extends Activity {
ListView lv;
private CardUI mCardView;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
// init CardView
mCardView = (CardUI) findViewById(R.id.cardsview);
mCardView.setSwipeable(false);
MyCard idag = new MyCard("Idag");
idag.getView(this);
// add one card, and then add another one to the last stack.
mCardView.addCard(new MyCard("Idag"));
lv = (ListView) findViewById(android.R.id.item_list);
String[] items = new String[] {"Item 1", "Item 2", "Item 3", "Item 4", "Item 5", "Item 6", "Item 7"};
ArrayAdapter<String> adapter =
new ArrayAdapter<String>(this, android.R.layout.select_dialog_singlechoice, items);
// lv.setAdapter(adapter);
// draw cards
mCardView.refresh();
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
getMenuInflater().inflate(R.menu.activity_main, menu);
return true;
}
}
card_ex.xml
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:orientation="vertical"
android:paddingBottom="16dp" >
<LinearLayout
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:gravity="center_vertical"
android:paddingLeft="8dp"
android:paddingRight="8dp" >
<TextView
android:id="@+id/title"
style="@style/CardTitle"
android:layout_width="0dp"
android:layout_height="wrap_content"
android:layout_weight="1"
android:text="title" />
</LinearLayout>
<View
android:layout_width="fill_parent"
android:layout_height="1dp"
android:layout_marginTop="4dp"
android:background="@color/stroke" />
<LinearLayout
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:background="@drawable/selectable_background_cardbank"
android:gravity="center_vertical"
android:padding="4dp" >
<ListView
android:id="@+id/item_list"
style="@style/CardText"
android:layout_width="wrap_content"
android:layout_height="400dp"
android:divider="@color/stroke"
android:dividerHeight="1dp"
android:ellipsize="end" />
</LinearLayout>
</LinearLayout>
activity_main.xml
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:background="@drawable/background"
android:id="@+id/cards">
<com.fima.cardsui.views.CardUI
android:id="@+id/cardsview"
android:layout_width="match_parent"
android:layout_height="match_parent" />
</RelativeLayout>
我想操作不在内容视图中的id“item_list”,而是在另一个xml文件“card_ex”中。
无论如何,我可以在当前设置的内容视图中获得一个 id 吗?