Question

我在listview中有五个项目和五个活动，现在我想打开单击listview的单个项目我的单个活动意味着Actvity1应该单击item1时打开，单击item2时应打开Activity2，单击item3时应打开Activity3。请告诉我我该怎么做。

谢谢

Answer 1

您可以根据onItemClick项目的位置OnItemClickListener ListView方法在ListView listView=(ListView) findViewById(R.id.listView); listView.setOnItemClickListener(new AdapterView.onItemClickListener() { @Override public void onItemClick(AdapterView<?> adapter, View view, int position, long arg) { if (position == 0) { Intent intent = new Intent(YourActivityName.this, Activity1.class); startActivity(intent); } else if (position == 1) { Intent intent = new Intent(YourActivityName.this, Activity2.class); startActivity(intent); } else if (position == 2) { Intent intent = new Intent(YourActivityName.this, Activity3.class); startActivity(intent); } else if (position == 3) { Intent intent = new Intent(YourActivityName.this, Activity4.class); startActivity(intent); } else if (position == 4) { Intent intent = new Intent(YourActivityName.this, Activity5.class); startActivity(intent); } } });方法中添加条件来实现：

public Telnet(Parcel in) {
    readFromParcel(in);
}

Answer 2

您可以从OnItemClickListener获取位置，如下所示：

  ListView lv=(ListView) findViewById(R.id.listview);
    lv.setOnItemClickListener(new AdapterView.OnItemClickListener() {
        @Override
        public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
            // position 0 -> activity one

        }
    });

或者您可以为每一行分配一个标记值，并使用view.getTag（）;

进行检查

希望这会有所帮助。

单击ListView的每个项目上的事件

2 个答案: