【发布时间】:2018-09-08 02:11:22
【问题描述】:
我创建了一个简单的listView,它从服务器获取数据,一切正常,但现在我想将listView(Item) 的值传递给另一个Activity
public void onViewBind(View view, Cursor cursor, ODataRow row) {
_list = new ArrayList<>();
OControls.setText(view, R.id.person_name, row.getString("name"));
// Retrive Code dept*/
Dep dept = new Dept(context, null);
ODataRow rows_c = dept.select(
new String[]{"code"}
).get(row.getInt("dept_id")-1);
_list.add(rows_c.getString("code"));
OControls.setText(view, R.id.dept_person, rows_c.getString("code"));
// Retrive name person*/
ResPartner partner = new ResPartner(context, null);
ODataRow rows = partner.select(
new String[]{"name"}
).get(row.getInt("person_id")-1);
OControls.setText(view, R.id.Type_sanction, rows.getString("name"));
}
我正在使用数组列表将值从第一种方法传递给第二个方法,但是当我单击时我想在列表视图中传递一个对象对应的行
@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long
id) {
final Intent intent;
intent = new Intent(view.getContext(), Detailsdept.class);
Bundle extras = new Bundle();
extras.putSerializable("ARRAYLIST",(Serializable)_list);
for(int i=0; i<_list.size(); i++) {
String x = _list.get(i);
extras.putString("code", x);
intent.putExtras(extras);
}
view.getContext().startActivity(intent);
Toast.makeText(view.getContext(), "id"+position, Toast.LENGTH_SHORT).show();
}
}
【问题讨论】:
-
我需要传递对象而不是删除
标签: java android listview android-intent bundle