我在WebViews
中存储了多个HashMap
,我需要能够随时显示/隐藏。
点击ListView
中的一行时,我需要显示与该行关联的WebView
并隐藏所有其他行。另一个WebViews
无法销毁,因为我不想卸载它们的内容。
这可能吗?
在一个WebView
中创建和存储Fragment
:
public WebView createWebView(Long id) {
WebView wv = new WebView(getActivity());
wv.setWebViewClient(new WebViewClient() {
@Override
public void onPageFinished(WebView view, String url) {
view.loadUrl("javascript:(registerCredentialsChangedCallback(function(credentials) {if (credentials.length >= 1) {window.location.href = 'callback:' + credentials;} else {alert('Error: callback set by registerCredentialsChangedCallback was given null string');}})) ()");
}
});
WebSettings webSettings = wv.getSettings();
webSettings.setJavaScriptEnabled(true);
webSettings.setUseWideViewPort(true);
webSettings.setLoadWithOverviewMode(true);
wv.loadUrl(makeViewerURL(id));
webViews.put(id, wv);
return wv;
}
onCreate
在另一个Fragment
中。我正在使用interface
将id通过Activity
传递到Fragment
应该在其中显示WebView
的地方:@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
View rootView = inflater.inflate(R.layout.fragment_directory, container, false);
ListView listView = (ListView) rootView.findViewById(R.id.list);
DatabaseHelper dbHelper = new DatabaseHelper(getActivity());
android.R.layout.simple_list_item_1, android.R.id.text1, values);
final SimpleCursorAdapter adapter = new SimpleCursorAdapter(getActivity(), android.R.layout.simple_list_item_1, dbHelper.getViewers(), new String[] { "description", "_id" }, new int[] { android.R.id.text1 }, CursorAdapter.FLAG_REGISTER_CONTENT_OBSERVER);
listView.setAdapter(adapter);
listView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> adapterView, View view, int i, long l) {
Long id = adapter.getItemId(i);
mCallback.onViewerSelected(id);
}
});
return rootView;
}
视图的XML布局应包含WebView:
<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:orientation="vertical"
android:background="#FFFFFF" >
<WebView xmlns:android="http://schemas.android.com/apk/res/android"
android:id="@+id/webview"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
/>
</RelativeLayout>
谢谢!
最佳答案
是的,有可能-但您需要具有ID和WebView
的键值对。
private Map<Long, View> mWebViews = new HashMap<Long, View>();
listView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> adapterView, View view, int i, long l) {
Long id = adapter.getItemId(i);
if(mViews.contains(id))
mWebView = mViews.get(id)
else
mWebView = createWebView();
}
});
然后,您只需要找到用户在列表中选择的相应
WebView
并将其显示。为了隐藏WebView
,您需要使用:mWebView.setVisibility(View.Gone);
附录:而不是XML文件中的webView,请使用frameLayout,例如:
<FrameLayout
android:id="@+id/web_container"
android:layout_width="fill_parent"
android:layout_height="420dip"
android:layout_gravity="center"
android:clickable="true"
android:enabled="true"
android:paddingLeft="10dip"
android:paddingRight="10dip" />
然后在代码中将WebView添加到Framelayout中:
FrameLayout mWebContainer = (FrameLayout) findViewById(R.id.web_container);
webView = new WebView(getApplicationContext());
mWebContainer.addView(webView);