更新:解决了!问题与我的Viewpager而不是WebView有关。
我正在尝试向WebView
内的Fragment
添加“返回”功能。但我无法弄清楚如何:
public final class TestFragment extends Fragment {
static WebView mWeb;
private View mContentView;
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState)
{
mContentView = inflater.inflate(R.layout.webview, null);
mWeb = (WebView)mContentView.findViewById(R.id.webview);
WebSettings settings = mWeb.getSettings();
settings.setJavaScriptEnabled(true);
settings.setSupportZoom(false);
mWeb.setScrollBarStyle(View.SCROLLBARS_INSIDE_OVERLAY);
mWeb.getSettings().setBuiltInZoomControls(false);
mWeb.loadUrl("myurl...");
mWeb.setOnKeyListener(new OnKeyListener(){
public boolean onKey(View v, int keyCode, KeyEvent event) {
if ((keyCode == KeyEvent.KEYCODE_BACK) && mWeb.canGoBack()) {
mWeb.goBack();
return true;
}
return false;
}
});
}
}
我也尝试过这样的事情:
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if ((keyCode == KeyEvent.KEYCODE_BACK) && mWeb.canGoBack()) {
mWeb.goBack();
return true;
}
return super.onKeyDown(keyCode, event);
}
另一个解决方案,但同样的问题:
@Override
public void onBackPressed()
{
if(webView.canGoBack())
webView.goBack();
else
super.onBackPressed();
}
任何想法如何使这个工作?
答案 0 :(得分:45)
也许是它的机器人限制。尝试使用handler。
public final class TestFragment extends Fragment {
static WebView mWeb;
private View mContentView;
private Handler handler = new Handler(){
@Override
public void handleMessage(Message message) {
switch (message.what) {
case 1:{
webViewGoBack();
}break;
}
}
};
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,Bundle savedInstanceState) {
mContentView = inflater.inflate(R.layout.webview, null);
mWeb = (WebView)mContentView.findViewById(R.id.webview);
WebSettings settings = mWeb.getSettings();
settings.setJavaScriptEnabled(true);
settings.setSupportZoom(false);
mWeb.setScrollBarStyle(View.SCROLLBARS_INSIDE_OVERLAY);
mWeb.getSettings().setBuiltInZoomControls(false);
mWeb.loadUrl("myurl...");
mWeb.setOnKeyListener(new OnKeyListener(){
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_BACK
&& event.getAction() == MotionEvent.ACTION_UP
&& mWeb.canGoBack()) {
handler.sendEmptyMessage(1);
return true;
}
return false;
}
});
}
private void webViewGoBack(){
mWeb.goBack();
}
}
答案 1 :(得分:21)
实际上你不能直接在片段里面做。可以在onBackPressed
中覆盖FragmentActivity
。你能做的是:
onBackPressed
。onBackPressed
时,请检查当前片段的实例是否为显示webview
的实例。fragment
webview
是否可以返回。 super
或任何您需要的内容编辑:
@Override
public void onBackPressed() {
Fragment webview = getSupportFragmentManager().findFragmentByTag("webview");
if (webview instanceof MyWebViewFragment) {
boolean goback = ((MyWebViewFragment)webview).canGoBack();
if (!goback)
super.onBackPressed();
}
}
答案 2 :(得分:16)
您可以查看以下代码:
webView.canGoBack();
webView.setOnKeyListener(new View.OnKeyListener() {
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_BACK
&& event.getAction() == MotionEvent.ACTION_UP
&& webView.canGoBack()) {
webView.goBack();
return true;
}
return false;
}
});
答案 3 :(得分:6)
在WebViewActivity.java
中,我添加了1个方法:
@Override
public void onBackPressed() {
WebViewFragment fragment = (WebViewFragment)
getSupportFragmentManager().findFragmentById(R.id.fragmentContainer);
if (fragment.canGoBack()) {
fragment.goBack();
} else {
super.onBackPressed();
}
}
在WebViewFragment.java
中,我添加了两种方法:
public boolean canGoBack() {
return mWebView.canGoBack();
}
public void goBack() {
mWebView.goBack();
}
答案 4 :(得分:3)
我的解决方案是片段我添加到公共方法
public static boolean canGoBack(){
return mWebView.canGoBack();
}
public static void goBack(){
mWebView.goBack();
}
然后从我打电话的活动
@Override
public void onBackPressed() {
if(webFragment.canGoBack()){
webFragment.goBack();
}else{
super.onBackPressed();
}
}
注意mwebview是静态的
答案 5 :(得分:2)
您可以通过以下方式执行此操作:
:
Activity
// Set WebView
public void setWebView(WebView web) {
this.web = web;
}
之后的网络片段中的:
ActivityCreated()
不要忘记从这些((Your_Activity) getActivity()).setWebView(webView);
设置webView
:
onCreateView()
答案 6 :(得分:1)
@OmidAmnivia回答你的应用是正确的崩溃解决方案
@Override
public void onBackPressed() {
if(webFragment.isInitialized && webFragment.canGoBack()){
webFragment.goBack();
}else{
super.onBackPressed();
}
}
您必须检查您的班级是否已初始化。
答案 7 :(得分:1)
@ RomanBlack的回答给了我正确的想法,但由于我们使用kotlin,我不得不稍微调整答案。
webView.setOnKeyListener { _, _, keyEvent ->
if (keyEvent.keyCode == KeyEvent.KEYCODE_BACK && !webView.canGoBack()) {
false
} else if (keyEvent.keyCode == KeyEvent.KEYCODE_BACK && keyEvent.action == MotionEvent.ACTION_UP) {
webView.goBack()
true
} else true
}
如果你想用回报来做,你必须添加如下内容:
return@setOnKeyListener true
答案 8 :(得分:1)
我创建了一个简单的界面:
public interface IOnBackPressed {
boolean onBackPressed();
}
在活动中:
public class MyActivity extends Activity {
@Override public void onBackPressed() {
Fragment fragment = getSupportFragmentManager().findFragmentById(R.id.main_container);
if (!(fragment instanceof IOnBackPressed) || !((IOnBackPressed) fragment).onBackPressed()) {
super.onBackPressed();
}
}
}
在片段中:
public class MyFragment extends Fragment implements IOnBackPressed {
@Override
public boolean onBackPressed() {
if (webview.canGoBack()) {
webview.goBack();
// backpress is not considered in the Activity
return true;
} else {
// activity will act normal
return false;
}
}
}
答案 9 :(得分:1)
这就是我在应用程序中所做的事情。我消耗了新闻发布事件,直到可以返回Web视图。一旦网络视图无法返回,我会向用户显示提示,如果他继续按回去,则应用程序将退出。
当webview无法返回时,它将为用户提供留在您应用中的机会。我觉得它更加用户友好:
//time passed between two back presses.
private val TIME_INTERVAL = 200
// variable to keep track of last back press
private var mBackPressed: Long = 0
webView!!.requestFocus()
webView.setOnKeyListener(View.OnKeyListener { v, keyCode, event ->
if (keyCode == KeyEvent.KEYCODE_BACK
&& event.action == MotionEvent.ACTION_UP
) {
if(webView.canGoBack()) {
//go back in previous page
webView.goBack()
return@OnKeyListener true
}
else
{
if (mBackPressed + TIME_INTERVAL > System.currentTimeMillis())
{ // dont consume back press and pass to super
return@OnKeyListener false
}
else {
// show hint for double back press
Toast.makeText(context, " Double Tap back button to exit the demo", Toast.LENGTH_SHORT).show();
mBackPressed = System.currentTimeMillis();
return@OnKeyListener true
}
}
}
return@OnKeyListener false
})
答案 10 :(得分:1)
BackPressedDispatcher有一个简单的方法
片段:
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
val callback = object : OnBackPressedCallback(true) {
override fun handleOnBackPressed() {
if(webView.canGoBack()){
webView.goBack()
} else {
isEnabled = false
requireActivity().onBackPressed()
}
}
}
requireActivity().onBackPressedDispatcher.addCallback(this,callback)
}
活动:
override fun onBackPressed() {
val fragment = supportFragmentManager.findFragmentByTag("WebViewFragment")
if (WebViewFragment::class.java.isInstance(fragment)) {
if (onBackPressedDispatcher.hasEnabledCallbacks()) {
onBackPressedDispatcher.onBackPressed()
return
}
super.onBackPressed()
}
}
也许这些代码可以改进,但是对我来说效果很好。有关更多信息,请访问here。
答案 11 :(得分:0)
首先回到片段
中 mContentView.setFocusableInTouchMode(true);
mContentView.requestFocus();
mContentView.setOnKeyListener( new OnKeyListener()
{
@Override
public boolean onKey( View v, int keyCode, KeyEvent event )
{
if( keyCode == KeyEvent.KEYCODE_BACK )
{
if (mWebView.canGoBack()) {
mWebView.goBack();
retune false;
} else {
return true;
}
}
return false;
}
} );
希望它能起作用。
答案 12 :(得分:0)
这对我来说很有效
public class FantasyFragment extends Fragment implements SwipeRefreshLayout.OnRefreshListener {
WebView webview;
SwipeRefreshLayout swipeLayout;
String currentUrl="https://www.stackoverflow.com/";
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
// Inflate the layout for this fragment
View root = inflater.inflate(R.layout.fragment_stadium, container, false);
swipeLayout = (SwipeRefreshLayout) root.findViewById(R.id.swipescreen);
swipeLayout.setOnRefreshListener(this);
return root;
}
@Override
public void onStart() {
super.onStart();
LoadWeb();
}
public void LoadWeb() {
webview = (WebView) getActivity().findViewById(R.id.webview786);
swipeLayout.setRefreshing(true);
webview.getSettings().setRenderPriority(WebSettings.RenderPriority.HIGH);
webview.getSettings().setCacheMode(WebSettings.LOAD_CACHE_ELSE_NETWORK);
webview.getSettings().setAppCacheEnabled(true);
WebSettings webSettings = webview.getSettings();
webSettings.setJavaScriptEnabled(true);
webSettings.setDatabaseEnabled(true);
webSettings.setLayoutAlgorithm(WebSettings.LayoutAlgorithm.NARROW_COLUMNS);
webSettings.setUseWideViewPort(true);
webSettings.setSavePassword(true);
webSettings.setSaveFormData(true);
webSettings.setEnableSmoothTransition(true);
webview.loadUrl(currentUrl);
webview.setWebViewClient(new WebViewClient() {
@Override
public void onReceivedError(WebView view, int errorCode, String description, String failingUrl) {
Snackbar.make(view, "Connection Error", Snackbar.LENGTH_LONG)
.setAction("Retry", new View.OnClickListener() {
@Override
public void onClick(View v) {
LoadWeb();
}
}).show();
}
@Override
public void onPageFinished(WebView view, String url) {
swipeLayout.setRefreshing(false);
currentUrl = url;
super.onPageFinished(view, url);
}
});
webview.canGoBack();
webview.setOnKeyListener(new View.OnKeyListener() {
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (keyCode == KeyEvent.KEYCODE_BACK
&& event.getAction() == MotionEvent.ACTION_UP
&& webview.canGoBack()) {
webview.goBack();
return true;
}
return false;
}
});
}
@Override
public void onRefresh() {
LoadWeb();
}
}