在网络视图中传递链接



当有人点击例如:eee.com,在我的Webview应用中打开它时,我正在尝试

公共类MainActivity扩展了AppCompatActivity{

public static WebView webView;
@RequiresApi(api = Build.VERSION_CODES.JELLY_BEAN_MR1)
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
webView = (WebView) findViewById(R.id.mainWebview);
webView.setWebViewClient(new WebViewClient());
WebSettings webSettings = webView.getSettings();
webSettings.setJavaScriptEnabled(true);
webView.loadUrl("https://www.eee.com/");
webView.setWebViewClient(new WebViewClient(){
@RequiresApi(api = Build.VERSION_CODES.N)
@Override
public boolean shouldOverrideUrlLoading(WebView view, String url) {
if(url.contains("eee.com")) {
view.loadUrl(url);
} else {
Intent i = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
startActivity(i);
}
return true;
}
});
}
@Override
protected void onResume() {
super.onResume();
Uri uri = this.getIntent().getData();
webView.loadUrl(uri.toString());
}

但是这个应用程序在打开它时崩溃了,我尝试了这个::

@Override
protected void onResume() {
super.onResume();
Uri uri = this.getIntent().getData();
if(getIntent().getData() != null){
webView.loadUrl(uri.toString());
}
}

应用程序工作,但在链接点击它只是打开应用程序而不通过链接

我自己用解决了这个问题

@Override
protected void onNewIntent(Intent intent)
{
super.onNewIntent(intent);
Uri x = intent.getData();
webView.loadUrl(x.toString());
}

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