[英]How to return to previous webview page from an activity
I have a WebView
and it will open another activity code in my Webview
client 我有一个WebView
,它将在Webview
客户端中打开另一个活动代码
if (Uri.parse(url).toString().contains(".mp3")) {
Intent intent = new Intent(MainActivity.this,Download.class);
intent.putExtra("Link",Uri.parse(url).toString());
intent.putExtra("Link2",myWebview.getOriginalUrl().toString());
startActivity(intent);
finish();
}
Now As this intent extra will pass the value of the url from where the next Download.class
is started 现在,由于此意图,额外的东西将从下一个Download.class
开始的地方传递url的值。
what I need is on onBackPressed()
it will open the url from where Download .class
was opened 我需要的是onBackPressed()
,它将从Download .class
位置打开URL。
This is my onBackPressed()
code in Download.class
这是我在Download.class
onBackPressed()
代码
Intent i = new Intent(Download.this, MainActivity.class);
i.setFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
startActivity(i);
finish();
Please refer bellow code code snippet. 请参阅下面的代码段。
@Override
public void onBackPressed() {
if(webView.canGoBack()){
webView.goBack();
} else {
super.onBackPressed();
}
}
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