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@Override
public void onReceivedError(WebView view, WebResourceRequest request, WebResourceError error) {
super.onReceivedError(view, request, error);
PtrCLog.d("WebFragment", "onReceivedError: " + "");
}
Diese Methode wurde nicht zurückgesetzt. Ich habe nachgesehen und festgestellt, dass sie API23 erfordert. . Kennt jemand andere Möglichkeiten, an 404 zu kommen?
phpcn_u15822017-05-16 13:28:53
new WebViewClient() {
@Override
public void onReceivedError(WebView view, int errorCode, String description, String failingUrl) {
super.onReceivedError(view, errorCode, description, failingUrl);
if (errorCode == 404) {
doSomething();
}
}
@Override
public void onReceivedError(WebView view, WebResourceRequest request, WebResourceError error) {
super.onReceivedError(view, request, error);
if (Build.VERSION.SDK_INT > Build.VERSION_CODES.M) {
int errorCode = error.getErrorCode();
if (errorCode == 404) {
doSomething();
}
}
}
};
滿天的星座2017-05-16 13:28:53
在onPageStarted里边跑一个AsyncTask,在AsyncTask里用OkHttpClient之类的Http客户端对需要访问的URL进行一次请求,取得code
class WebViewStatusRequester extends AsyncTask<String, String, Integer> {
@Override
protected void onPreExecute() {
super.onPreExecute();
web.setVisibility(View.GONE);
}
@Override
protected void onPostExecute(Integer result) {
super.onPostExecute(result);
if(result == 1) {
web.setVisibility(View.VISIBLE);
} else if(result == 0) {
showLoadFail();
}
}
@Override
protected Integer doInBackground(String... params) {
String url = params[0];
if(url.substring(0, 4).equals("file") == false) {
try {
OkHttpClient client = new OkHttpClient();
Request request = new Request.Builder().url(url).build();
Response response = client.newCall(request).execute();
if(response.isSuccessful()) {
return 1;
} else {
Log.i("TAG", "fail code:" + response.code());
return 0;
}
} catch (IOException e) {
e.printStackTrace();
}
return 0;
}
return 1;
}
}
在这里实际上是Webview和OkHttpClient进行了加载,只是AsyncTask跑的时候把WebView隐藏起来了,确认是200了就显示,不是200就显示加载失败的页面