問(wèn)題描述

我在使用

@Overridepublic void onReceivedError(WebView view, WebResourceRequest request, WebResourceError error) { super.onReceivedError(view, request, error); PtrCLog.d('WebFragment', 'onReceivedError: ' + '');}

這個(gè)方法并沒(méi)有被回掉查看了一下說(shuō)是需要API23 才可以。。有大佬知道有別的方法獲取404嗎？

問(wèn)題解答

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(); }} }};回答2：

在onPageStarted里邊跑一個(gè)AsyncTask，在AsyncTask里用OkHttpClient之類(lèi)的Http客戶(hù)端對(duì)需要訪(fǎng)問(wèn)的URL進(jìn)行一次請(qǐng)求，取得code

class WebViewStatusRequester extends AsyncTask<String, String, Integer> {@Overrideprotected void onPreExecute() { super.onPreExecute(); web.setVisibility(View.GONE);}@Overrideprotected void onPostExecute(Integer result) { super.onPostExecute(result); if(result == 1) {web.setVisibility(View.VISIBLE); } else if(result == 0) {showLoadFail(); }}@Overrideprotected 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;} }

在這里實(shí)際上是Webview和OkHttpClient進(jìn)行了加載，只是AsyncTask跑的時(shí)候把WebView隱藏起來(lái)了，確認(rèn)是200了就顯示，不是200就顯示加載失敗的頁(yè)面