android 下webview 如何判断404？

问题描述

我在使用

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

这个方法并没有被回掉查看了一下说是需要API23 才可以。。有大佬知道有别的方法获取404吗？

问题解答

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> {@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