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Android:WebView shouldInterceptRequest 不在 WebView 中添加 RequestProperties

转载 作者:塔克拉玛干 更新时间:2023-11-02 21:36:28 28 4
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我正在使用 shouldInterceptRequest 拦截来自 webview 的请求

下面是我返回 WebResourceResponse 的代码

@TargetApi(Build.VERSION_CODES.LOLLIPOP)
private static WebResourceResponse handleRequestViaUrlOnly(WebResourceRequest webResourceRequest){
String url = webResourceRequest.getUrl().toString();
Log.i("intercepting req....!!!", url);
String ext = MimeTypeMap.getFileExtensionFromUrl(url);
String mime = MimeTypeMap.getSingleton().getMimeTypeFromExtension(ext);

try {
HttpURLConnection conn = (HttpURLConnection) new URL(url).openConnection();
conn.setRequestProperty("Sample-Header", "hello");
conn.setDoOutput(true);
conn.setDoInput(true);
conn.setUseCaches(false);
return new WebResourceResponse(mime, "UTF-8", conn.getInputStream());
} catch (IOException e) {
e.printStackTrace();
}

return null;
}

我在我的 CustomWebViewClient 中调用这个方法

class CustomWebViewClient extends WebViewClient {

@TargetApi(Build.VERSION_CODES.LOLLIPOP)
@Override
public WebResourceResponse shouldInterceptRequest(WebView view, WebResourceRequest request) {
return handleRequestViaUrlOnly(request);
}
}

但是,当我在 chrome://inspect/#devices 中检查来自 WebView 远程调试器的请求 header 时。

我添加的额外 RequestProperty 不存在。

conn.setRequestProperty("Sample-Header", "hello");

WebView 的请求 header 中不存在 Sample-Header。

我错过了什么吗?我将不胜感激。

最佳答案

所以问题是,当您传递 conn.getInputStream() 时,它只提供数据。可以通过 conn.getHeaderFields() 提取响应 header 。此外,除非服务器支持它并且CORS,否则您将无法取回额外的 header 。没有参与。这是连接的 wireshark 输出

GET /~fdc/sample.html HTTP/1.1
Sample-Header: hello
Content-Type: application/x-www-form-urlencoded
User-Agent: Dalvik/2.1.0 (Linux; U; Android 7.1; Android SDK built for x86_64 Build/NPF26K)
Host: www.columbia.edu
Connection: Keep-Alive
Accept-Encoding: gzip
Content-Length: 0

HTTP/1.1 200 OK
Date: Wed, 01 Mar 2017 09:06:58 GMT
Server: Apache
Last-Modified: Thu, 22 Apr 2004 15:52:25 GMT
Accept-Ranges: bytes
Vary: Accept-Encoding,User-Agent
Content-Encoding: gzip
Content-Length: 8664
Keep-Alive: timeout=15, max=99
Connection: Keep-Alive
Content-Type: text/html

如您所见,响应中没有 Sample-Header: hello header 。

下面是从响应构建 WebResourceResponse header 并将您的自定义 header 附加到它的简单代码:

webView.setWebViewClient(new WebViewClient() {
private Map<String, String> convertResponseHeaders(Map<String, List<String>> headers) {
Map<String, String> responseHeaders = new HashMap<>();
responseHeaders.put("Sample-Header", "hello");

for (Map.Entry<String, List<String>> item : headers.entrySet()) {
List<String> values = new ArrayList<String>();

for (String headerVal : item.getValue()) {
values.add(headerVal);
}
String value = StringUtil.join(values, ",");
Log.e(TAG, "processRequest: " + item.getKey() + " : " + value);

responseHeaders.put(item.getKey(), value);
}

return responseHeaders;
}

@RequiresApi(api = Build.VERSION_CODES.LOLLIPOP)
@Override
public WebResourceResponse shouldInterceptRequest(WebView view, WebResourceRequest request) {
final String method = request.getMethod();
final String url = request.getUrl().toString();
Log.d(TAG, "processRequest: " + url + " method " + method);
String ext = MimeTypeMap.getFileExtensionFromUrl(url);
String mime = MimeTypeMap.getSingleton().getMimeTypeFromExtension(ext);

try {
HttpURLConnection conn = (HttpURLConnection)new URL(url).openConnection();
conn.setRequestMethod(method);
conn.setRequestProperty("Sample-Header", "hello");
conn.setDoInput(true);
conn.setUseCaches(false);

Map<String, String> responseHeaders = convertResponseHeaders(conn.getHeaderFields());

responseHeaders.put("Sample-Header", "hello");

return new WebResourceResponse(
mime,
conn.getContentEncoding(),
conn.getResponseCode(),
conn.getResponseMessage(),
responseHeaders,
conn.getInputStream()
);

} catch (Exception e) {
Log.e(TAG, "shouldInterceptRequest: " + e);
}
return null;
}
});

关于Android:WebView shouldInterceptRequest 不在 WebView 中添加 RequestProperties,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41869163/

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