向WebView资源请求添加自定义标头 - android

Question

我需要为来自WebView的每个请求添加自定义标头.我知道loadURL有参数extraHeaders,但这些只适用于初始请求.所有后续请求都不包含标头.我查看了所有覆盖内容WebViewClient,但没有任何内容允许向资源请求添加标头 - onLoadResource(WebView view, String url).任何帮助都会很精彩.

谢谢,雷

Answer 1

尝试

loadUrl(String url, Map<String, String> extraHeaders)

要将标头添加到资源加载请求,请创建自定义WebViewClient并覆盖:

API 24+:
WebResourceResponse shouldInterceptRequest(WebView view, WebResourceRequest request)
or
WebResourceResponse shouldInterceptRequest(WebView view, String url)

Answer 2

您需要使用WebViewClient.shouldInterceptRequest拦截每个请求

每次拦截时,您需要获取网址,自己发出此请求,然后返回内容流:

WebViewClient wvc = new WebViewClient() {
    @Override
    public WebResourceResponse shouldInterceptRequest(WebView view, String url) {

        try {
            DefaultHttpClient client = new DefaultHttpClient();
            HttpGet httpGet = new HttpGet(url);
            httpGet.setHeader("MY-CUSTOM-HEADER", "header value");
            httpGet.setHeader(HttpHeaders.USER_AGENT, "custom user-agent");
            HttpResponse httpReponse = client.execute(httpGet);

            Header contentType = httpReponse.getEntity().getContentType();
            Header encoding = httpReponse.getEntity().getContentEncoding();
            InputStream responseInputStream = httpReponse.getEntity().getContent();

            String contentTypeValue = null;
            String encodingValue = null;
            if (contentType != null) {
                contentTypeValue = contentType.getValue();
            }
            if (encoding != null) {
                encodingValue = encoding.getValue();
            }
            return new WebResourceResponse(contentTypeValue, encodingValue, responseInputStream);
        } catch (ClientProtocolException e) {
            //return null to tell WebView we failed to fetch it WebView should try again.
            return null;
        } catch (IOException e) {
             //return null to tell WebView we failed to fetch it WebView should try again.
            return null;
        }
    }
}

Webview wv = new WebView(this);
wv.setWebViewClient(wvc);

如果您的最低API目标是21级,则可以使用新的shouldInterceptRequest,它会为您提供其他请求信息(例如标题),而不仅仅是URL.

Answer 3

也许我的反应很晚,但它涵盖了API 以下和21级以上.

要添加标头,我们应该拦截每个请求并使用所需的标头创建新的请求.

因此我们需要覆盖在两种情况下调用的shouldInterceptRequest方法:1.用于API直到21级; 2.对于API级别21+

    webView.setWebViewClient(new WebViewClient() {

        // Handle API until level 21
        @SuppressWarnings("deprecation")
        @Override
        public WebResourceResponse shouldInterceptRequest(WebView view, String url) {

            return getNewResponse(url);
        }

        // Handle API 21+
        @TargetApi(Build.VERSION_CODES.LOLLIPOP)
        @Override
        public WebResourceResponse shouldInterceptRequest(WebView view, WebResourceRequest request) {

            String url = request.getUrl().toString();

            return getNewResponse(url);
        }

        private WebResourceResponse getNewResponse(String url) {

            try {
                OkHttpClient httpClient = new OkHttpClient();

                Request request = new Request.Builder()
                        .url(url.trim())
                        .addHeader("Authorization", "YOU_AUTH_KEY") // Example header
                        .addHeader("api-key", "YOUR_API_KEY") // Example header
                        .build();

                Response response = httpClient.newCall(request).execute();

                return new WebResourceResponse(
                        null,
                        response.header("content-encoding", "utf-8"),
                        response.body().byteStream()
                );

            } catch (Exception e) {
                return null;
            }

        }
   });

如果应该处理响应类型,您可以更改

        return new WebResourceResponse(
                null, // <- Change here
                response.header("content-encoding", "utf-8"),
                response.body().byteStream()
        );

至

        return new WebResourceResponse(
                getMimeType(url), // <- Change here
                response.header("content-encoding", "utf-8"),
                response.body().byteStream()
        );

并添加方法

        private String getMimeType(String url) {
            String type = null;
            String extension = MimeTypeMap.getFileExtensionFromUrl(url);

            if (extension != null) {

                switch (extension) {
                    case "js":
                        return "text/javascript";
                    case "woff":
                        return "application/font-woff";
                    case "woff2":
                        return "application/font-woff2";
                    case "ttf":
                        return "application/x-font-ttf";
                    case "eot":
                        return "application/vnd.ms-fontobject";
                    case "svg":
                        return "image/svg+xml";
                }

                type = MimeTypeMap.getSingleton().getMimeTypeFromExtension(extension);
            }

            return type;
        }

Answer 4

如前所述,您可以这样做:

 WebView  host = (WebView)this.findViewById(R.id.webView);
 String url = "<yoururladdress>";

 Map <String, String> extraHeaders = new HashMap<String, String>();
 extraHeaders.put("Authorization","Bearer"); 
 host.loadUrl(url,extraHeaders);

我测试了这个,并使用MVC控制器,我扩展了授权属性以检查标头,标头就在那里.

Answer 5

这对我有用:

1. 首先,您需要创建方法,该方法将返回您要添加到请求的标头:

   private Map<String, String> getCustomHeaders()
   {
       Map<String, String> headers = new HashMap<>();
       headers.put("YOURHEADER", "VALUE");
       return headers;
   }
   

2. 其次,您需要创建WebViewClient:

   private WebViewClient getWebViewClient()
   {
   
       return new WebViewClient()
       {
   
       @Override
       @TargetApi(Build.VERSION_CODES.LOLLIPOP)
       public boolean shouldOverrideUrlLoading(WebView view, WebResourceRequest request)
       {
           view.loadUrl(request.getUrl().toString(), getCustomHeaders());
           return true;
       }
   
       @Override
       public boolean shouldOverrideUrlLoading(WebView view, String url)
       {
           view.loadUrl(url, getCustomHeaders());
           return true;
       }
   };
   }
   

3. 将WebViewClient添加到WebView:

   webView.setWebViewClient(getWebViewClient());
   

希望这可以帮助.

Answer 6

这是使用 HttpUrlConnection 的实现：

class CustomWebviewClient : WebViewClient() {
    private val charsetPattern = Pattern.compile(".*?charset=(.*?)(;.*)?$")

    override fun shouldInterceptRequest(view: WebView, request: WebResourceRequest): WebResourceResponse? {
        try {
            val connection: HttpURLConnection = URL(request.url.toString()).openConnection() as HttpURLConnection
            connection.requestMethod = request.method
            for ((key, value) in request.requestHeaders) {
                connection.addRequestProperty(key, value)
            }

            connection.addRequestProperty("custom header key", "custom header value")

            var contentType: String? = connection.contentType
            var charset: String? = null
            if (contentType != null) {
                // some content types may include charset => strip; e. g. "application/json; charset=utf-8"
                val contentTypeTokenizer = StringTokenizer(contentType, ";")
                val tokenizedContentType = contentTypeTokenizer.nextToken()

                var capturedCharset: String? = connection.contentEncoding
                if (capturedCharset == null) {
                    val charsetMatcher = charsetPattern.matcher(contentType)
                    if (charsetMatcher.find() && charsetMatcher.groupCount() > 0) {
                        capturedCharset = charsetMatcher.group(1)
                    }
                }
                if (capturedCharset != null && !capturedCharset.isEmpty()) {
                    charset = capturedCharset
                }

                contentType = tokenizedContentType
            }

            val status = connection.responseCode
            var inputStream = if (status == HttpURLConnection.HTTP_OK) {
                connection.inputStream
            } else {
                // error stream can sometimes be null even if status is different from HTTP_OK
                // (e. g. in case of 404)
                connection.errorStream ?: connection.inputStream
            }
            val headers = connection.headerFields
            val contentEncodings = headers.get("Content-Encoding")
            if (contentEncodings != null) {
                for (header in contentEncodings) {
                    if (header.equals("gzip", true)) {
                        inputStream = GZIPInputStream(inputStream)
                        break
                    }
                }
            }
            return WebResourceResponse(contentType, charset, status, connection.responseMessage, convertConnectionResponseToSingleValueMap(connection.headerFields), inputStream)
        } catch (e: Exception) {
            e.printStackTrace()
        }
        return super.shouldInterceptRequest(view, request)
    }

    private fun convertConnectionResponseToSingleValueMap(headerFields: Map<String, List<String>>): Map<String, String> {
        val headers = HashMap<String, String>()
        for ((key, value) in headerFields) {
            when {
                value.size == 1 -> headers[key] = value[0]
                value.isEmpty() -> headers[key] = ""
                else -> {
                    val builder = StringBuilder(value[0])
                    val separator = "; "
                    for (i in 1 until value.size) {
                        builder.append(separator)
                        builder.append(value[i])
                    }
                    headers[key] = builder.toString()
                }
            }
        }
        return headers
    }
}

请注意，这不适用于 POST 请求，因为 WebResourceRequest 不提供 POST 数据。有一个请求数据 - WebViewClient 库，它使用 JavaScript 注入解决方法来拦截 POST 数据。