如何将带有“重定向”代码的WebResourceResponse传递到WebViewClient.shouldInterceptRequest中的Android WebView？

Question

为了访问响应头（获取特定头值），我们必须在WebView中拦截HTTP请求，自己执行HTTP资源的下载并返回结果作为实例WebResourceResponse：

public WebResourceResponse shouldInterceptRequest(WebView view, WebResourceRequest request)
{
    return new WebResourceResponse("text/plain", "UTF-8", 302, ...); // 302 is invalid (not supported) value
}

但不接受300-399 范围内的代码：

2019-12-02 16:48:05.710 1812-1903 W/System.err: java.lang.IllegalArgumentException: statusCode can't be in the [300, 399] range.
2019-12-02 16:48:05.712 1812-1903 W/System.err:     at android.webkit.WebResourceResponse.setStatusCodeAndReasonPhrase(WebResourceResponse.java:134)
2019-12-02 16:48:05.712 1812-1903 W/System.err:     at android.webkit.WebResourceResponse.<init>(WebResourceResponse.java:76)
...
2019-12-02 16:48:05.714 1812-1903 W/System.err:     at xl.a(PG:56)
2019-12-02 16:48:05.714 1812-1903 W/System.err:     at aeW.a(PG:9)
2019-12-02 16:48:05.714 1812-1903 W/System.err:     at org.chromium.android_webview.AwContentsBackgroundThreadClient.shouldInterceptRequestFromNative(PG:11)
2019-12-02 16:48:05.718 1812-1903 A/chromium: [FATAL:jni_android.cc(256)] Please include Java exception stack in crash report

有没有正确的方法让 WebView 接受重定向响应？

附言。我能够使用 Java 反射来解决它，并将值直接注入到字段中（不是通过带有验证的 ctor），但它似乎非常脆弱（尽管它实际上有效）。

Answer 1

我遇到了同样的问题。

解决方案1：加载新的url

public WebResourceResponse shouldInterceptRequest(WebView view, WebResourceRequest request) {
   String statusCode = 302; //get it from your response;
   if (statusCode >= 300 && statusCode <= 399) {
       final String newUrl = "https://example.com";
       final myWebview = view;
       view.post(new Runnable() {
           @Override public void run() {
               myWebview.loadUrl(newUrl);
           }
       });
       WebResourceResponse nullRes = new WebResourceResponse("text/html", "utf-8", new ByteArrayInputStream("".getBytes()));
       return nullRes;
   }
}

解决方案2：返回html模板

public WebResourceResponse shouldInterceptRequest(WebView view, WebResourceRequest request) {
    String statusCode = 302; //get it from your response;
    if (statusCode >= 300 && statusCode <= 399) {
        String newUrl = "https://example.com";
        String content = "<script>location.href = '" + newUrl + "'</script>";
        WebResourceResponse redirectRes = new WebResourceResponse("text/html", "utf-8", new ByteArrayInputStream(content.getBytes()));
        return redirectRes;
    }
}