Android的Webview在截取WebResourceResponse时怎样正常处理页面redirect重定向

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在研究Android WebView的时候，有时需要截取页面响应，修改些信息再返回给WebView处理，期间发现如果页面返回302重定向到一个新的链接，而你正常返回302会有一个错误提示:

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statusCode can't be in the [300, 399] range.

你说气人不气人…

所以只有曲线救国了，经过测试有2种方法可以实现：

方法1：直接让Webview加载新的链接：
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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;
   }
}
方法2：修改WebResourceResponse，返回一个用js跳转新链接的html模板
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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;
    }
}

这两个方法我也在Stackoverflow上做了解答：https://stackoverflow.com/questions/59965544/how-to-pass-webresourceresponse-with-redirect-code-to-android-webview-in-webvi/64605504#64605504