Question

我需要在我的Android应用程序中检查请求是否是POST或GET中的shouldInterceptRequest。请参阅以下代码：

public class CustomWebViewClient extends WebViewClient {

    ...

    @Override
    public WebResourceResponse shouldInterceptRequest(WebView view, String url) {
        if ("request is POST")
            Log.d("CustomWebViewClient", "request is a POST");
        else if ("request is GET")
            Log.d("CustomWebViewClient", "request is a GET");

        ...
    }
}

是否可以在WebViewClient的范围内确定这一点？

Answer 1

可以通过扩展WebViewClient来实现，但它可能涉及比预期更多的工作。 WebViewClient中的回调方法由JNI调用，您无法调用它来获取标题和方法，所以最好的办法是使用JavaScript。

此解决方案基于kristof对http://code.google.com/p/android/issues/detail?id=9122#c21

的评论

1。创建一个名为post_interceptor.js的文件并将其放在res / raw

中

post_interceptor.js

HTMLFormElement.prototype._submit = HTMLFormElement.prototype.submit;
HTMLFormElement.prototype.submit = interceptor;

window.addEventListener('submit', function(e) {
    interceptor(e);
}, true);

function interceptor(e) {
var frm = e ? e.target : this;
    interceptor_onsubmit(frm);
    frm._submit();
}

function interceptor_onsubmit(f) {
    var jsonArr = [];
    for (i = 0; i < f.elements.length; i++) {
        var parName = f.elements[i].name;
        var parValue = f.elements[i].value;
        var parType = f.elements[i].type;

        jsonArr.push({
            name : parName,
            value : parValue,
            type : parType
        });
    }

    window.interception.customSubmit(JSON.stringify(jsonArr),
            f.attributes['method'] === undefined ? null : f.attributes['method'].nodeValue,
            f.attributes['enctype'] === undefined ? null : f.attributes['enctype'].nodeValue);
}

lastXmlhttpRequestPrototypeMethod = null;
XMLHttpRequest.prototype.reallyOpen = XMLHttpRequest.prototype.open;
XMLHttpRequest.prototype.open = function(method, url, async, user, password) {
    lastXmlhttpRequestPrototypeMethod = method;
    this.reallyOpen(method, url, async, user, password);
};
XMLHttpRequest.prototype.reallySend = XMLHttpRequest.prototype.send;
XMLHttpRequest.prototype.send = function(body) {
    window.interception.customAjax(lastXmlhttpRequestPrototypeMethod, body);
    lastXmlhttpRequestPrototypeMethod = null;
    this.reallySend(body);
};

2。创建一个名为JavascriptPostIntercept

的Java类

根据需要更改包/类名。

JavascriptPostIntercept.java

public class JavascriptPostIntercept {

    public interface JavascriptPostInterceptInterface {
        public void nextMessageIsAjaxRequest(AjaxRequestContents contents);
        public void nextMessageIsFormRequest(FormRequestContents contents);
    }

    private static String sInterceptHeader;

    private JavascriptPostInterceptInterface mClient;

    public static String getInterceptHeader() {
        if (sInterceptHeader == null) {
            // Assuming you have your own stream to string implementation
            sInterceptHeader = StringUtils.readInputStream(
                Resources.getSystem().openRawResource(R.raw.post_interceptor));
        }
        return sInterceptHeader;
    }

    public static class AjaxRequestContents {
        private String mMethod;
        private String mBody;

        public AjaxRequestContents(String method, String body) {
            mMethod = method;
            mBody = body;
        }

        public String getMethod() {
            return mMethod;
        }

        public String getBody() {
            return mBody;
        }
    }

    public static class FormRequestContents {
        private String mJson;
        private String mMethod;
        private String mEnctype;

        public FormRequestContents(String json, String method, String enctype) {
            mJson = json;
            mMethod = method;
            mEnctype = enctype;
        }

        public String getJson() {
            return mJson;
        }

        public String getMethod() {
            return mMethod;
        }

        public String getEnctype() {
            return mEnctype;
        }
    }

    public JavascriptPostIntercept(JavascriptPostInterceptInterface client) {
        mClient = client;
    }

    @JavascriptInterface
    public void customAjax(final String method, final String body) {
        mClient.nextMessageIsAjaxRequest(new AjaxRequestContents(method, body));
    }

    @JavascriptInterface
    public void customSubmit(String json, String method, String enctype) {
        mClient.nextMessageIsFormRequest(new FormRequestContents(json, method, enctype));
    }
}

3。创建WebViewClient子类

下面的代码只获取最新请求的HTTP方法，这看起来就足以满足您的要求，但很明显，AjaxRequestContents和FormSubmitContents上的其他方法可以让您访问帖子正文和其他内容（如果需要）它

class MyWebViewClient extends WebViewClient implements JavascriptPostIntercept.JavascriptPostInterceptInterface {
    private String mLastRequestMethod = "GET";

    /// evaluate post_interceptor.js after the page is loaded
    @Override
    public void onPageFinished(WebView view, String url) {
        view.loadUrl("javascript: " + JavascriptPostIntercept.getInterceptHeader());
    }

    @TargetApi(11)
    @Override
    public WebResourceResponse shouldInterceptRequest(WebView view, String url) {
        if (mLastRequestMethod.equals("POST")) {
            // do stuff here...
        } else if (mLastRequestMethod.equals("GET")) {
            // do other stuff here...
        }
        // return something here...
    }

    @Override
    public void nextMessageIsAjaxRequest(JavascriptPostIntercept.AjaxRequestContents contents) {
        mLastRequestMethod = contents.getMethod();
    }

    @Override
    public void nextMessageIsFormRequest(JavascriptPostIntercept.FormRequestContents contents) {
        mLastRequestMethod = contents.getMethod();
    }
}

4。创建适当的JS-Java链接

MyWebViewClient webViewClient = new MyWebViewClient();
mWebView.setWebViewClient(webViewClient);
mWebView.addJavascriptInterface(new JavascriptPostIntercept(webViewClient), "interception");

Answer 2

覆盖shouldInterceptRequest(WebView view, WebResourceRequest request)的{{1}}方法。

例如

WebViewClient

Android - 检查请求是GET还是POST

2 个答案: