17

我正在使用 HttpRequestBase,我想在使用之前将请求完全记录到日志文件中。

默认 toString 只返回请求行,我想打印所有标题、参数、请求正文等...

有没有办法这样做?

4

2 回答 2

17

HttpRequestBase对象(HttpGetHttpPost等)包含有关标头、参数的信息,实现类包含主体,但它实际上并未序列化为String. 当HttpClient实际发送请求时会发生这种情况。

您可以使用 http 组件日志记录配置

或者您可以调用适当的方法并自己做。

HttpRequestBase base = new HttpGet("www.google.com");
Header[] headers = base.getAllHeaders();
// iterate and print

对于正文,您需要强制转换为您的实现类并获取HttpEntity,如果它有的话。

HttpEntity entity = ((HttpPost)base).getEntity(); // example

并打印它(它的InputStream内容)。注意:这可能会消耗实体。

完整示例

HttpPost post = new HttpPost("www.google.com");
post.setHeader(new BasicHeader("User-Agent", "random client"));
HttpEntity entity = new StringEntity("yellaworld");
post.setEntity(entity);
Header[] headers = post.getAllHeaders();
String content = EntityUtils.toString(entity);

System.out.println(post.toString());
for (Header header : headers) {
    System.out.println(header.getName() + ": " + header.getValue());
}
System.out.println();
System.out.println(content);

印刷

POST www.google.com HTTP/1.1
User-Agent: random client

yellaworld
于 2013-09-11T14:48:44.097 回答
2

这有效

private void printRequest() {
            System.out.println("receive " + httpRequest.getMethod() +" notification for "+ httpRequest.getRequestURI());


            System.out.println(" \n\n Headers");

            Enumeration headerNames = httpRequest.getHeaderNames();
            while(headerNames.hasMoreElements()) {
                String headerName = (String)headerNames.nextElement();
                System.out.println(headerName + " = " + httpRequest.getHeader(headerName));
            }

            System.out.println("\n\nParameters");

            Enumeration params = httpRequest.getParameterNames();
            while(params.hasMoreElements()){
                String paramName = (String)params.nextElement();
                System.out.println(paramName + " = " + httpRequest.getParameter(paramName));
            }

            System.out.println("\n\n Row data");
            System.out.println(extractPostRequestBody(httpRequest));
        }

        static String extractPostRequestBody(HttpServletRequest request) {
            if ("POST".equalsIgnoreCase(request.getMethod())) {
                Scanner s = null;
                try {
                    s = new Scanner(request.getInputStream(), "UTF-8").useDelimiter("\\A");
                } catch (IOException e) {
                    e.printStackTrace();
                }
                return s.hasNext() ? s.next() : "";
            }
            return "";
        }
于 2015-05-29T19:27:28.170 回答