Question

我有一个涉及httprequest的错误，有时会发生这种错误，因此我想在发生这种情况时记录HttpGet和HttpPost请求的内容。

所以，让我们说，我像这样创建HttpGet：

HttpGet g = new HttpGet();
g.setURI(new URI("http://www.google.com"));
g.setHeader("test", "hell yeah");

这是我想要的字符串表示：

GET / HTTP/1.1
Host: www.google.com
test: hell yeah

根据帖子请求，我还想获取内容字符串。

在java for android中最简单的方法是什么？

Answer 1

您可以使用以下方式打印请求类型：

request.getMethod();

您可以打印所有标题：

Enumeration<String> headerNames = request.getHeaderNames();
while(headerNames.hasMoreElements()) {
  String headerName = headerNames.nextElement();
  System.out.println("Header Name - " + headerName + ", Value - " + request.getHeader(headerName));
}

要打印所有请求参数，请使用：

Enumeration<String> params = request.getParameterNames(); 
while(params.hasMoreElements()){
 String paramName = params.nextElement();
 System.out.println("Parameter Name - "+paramName+", Value - "+request.getParameter(paramName));
}

request是HttpServletRequest

的实例

您可以根据需要美化输出。

Answer 2

有助于记录的更多细节

    String client = request.getRemoteAddr();
    logger.info("###### requested client: {} , Session ID : {} , URI :" + request.getMethod() + ":" + request.getRequestURI() + "", client, request.getSession().getId());

    Map params = request.getParameterMap();
    Iterator i = params.keySet().iterator();
    while (i.hasNext()) {
        String key = (String) i.next();
        String value = ((String[]) params.get(key))[0];
        logger.info("###### Request Param Name : {} , Value :  {} ", key, value);
    }

Answer 3

这应该对调试更有帮助。 @Juned Ahsan的回答不会指定完整的URL，也不会打印多个标题/参数。

private String httpServletRequestToString(HttpServletRequest request) {
    StringBuilder sb = new StringBuilder();

    sb.append("Request Method = [" + request.getMethod() + "], ");
    sb.append("Request URL Path = [" + request.getRequestURL() + "], ");

    String headers =
        Collections.list(request.getHeaderNames()).stream()
            .map(headerName -> headerName + " : " + Collections.list(request.getHeaders(headerName)) )
            .collect(Collectors.joining(", "));

    if (headers.isEmpty()) {
        sb.append("Request headers: NONE,");
    } else {
        sb.append("Request headers: ["+headers+"],");
    }

    String parameters =
        Collections.list(request.getParameterNames()).stream()
            .map(p -> p + " : " + Arrays.asList( request.getParameterValues(p)) )
            .collect(Collectors.joining(", "));             

    if (parameters.isEmpty()) {
        sb.append("Request parameters: NONE.");
    } else {
        sb.append("Request parameters: [" + parameters + "].");
    }

    return sb.toString();
}

Answer 4

如果您想要内容字符串，并且此字符串没有参数，则可以使用

    String line = null;
    BufferedReader reader = request.getReader();
    while ((line = reader.readLine()) != null){
        System.out.println(line);
    }

Answer 5

万一有人也想像我一样甩掉回应。我避免抛弃响应主体。以下代码只是转储StatusCode和Headers。

static private String dumpResponse(HttpServletResponse resp){
    StringBuilder sb = new StringBuilder();

    sb.append("Response Status = [" + resp.getStatus() + "], ");
    String headers = resp.getHeaderNames().stream()
                    .map(headerName -> headerName + " : " + resp.getHeaders(headerName) )
                    .collect(Collectors.joining(", "));

    if (headers.isEmpty()) {
        sb.append("Response headers: NONE,");
    } else {
        sb.append("Response headers: "+headers+",");
    }

    return sb.toString();
}

Answer 6

通过流在一行中重写@Juned Ahsan解决方案（标头的处理方式相同）：

public static String printRequest(HttpServletRequest req) {
    String params = StreamSupport.stream(
            ((Iterable<String>) () -> req.getParameterNames().asIterator()).spliterator(), false)
            .map(pName -> pName + '=' + req.getParameter(pName))
            .collect(Collectors.joining("&"));
    return req.getRequestURI() + '?' + params;
}

另请参阅how to convert an iterator to a stream解决方案。

如何打印httprequest请求的内容？

6 个答案: