Question

HttpExchange对象的getRequestBody方法返回一个InputStream。正确阅读“身体”还有很多工作要做。它是一个Java库+对象+方法又领先一步并将主体（在服务器端）作为即用型Java字符串返回？

谢谢！

Answer 1

InputStreamReader isr =  new InputStreamReader(t.getRequestBody(),"utf-8");
BufferedReader br = new BufferedReader(isr);

// From now on, the right way of moving from bytes to utf-8 characters:

int b;
StringBuilder buf = new StringBuilder(512);
while ((b = br.read()) != -1) {
    buf.append((char) b);
}

br.close();
isr.close();

// The resulting string is: buf.toString()
// and the number of BYTES (not utf-8 characters) from the body is: buf.length()

Answer 2

如果您使用的是Spring MVC，则可以对类型为String的方法参数使用@RequestBody注释。例如。

@RequestMapping(value = "/something", method = RequestMethod.POST)
public void doSomething(@RequestBody String requestBodyString) {
    // does something..
}

Answer 3

您可以使用Commons IO的org.apache.commons.io.IOUtils.toString(InputStream, String)在一行中执行此操作。（虽然它可能不适用于HTTP keep-alive）

编辑：

如果你想直接使用JSON，有一堆Web服务堆栈可以为你进行解组。尝试

春天：http://www.cribbstechnologies.com/2011/04/08/spring-mvc-ajax-web-services-part-2-attack-of-the-json-post/

CXF / JAX-RS：http://cxf.apache.org/docs/jax-rs-data-bindings.html#JAX-RSDataBindings-JSONsupport

Answer 4

你试过这个吗？

 InputStreamReader isr =  new InputStreamReader(exchange.getRequestBody(),"utf-8");
 BufferedReader br = new BufferedReader(isr);
 String value = br.readLine();

Answer 5

在HttpHandler中：

InputStreamReader isr = new InputStreamReader(he.getRequestBody(), "utf-8");
        BufferedReader br = new BufferedReader(isr);

        int b;
        StringBuilder buf = new StringBuilder();
        while ((b = br.read()) != -1) {
            buf.append((char) b);
        }

        br.close();
        isr.close();
System.out.println(buf.toString());

如何在服务器端将HTTP POST请求主体作为Java String？

5 个答案: