如何在不读取整个请求的情况下发送响应？

Question

我正在基于Tomcat servlet和NIO创建服务。输入时有大的XML请求（约100 MB），通过HTML POST方法发送。我只想流前8千磅，然后立即发送响应到客户端。

public class A extends HttpServlet {
    @Override
    protected void service(HttpServletRequest rq, HttpServletResponse rs) {
        byte[] buffer = new byte[1024*8];
        try {
            rq.getInputStream().read(buffer);
            rs.setContentType("text/plain");
            rs.getOutputStream().write("Some Response".getBytes());
        } catch (IOException e) {
            e.printStackTrace();
        } 
     }
}

当我尝试发送小请求（内容中只有几行）时，套接字工作正常。

2016-02-01 10:44:52 Http11NioProtocol[DEBUG]套接字：[org.apache.tomcat.util.net.NioEndpoint$KeyAttachment@74f19fed：组织。阿帕奇。公猫util。网NioChannel@c478210：java。nio。频道。SocketChannel[connected local=/0:0:0:0:0:0:1:8080 remote=/0:0:0:0:0:0:0:0:1:63943]]，状态为：[OPEN\u READ]，状态为：[OPEN]

但，若我尝试发送更大的请求（超过100MB），客户端并没有响应。

2016-02-01 10:48:42 Http11NioProtocol[DEBUG]套接字：[org.apache.tomcat.util.net.NioEndpoint$KeyAttachment@2b36c88f：组织。阿帕奇。公猫util。网NioChannel@25f12241：java。nio。频道。SocketChannel[connected local=/0:0:0:0:0:0:1:8080 remote=/0:0:0:0:0:0:0:0:1:64079]]，状态为in:[OPEN\u READ]，状态为out:[CLOSED]

2016-02-01 10:48:42超文本传输协议-Nio-8080-exec-3

在我读取整个输入流请求之前，Tomcat不想打开套接字（State out： CLOSED）。

是否可以在不读取整个请求的情况下将响应发送到客户端？根据规范，我能够在请求的前8 KiB上找到有趣的me信息。

Answer 1

你的代码只是请求的一部分，

private static String convertStreamToString(InputStream is) throws IOException{
        int i = 0;
        byte[] buff = new byte[1024];
        StringBuilder sb = new StringBuilder();
        while (i != -1) {            
            i = is.read(buff);
            if(i != -1){
            sb.append(new String(buff,0,i));
            }
        }
        return sb.toString();
    }

所以，在你的代码中，它将是

public class A extends HttpServlet {
    @Override
    protected void service(HttpServletRequest rq, HttpServletResponse rs) {

        try {
            String response = convertStreamToString(rq.getInputStream());   
            rs.getOutputStream().write(response.getBytes());
        } catch (IOException e) {
            e.printStackTrace();
        } 
     }
     private static String convertStreamToString(InputStream is) throws IOException{
        int i = 0;
        byte[] buff = new byte[1024];
        StringBuilder sb = new StringBuilder();
        while (i != -1) {            
            i = is.read(buff);
            if(i != -1){
                sb.append(new String(buff,0,i));
            }
        }
        return sb.toString();
    }

}

或者您可以使用Scanner直接将流转换为字符串

    Scanner scanner = new Scanner(inputStream);
    StringBuilder sb = new StringBuilder();
    while (scanner.hasNext()) {            
        sb.append(scanner.next());
    }

因此，在您的代码中应该是

public class A extends HttpServlet {
    @Override
    protected void service(HttpServletRequest rq, HttpServletResponse rs) {

        try {
             Scanner scanner = new Scanner(rq.getInputStream());
             StringBuilder sb = new StringBuilder();
             while (scanner.hasNext()) {            
                 sb.append(scanner.next());
             }
            rs.getOutputStream().write(sb.toString().getBytes());
        } catch (IOException e) {
            e.printStackTrace();
        } 
     }

}

此代码

byte[] buffer = new byte[1024*8];
rq.getInputStream().read(buffer);

只读取1024个字节的请求并将其存储到缓冲区数组（1024个字节），方法读取将返回读取的字节数，因为根本不会读取1024个字节，如果请求只发送200个字节，824个字节将无用。因此，如果请求只发送200字节

int x = rq.getInputStream().read(buffer);

这个x将是200，所以要完全阅读，你需要做一个循环

// integer to read
int bytesRead = 0;
// loop until bytesRead is -1 (end of request), -1 means there are no bytes to read, so it will read until there are no bytes to read
while(bytesRead != -1){
    bytesRead = rq.getInputStream().read(buffer);
    if(bytesRead != -1){
    // process the bytes here , but only process bytes from 0 to bytesRead
     }
}