仅在某些请求Java上执行代码

Question

我正在为课程构建一个小API,老师为我们提供了一个指向教程的链接,该教程提供了一个实现Runnable的简单Web服务器.

我已经编写了一些代码来解析参数的参数(或者至少让我得到请求字符串)和一些代码将返回一些简单的xml.

但是我认为像favicon这样的某些请求会被发送,我认为它会弄乱我的代码.我把它包裹在一个if else中,但它似乎没有起作用.

=====已编辑因为基本字符串比较错误=====

package server;

import java.io.IOException;


import java.io.InputStream;
import java.io.OutputStream;
import java.net.Socket;
import java.util.*;
import java.io.*; 
import java.net.*;
import parkinglots.*;

/**
import our API 
*/
public class WorkerRunnable implements Runnable{

protected Socket clientSocket = null;
protected String serverText   = null;

public WorkerRunnable(Socket clientSocket, String serverText) {
    this.clientSocket = clientSocket;
    this.serverText   = serverText;
}

public Boolean authenticateAPI(String key){
    //Authenticate Key against Stored Keys

    //TODO: Create Stored Keys and Compare
    return true;
}

public void run() {
    try {
        InputStream input = clientSocket.getInputStream();
        OutputStream output = clientSocket.getOutputStream();
        long time = System.currentTimeMillis();


        //TODO: Parse args and output different formats and Authentication
        //Parse URL Arguments
        BufferedReader in = new BufferedReader( 
                new InputStreamReader(clientSocket.getInputStream(), "8859_1"));

        String request = in.readLine();

        //Server gets Favicon Request so skip that and goto args
        System.out.println(request);
        if ( !"GET /favicon.ico HTTP/1.1".equals(request) && !"GET / HTTP/1.1".equals(request) && !"".equals(request) ){

            String format = "", apikey ="";

            System.out.println("I am Here");

            String request_location = request.split(" ")[1];
            String request_args = request_location.replace("/","");
            request_args = request_args.replace("?","");
            String[] queries = request_args.split("&");

            System.out.println(queries[0]);

            for ( int i = 0; i < queries.length; i++ ){
                if( queries[i].equals("format") ){
                    format = queries[i].split("=")[1];
                }
                else if( queries[i].equals("apikey") ){
                    apikey = queries[i].split("=")[1];
                }
            }

            if( apikey.equals("") ){
                apikey = "None";
            }

            if( format.equals("") ){
                format = "xml";
            }

            System.out.println("format: "+format);

            Boolean auth = authenticateAPI(apikey);

            if ( auth ){
                if ( format.equals("xml")){
                    // Retrieve XML Document
                    String xml = LotFromDB.getParkingLotXML();
                    output.write((xml).getBytes());
                }else{
                    //Retrieve JSON
                    String json = LotFromDB.getParkingLotJSON();
                    output.write((json).getBytes());
                }
            }else{
                output.write(("Access Denied - User is Not Authenticated").getBytes());
            }
        }else{
            output.write(("Access Denied Must Pass API Key").getBytes());
        }

        output.close();
        input.close();
        System.out.println("Request processed: " + time);
    } catch (IOException e) {
        //report exceptions
        e.printStackTrace();
    }
}
}

控制台输出我得到

GET /?format=json HTTP/1.1
I am Here
format=json
json
format: xml

它总是返回XML.

这是我第一次接触编写Web服务器和处理Java中的网络,这让我很沮丧,所以这里的任何建议都非常感激.

在得到其他人的帮助后,原来是这里的一部分.我正在比较字符串的错误部分,所以它永远不匹配

                if( queries[i].split("=")[0].equals("format") ){
                    format = queries[i].split("=")[1];
                }
                else if( queries[i].split("=")[0].equals("apikey") ){
                    apikey = queries[i].split("=")[1];
                }

Answer 1

对于所有你的字符串比较,你需要使用equals而不是 ==

示例:if( format.equals("xml"))在条件块中还检查是否request为null

if (request != null && !"GET /favicon.ico HTTP/1.1".equals(request) 
  && !"GET / HTTP/1.1".equals(request) &&  !"".equals(request) )