Question

Apache telnet使用InputStream，我希望read作为字符串（或类似字符串的数组）。

如何将InputStream转换为更容易（对我来说）处理的事情，StringBuffer或类似事件。

Answer 1

public static String toString(InputStream input)
                       throws IOException

来自doc：

此方法在内部缓冲输入，因此无需使用的BufferedInputStream。

Answer 2

您无法将InputStream转换为StringBuffer。但是如果您想使用StringBuffer，那么您可以将InputStream的输入作为int读取，并将其附加到StringBuffer，如下所示。

int i=fis.read(); //it will read Character from stream and returns ascii value of char.
StringBuffer sb=new StringBuffer();
sb.append((char)i); //This will cast int into character and append to StringBuffer.

Answer 3

这就在那里：

package teln;

import static java.lang.System.out;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.net.SocketException;
import org.apache.commons.net.telnet.TelnetClient;

public final class ConnectMUD {

    private static String EOL = "/[#]/";
    private static TelnetClient tc;

    public static void main(String[] args) throws SocketException, IOException {
        tc = new TelnetClient();
        tc.connect("some_mud.com", 123);
        readLines(tc.getInputStream());
    }

    private static void readLines(InputStream in) throws IOException {

        InputStreamReader is = new InputStreamReader(in);
        StringBuilder sb = new StringBuilder();
        BufferedReader br = new BufferedReader(is);
        String read = br.readLine();

        while (read != null) {
            out.println(read);
            sb.append(read);
            read = br.readLine();
            parseLine(read);
        }
    }

    private static void parseLine(String read) {
        login(read);
    }

    private static void login(String read) {
        //hmm, how do you know when you
        //get some funky, non-standard login?
        //look for an EOL of some sort?
    }
}

那里有一些非常大的字符串，但我只关注功能。只是尝试开始解析播放da MUD的行。

将InputStream转换为String（telnet）

3 个答案: