Java常用逻辑-FtpPath的处理

import java.net.URI;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

/**
 * @author wanghao
 */
public class Test006 {

    public static void main(String[] args) {
        String ftpPath = "ftp://Huser:wang@127.0.0.1/xml/212321.xml";
        readFtpPath1(ftpPath);
        readFtpPath2(ftpPath);

    }

    /**
     * 使用URI类获取ftp地址的信息
     * @param ftpPath   ftp地址
     */
    private static void readFtpPath1(String ftpPath){
        try {
            URI uri = new URI(ftpPath);
            //ftp
            String ftp = uri.getScheme();
            //Huser:password
            String userPwd = uri.getUserInfo();
            //127.0.0.1
            String host = uri.getHost();
            //21
            int port = uri.getPort()>0 ? uri.getPort():21;
            // /xml/212321.xml
            String path = uri.getPath();
            System.out.println(uri.toString());
        } catch (Exception e) {
            throw new RuntimeException(e);
        }
    }

    /**
     * 使用正则获取ftp地址里面的信息
     * @param ftpPath   ftp地址
     */
    private static void readFtpPath2(String ftpPath) {
        String pattern = "(ftp://)(.*?)(:)(.*?)(@)(.*?)(/.*)";
        Pattern regex = Pattern.compile(pattern);
        Matcher matcher = regex.matcher(ftpPath);
        if (matcher.find()) {
            String name = matcher.group(2);
            String password = matcher.group(4);
            //127.0.0.1:21
            String ipPort = matcher.group(6);
            String path = matcher.group(7);
        }else{
            System.out.println("ftpPath error ");
        }
    }
}