在Java中打开URL以获取内容 [英] Open URL in Java to get the content

问题描述

我正在寻找机会在java中打开网址。

I´m searching for a opportunity to open a url in java.

URL url = new URL("http://maps.google.at/maps?saddr=4714&daddr=Marchtrenk&hl=de");
    InputStream is = url.openConnection().getInputStream();

    BufferedReader reader = new BufferedReader( new InputStreamReader( is )  );

    String line = null;
    while( ( line = reader.readLine() ) != null )  {
       System.out.println(line);
    }
    reader.close();

我找到了这种方式。

I在我的程序中添加它并发生以下错误。

I added it in my program and the following error occurred.

The method openConnection() is undefined for the type URL

（通过url.openConnection（））

(by url.openConnection())

什么是我的问题？

我使用带有servlet的tomcat-server，...

I use a tomcat-server with servlets, ...

推荐答案

public class UrlContent{
    public static void main(String[] args) {

        URL url;

        try {
            // get URL content

            String a="http://localhost:8080/TestWeb/index.jsp";
            url = new URL(a);
            URLConnection conn = url.openConnection();

            // open the stream and put it into BufferedReader
            BufferedReader br = new BufferedReader(
                               new InputStreamReader(conn.getInputStream()));

            String inputLine;
            while ((inputLine = br.readLine()) != null) {
                    System.out.println(inputLine);
            }
            br.close();

            System.out.println("Done");

        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

    }
}

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