如何使用JAVA从互联网下载zip文件并将其保存在特定文件夹中?

时间:2012-01-13 10:19:59

标签: java url exception

这是我的代码:

public static void downloadZipFile() {
    String saveTo = "C:\\Users\\aria\\Downloads\\Temp";
    try {
        URL url = new URL("http://www.bcfi.be/download/files/R1112B2_BcfiHtm.zip");
        URLConnection conn = url.openConnection();
        InputStream in = conn.getInputStream();
        FileOutputStream out = new FileOutputStream(saveTo + "BcfiHtm.zip");
        byte[] b = new byte[1024];
        int count;
        while ((count = in.read(b)) >= 0) {
            out.write(b, 0, count);
        }
        out.flush(); out.close(); in.close();                   

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

**当我编译它时我得到以下错误,但如果我直接在浏览器中使用url evrything就可以了。

我该如何解决?或者还有其他方法可以下载zip文件吗?**

java.net.UnknownHostException: www.bcfi.be
at java.net.PlainSocketImpl.connect(PlainSocketImpl.java:195)
at java.net.SocksSocketImpl.connect(SocksSocketImpl.java:366)
at java.net.Socket.connect(Socket.java:529)
at java.net.Socket.connect(Socket.java:478)
at sun.net.NetworkClient.doConnect(NetworkClient.java:163)
at sun.net.www.http.HttpClient.openServer(HttpClient.java:395)
at sun.net.www.http.HttpClient.openServer(HttpClient.java:530)
at sun.net.www.http.HttpClient.<init>(HttpClient.java:234)
at sun.net.www.http.HttpClient.New(HttpClient.java:307)
at sun.net.www.http.HttpClient.New(HttpClient.java:324)
at sun.net.www.protocol.http.HttpURLConnection.getNewHttpClient(HttpURLConnection.java:970)
at sun.net.www.protocol.http.HttpURLConnection.plainConnect(HttpURLConnection.java:911)
at sun.net.www.protocol.http.HttpURLConnection.connect(HttpURLConnection.java:836)
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1172)
at be.azvub.ext.prismaFlex.Exterahelp.download.DownloadFile.downloadZipFile(DownloadFile.java:72)
at be.azvub.ext.prismaFlex.Exterahelp.download.DownloadFile.main(DownloadFile.java:37)

2 个答案:

答案 0 :(得分:7)

从Java Docs引用:

  

抛出以指示主机的IP地址不能   确定。

确保您的程序未被防火墙代理阻止。

<强>更新

要配置代理,请按照Peter Liljenberg的建议:

  

您可以将代理信息传递给openConnection调用   你的代码是这样的:

     

代理代理=新代理(Proxy.Type.HTTP,new   如果InetSocketAddress( “亲”,9999)); URLConnection conn =   url.openConnection(代理);

答案 1 :(得分:6)

由于您在代理服务器后面,您可以尝试一些不同的方法:

1)启动时将代理信息添加到JVM:

java -Dhttp.proxyHost=proxyhostURL -Dhttp.proxyPort=proxyPortNumber 
-Dhttp.proxyUser=someUserName -Dhttp.proxyPassword=somePassword javaClassToRun

在你的情况下,它可能是:

java -Dhttp.proxyHost=pro -Dhttp.proxyPort=9999  javaClassToRun

2)您可以将代理信息传递给代码中的openConnection调用,如下所示:

Proxy proxy = new Proxy(Proxy.Type.HTTP, new InetSocketAddress("pro",9999));
URLConnection conn = url.openConnection(proxy);