在一个applet读取文件

问题描述

您好我想读出了位于服务器上的文件。
我由一个参数来获取文件路径

 ＆LT; PARAM NAME = fileToRead值=htt​​p://someserver.de/file.txt＆GT;
 

当我现在开始小程序出现下列错误

产生的原因：java.lang.IllegalArgumentException异常：URI方案不是文件

有人可以给我一个提示？

  BufferedReader中的文件;
                        字符串strFile =新的String（的getParameter（fileToRead））;                        网址URL =新的URL（strFile）;
                        的URI的uri = url.toURI（）;
                        尝试{                            文件theFile =新的文件（URI）;
                            文件=新的BufferedReader（新的FileReader（新文件（URI）））;                        字符串输入=;                            而（（输入= file.readLine（））！= NULL）{
                               字词。（输入）;
                            }
                        }赶上（IOException异常前）{
                          。Logger.getLogger（Hedgeman.class.getName（））日志（Level.SEVERE，空，前）;
                        }
 

解决方案

您正在尝试为一个文件，这东西不符合文件打开：// URI，因为错误提示

如果你想使用一个网址，我建议你只使用url.openStream（）这应该是简单的。

Hi there I want to read out a file that lies on the server.I get the path to the file by a parameter

<PARAM name=fileToRead value="http://someserver.de/file.txt">

when I now start the applet following error occurs

Caused by: java.lang.IllegalArgumentException: URI scheme is not "file"

Can someone give me a hint?

BufferedReader file;
                        String strFile = new String(getParameter("fileToRead"));

                        URL url = new URL(strFile);
                        URI uri = url.toURI();
                        try {

                            File theFile = new File(uri);
                            file = new BufferedReader(new FileReader(new File(uri)));

                        String input = "";

                            while ((input = file.readLine()) != null) {
                               words.add(input);
                            }
                        } catch (IOException ex) {
                          Logger.getLogger(Hedgeman.class.getName()).log(Level.SEVERE, null, ex);
                        }

解决方案

You are trying open as a file, something which doesn't follow the file:// uri, as the error suggests.

If you want to use a URL, I suggest you just use url.openStream() which should be simpler.

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