java – 将文件作为命令行参数传递并读取其行
发布时间:2020-12-15 05:17:02 所属栏目:Java 来源:网络整理
导读:这是我在互联网上找到的用于读取文件行的??代码,我也使用 eclipse,并在其参数字段中将文件名称作为SanShin.txt传递.但它会打印: Error: textfile.txt (The system cannot find the file specified) 码: public class Zip { public static void main(String
这是我在互联网上找到的用于读取文件行的??代码,我也使用
eclipse,并在其参数字段中将文件名称作为SanShin.txt传递.但它会打印:
Error: textfile.txt (The system cannot find the file specified) 码: public class Zip { public static void main(String[] args){ try{ // Open the file that is the first // command line parameter FileInputStream fstream = new FileInputStream("textfile.txt"); BufferedReader br = new BufferedReader(new InputStreamReader(fstream)); String strLine; //Read File Line By Line while ((strLine = br.readLine()) != null) { // Print the content on the console System.out.println (strLine); } //Close the input stream in.close(); }catch (Exception e){//Catch exception if any System.err.println("Error: " + e.getMessage()); } } } 请帮我解释为什么会出现这个错误. 解决方法... // command line parameter if(argv.length != 1) { System.err.println("Invalid command line,exactly one argument required"); System.exit(1); } try { FileInputStream fstream = new FileInputStream(argv[0]); } catch (FileNotFoundException e) { // TODO Auto-generated catch block e.printStackTrace(); } // Get the object of DataInputStream ... > java -cp ... Zip pathtotest.file (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |