Input File Path as argument to program
File Path : E:\TestCode\Test file space\abc.xml
Code : This code will accept the above file path as argument.
package com.org;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
public class FileSepratorTest {
public static void main(String[] args) {
String filePath = args[0]; // It will take file path as E:\TestCode\Test file space\abc.xml
try {
System.out.println("File Path :"+filePath); // printing file path
FileInputStream file = new FileInputStream(new File(filePath));
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
Exception occurs when execute this code with space separated folder name
File Path :E:\TestCode\Test
java.io.FileNotFoundException: E:\TestCode\Test (The system cannot find the file specified)
at java.io.FileInputStream.open(Native Method) at java.io.FileInputStream.(Unknown Source) at com.org.FileSepratorTest.main(FileSepratorTest.java:16)
它没有显示整个文件路径，我知道我需要将此作为参数，但我想执行 jar 文件，jar 文件路径指定如下
E:\TestCode\executable_jar>java -cp E:\TestCode\executable_jar\abc_lib -jar redmas- migration.jar E:\TestCode\Migration Letest File\abc.xml** **in my application.
最佳答案
用 "- 双引号将文件路径括起来
E:\TestCode\executable_jar>java -cp E:\TestCode\executable_jar\abc_lib -jar redmas- migration.jar "E:\TestCode\Migration Letest File\abc.xml"
关于java - 从命令行执行 Jar 文件，文件夹名称中包含空格，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24796467/