我试图读取在CLASSPATH系统变量中设置的文本文件。不是用户变量。
我试图获得输入流文件如下:
将文件目录(D:\myDir)放在CLASSPATH中，然后尝试如下操作:
InputStream in = this.getClass().getClassLoader().getResourceAsStream("SomeTextFile.txt");
InputStream in = this.getClass().getClassLoader().getResourceAsStream("/SomeTextFile.txt");
InputStream in = this.getClass().getClassLoader().getResourceAsStream("//SomeTextFile.txt");
将文件(D:\myDir\SomeTextFile.txt)的完整路径放在CLASSPATH中，并尝试上面的3行代码。
但不幸的是，他们都没有工作，我总是得到null到我的输入流。
当前回答
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
public class ReadFile
{
/**
* * feel free to make any modification I have have been here so I feel you
* * * @param args * @throws InterruptedException
*/
public static void main(String[] args) throws InterruptedException {
// thread pool of 10
File dir = new File(".");
// read file from same directory as source //
if (dir.isDirectory()) {
File[] files = dir.listFiles();
for (File file : files) {
// if you wanna read file name with txt files
if (file.getName().contains("txt")) {
System.out.println(file.getName());
}
// if you want to open text file and read each line then
if (file.getName().contains("txt")) {
try {
// FileReader reads text files in the default encoding.
FileReader fileReader = new FileReader(
file.getAbsolutePath());
// Always wrap FileReader in BufferedReader.
BufferedReader bufferedReader = new BufferedReader(
fileReader);
String line;
// get file details and get info you need.
while ((line = bufferedReader.readLine()) != null) {
System.out.println(line);
// here you can say...
// System.out.println(line.substring(0, 10)); this
// prints from 0 to 10 indext
}
} catch (FileNotFoundException ex) {
System.out.println("Unable to open file '"
+ file.getName() + "'");
} catch (IOException ex) {
System.out.println("Error reading file '"
+ file.getName() + "'");
// Or we could just do this:
ex.printStackTrace();
}
}
}
}
}
}
其他回答
你必须把你的“系统变量”放在Java类路径上。
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
public class ReadFile
{
/**
* * feel free to make any modification I have have been here so I feel you
* * * @param args * @throws InterruptedException
*/
public static void main(String[] args) throws InterruptedException {
// thread pool of 10
File dir = new File(".");
// read file from same directory as source //
if (dir.isDirectory()) {
File[] files = dir.listFiles();
for (File file : files) {
// if you wanna read file name with txt files
if (file.getName().contains("txt")) {
System.out.println(file.getName());
}
// if you want to open text file and read each line then
if (file.getName().contains("txt")) {
try {
// FileReader reads text files in the default encoding.
FileReader fileReader = new FileReader(
file.getAbsolutePath());
// Always wrap FileReader in BufferedReader.
BufferedReader bufferedReader = new BufferedReader(
fileReader);
String line;
// get file details and get info you need.
while ((line = bufferedReader.readLine()) != null) {
System.out.println(line);
// here you can say...
// System.out.println(line.substring(0, 10)); this
// prints from 0 to 10 indext
}
} catch (FileNotFoundException ex) {
System.out.println("Unable to open file '"
+ file.getName() + "'");
} catch (IOException ex) {
System.out.println("Error reading file '"
+ file.getName() + "'");
// Or we could just do this:
ex.printStackTrace();
}
}
}
}
}
}
当使用Spring框架时(作为实用工具或容器的集合—您不需要使用后一种功能)，您可以轻松地使用Resource抽象。
Resource resource = new ClassPathResource("com/example/Foo.class");
通过Resource接口，您可以以InputStream、URL、URI或File的形式访问资源。将资源类型更改为文件系统资源，只需更改实例即可。
我使用webshpere应用服务器和我的Web模块是建立在Spring MVC。测试。属性位于资源文件夹中，我尝试使用以下方法加载此文件:
.getResourceAsStream .getClassLoader this.getClass()()(“Test.properties”); this.getClass () .getResourceAsStream (" / Test.properties ");
上面的代码都没有加载文件。
但是在下面代码的帮助下，属性文件被成功加载:
.getResourceAsStream .getContextClassLoader Thread.currentThread()()(“Test.properties”);
感谢用户“user1695166”。
我的回答与问题中所问的不完全一样。相反，我给出了一个解决方案，我们可以很容易地从我们的项目类路径读取文件到java应用程序。
例如，假设配置文件example.xml位于如下路径
com.myproject.config.dev
我们的Java可执行类文件在下面的路径:-
com.myproject.server.main
now just check in both the above path which is the nearest common directory/folder from where you can access both dev and main directory/folder (com.myproject.server.main - where our application’s java executable class is existed) – We can see that it is myproject folder/directory which is the nearest common directory/folder from where we can access our example.xml file. Therefore from a java executable class resides in folder/directory main we have to go back two steps like ../../ to access myproject. Now following this, see how we can read the file:-
package com.myproject.server.main;
class Example {
File xmlFile;
public Example(){
String filePath = this.getClass().getResource("../../config/dev/example.xml").getPath();
this.xmlFile = new File(filePath);
}
public File getXMLFile() {
return this.xmlFile;
}
public static void main(String args[]){
Example ex = new Example();
File xmlFile = ex.getXMLFile();
}
}
