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。
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();
}
}
要将文件的内容从类路径读入String,你可以使用这个:
private String resourceToString(String filePath) throws IOException, URISyntaxException
{
try (InputStream inputStream = this.getClass().getClassLoader().getResourceAsStream(filePath))
{
return IOUtils.toString(inputStream);
}
}
注意:
IOUtils是Commons IO的一部分。
这样叫它:
String fileContents = resourceToString("ImOnTheClasspath.txt");