使用java中的File.listFiles（）在Linux环境中的项目目录中列出文件

Question

String path = ".";
String files="";
File folder = new File(path);
File[] listOfFiles = folder.listFiles(); 

for (int i = 0; i < listOfFiles.length; i++) 
{
  if (listOfFiles[i].isFile()) 
  {
     files = listOfFiles[i].getName();
     System.out.println("files::::::::::::"+files);
  }
}

Say the above java file I saved under the path D:/spike/FileList.java .

In the above code if I run in windows platform , it will list the files under the ' spike ' directory.

But when I keep this in Linux environment say the path is usr/local/apache/webapps/webtest/src/FileList.java

The result I get is files under root directory.

What I require is under the project root folder, ie in above case under the webtest directory.

How can I do the same. My requirement is I need to first list the files and then from the list of files I need to read sample.properties file.

I know we can hard code the path to get the same. But without hard coding how can I get the list of files under the project root folder of webapps ie, under webtest folder in my case.

I also tried by reading the environmental variables. But the problem here is my apache folder owner is spike and not root. So when I execute System.getenv() what I get is only whatever variables the user spike has set.

But when I execute System.getenv() from any other folder whose owner is root, then I get the complete environmental variables.

So is there any way I can get the project root folder by using the above java code snippet without hard coding the path?

By the way this is a web application deployed in tomcat. First the app will read the details from the server.properties file. But Im not supposed to hard code the path as the path changes from system to system. So my intention is that the code read the properties file from the project starting folder.

Answer 1

This is not true, your code on Linux prints the files in the current folder. I just tried it.

As to tomcat, see here What determines the current working directory of Tomcat Java process? and/or look for similar information on the Tomcat site.

Answer 2

You can try with the current working directory:

String path = System.getProperty("user.dir");

But this depends where exactly is the CWD, eg how was this application started?

---

In the context of a web application, you can do (as long as you have the ServletContext object):

ServletContext application = ...
String path = application.getRealPath("/");

This will give you the root of the web application, you can navigate to the required directory from there.

Answer 3

If your file sample.properties is located in the root folder, and actually is a properties-file I would try this loading the file and printing the content:

    String filename = "sample.properties";
    Properties properties = new Properties();
    properties.load(getClass().getClassLoader().getResource(filename).openStream());
    properties.list(System.out);

getClass.getClassLoader() will get the location of the rootfolder. If you skip getClassLoader you will get the folder of the package in your class.

(If you are running this in a standalone java app the properties file needs to be located in the classes folder.)

Answer 4

Thanks to all your valuable suggestions & I was able to solve it. The problem was, in linux Im starting tomcat server using services like 'service tomcat start' commands. These services are in root folder. So when start tomcat using them and check my current working directory I get the root folder as PWD and not my project folder. So what i did is I started tomcat using startup.sh in $TOMCAT_DIR/bin. Now my current PWD returns $TOMCAT_DIR/bin. Through this I changed the path according to my requirement and was able to access the properties files. But now my problem is I need to restart tomcat server via my application which is inside webapps. So I put this script given below:

restart.sh

cd ../../bin/
./shutdown.sh
wait
sleep 2s  
./startup.sh
wait
sleep 2s
read -p "Close the terminal"

As per above script my tomcat should succesfully restart. But its not doing so. I also tried running manually the same above script. In that case it will work provided I dont close the terminal. When I close the terminal again I lose contact with the server. Any idea why it might be causing. Btw for executing the above script from my java code the below is steps I followed:

String path = "../webapps/webtest/";
try {
    Runtime.getRuntime().exec(path+"restart.sh");
} catch (IOException e) {
    System.out.println("Exception while running script: "+e);
}