在Maven项目中保存用户状态

Question

I am writing a javaFX project with netbeans IDE and maven.

I am saving user edits in a property file and i loading it when the application start up and i want to update it when application shuts down to reuse it next time.

So I am reading the property file as below:

public static Properties propConfig = new Properties();

InputStream input;
input = Config.class.getClassLoader().getResourceAsStream("config/displayConfig.properties");
propConfig.load(input);

which works fine..

but i don't know how to update the property file :( as

    output =  new FileOutputStream( new File(Config.class.getClassLoader().getResource("config/displayConfig.properties").toURI()) );

is not working since it reads the resource file from jar

jar:file:/D:/freelance%20projects/01%20school%20tool%20bar/mavenprojectFX/target/racer40-1.0-SNAPSHOT.jar!/config/displayConfig.properties

Answer 1

Normally it's not recommended to update files in JARs. Use local files instead, eg:

Path userDirPath = Paths.get(System.getProperty("user.home"), ".<myAppSymbolicName>", "<myAppVersion>");
Path configDirPath =  userDirPath.resolve("config"); 
Path displayConfigFilePath = configDirPath.resolve("displayConfig.properties");
// read file

Answer 2

well i have done a workaround, since I can only read files from jar i have defined a new properties file and i have defined the configuration path in it.

i have modified resources in POM.xml to set filtering true

    <resources>
        <resource>
            <directory>src/main/resources</directory>
            <filtering>true</filtering>
        </resource>
    </resources>

i have placed the path property file in "src/main/resources/config" folder

the path file contains

Path=${project.build.directory}

and i am reading it as the following

InputStream pathStream
                = Config.class.getClassLoader().getResourceAsStream("config/path.properties");

        Properties pathProperties = new Properties();
        pathProperties.load(pathStream);
        path = pathProperties.get("Path").toString().replace("target", "");
        System.out.println(pathProperties.get("Path"));
        pathStream.close();

then i am using 'path' variable as the following

OutputStream  output = new FileOutputStream(new File(path + displayConfig.properties"));

so I am now putting the configuration file "displayConfig.properties" in the same path of jar.