如何从maven命令行读取环境变量并选择正确的属性文件

<build>
  <plugins>
    <plugin>
      <groupId>org.apache.maven.plugins</groupId>
      <artifactId>maven-resources-plugin</artifactId>
      <version>3.2.0</version>
      <configuration>
        <encoding>UTF-8</encoding>
      </configuration>
      <executions>
        <execution>
          <id>copy-resources</id>
          <phase>validate</phase>
          <goals>
            <goal>copy-resources</goal>
          </goals>
          <configuration>
            <outputDirectory>${project.build.directory}/classes</outputDirectory>
            <resources>
              <resource>
                <directory>src/main/resources</directory>
                <filtering>true</filtering>
              </resource>
            </resources>
          </configuration>
        </execution>
      </executions>
    </plugin>
  </plugins>
</build>

import java.io.FileInputStream;
import java.io.IOException;
import java.util.Properties;

public class AppConfig {
    private static final String PROPERTY_FILE = "config.properties";

    public static void main(String[] args) {
        Properties properties = new Properties();
        try {
            properties.load(new FileInputStream(PROPERTY_FILE));
            String databaseUrl = properties.getProperty("database.url");
            // 使用读取到的配置信息进行后续操作
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}