访问子项目Spring Boot中的资源

Question

我有1个根项目和3个模块(api，模型，存储）在其中。以下是项目结构：

**root**
--**api**
----src
------main
--------java
----------Application.java
--------resources
----------data.csv
----build.gradle
--**model**
----src
----build.gradle
--**storage**
----src
----build.gradle
build.gradle
settings.gradle

在我的application.java中，我试图从参考资料中读取CSV文件：

    @SpringBootApplication
    @EnableAutoConfiguration
    @EnableJpaRepositories
    @EnableSolrRepositories
    public class MyApp{

        public static void main(String[] args) throws IOException {
            SpringApplication.run(MatMatchApp.class);
            ClassPathResource res = new ClassPathResource("classpath:data.csv");
            String path =res.getPath();
            File csv = new File(path);
            InputStream stream = new FileInputStream(csv);
        }
    }

Caused by: java.io.FileNotFoundException: data.csv (The system cannot find the file specified)
    at java.io.FileInputStream.open0(Native Method) ~[na:1.8.0_101]
    at java.io.FileInputStream.open(FileInputStream.java:195) ~[na:1.8.0_101]
    at java.io.FileInputStream.<init>(FileInputStream.java:138) ~[na:1.8.0_101]

File file = new File(getClass().getResource("data.csv").getFile());

解决了此代码工作良好：

InputStream is = new ClassPathResource("/example.csv").getInputStream();

要了解更多详细信息，请检查以下答案：作为jar运行时未找到classpath资源

Answer 1

我测试了一个普通项目，您可以在这里看到spring-boot-resource-access

您可能会错过文件前面的/。

ClassPathResource res = new ClassPathResource("classpath:/data.csv");

或

File file = new File(getClass().getResource("/data.csv").getFile());

测试应用程序时，您必须从实例类中找到类路径，例如ConfigurableApplicationContext

public static void main(String[] args) throws URISyntaxException {
    ConfigurableApplicationContext context = SpringApplication.run(DemoApplication.class);
    File csv = new File(context.getClass().getResource("/application.properties").toURI());
    System.out.println(csv.getAbsolutePath());
    System.out.println(String.format("does file exists? %s", csv.exists()));
}