3

我将在我的 Springboot 应用程序的资源文件夹中读取一个可用的文件。我已经使用 ResourceLoader 来做到这一点。但是当我尝试执行我的应用程序时，我得到了 FileNotFoundException。

Error creating bean with name 'demoController': Invocation of init method failed; nested exception is java.io.FileNotFoundException: class path resource [schema.graphql] cannot be resolved to absolute file path because it does not reside in the file system: jar:file:/home/dilan/Projects/demo/target/demo.jar!/BOOT-INF/classes!/schema.graphql
    at org.springframework.beans.factory.annotation.InitDestroyAnnotationBeanPostProcessor.postProcessBeforeInitialization(InitDestroyAnnotationBeanPostProcessor.java:137) ~[spring-beans-4.3.9.RELEASE.jar!/:4.3.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.applyBeanPostProcessorsBeforeInitialization(AbstractAutowireCapableBeanFactory.java:409) ~[spring-beans-4.3.9.RELEASE.jar!/:4.3.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1620) ~[spring-beans-4.3.9.RELEASE.jar!/:4.3.9.RELEASE]
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.doCreateBean(AbstractAutowireCapableBeanFactory.java:555) ~[spring-beans-4.3.9.RELEASE.jar!/:4.3.9.RELEASE]

下面是我的代码

 @Autowired
private ResourceLoader resourceLoader;

final Resource fileResource = resourceLoader.getResource("classpath:schema.graphql");
File schemaFile = fileResource.getFile();
TypeDefinitionRegistry definitionRegistry = new SchemaParser().parse(schemaFile);
RuntimeWiring wiring = buildRuntimeWiring();

谁能帮我解决这个问题

在此处输入图像描述

4

2 回答 2

4

你可以使用ClassPathResource

像这样：

InputStream inputStream = new ClassPathResource("schema.graphql").getInputStream();

或者

File file = new ClassPathResource("schema.graphql").getFile();
于 2017-10-04T11:39:07.533 回答
0

尝试使用下面的代码通过 spring boot 访问 ressources 文件夹中的文件

 File file = new ClassPathResource("schema.graphql").getFile();
于 2017-10-04T11:37:53.773 回答