spring Spring中application.properties中指定相对路径
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Specifying relative path in application.properties in Spring
提问by Somasundaram Sekar
Is there a way we can lookup file resources using relative path in application.properties file in Spring boot application as specified below
有没有一种方法可以使用如下指定的 Spring boot 应用程序中 application.properties 文件中的相对路径查找文件资源
spring.datasource.url=jdbc:hsqldb:file:${project.basedir}/db/init
采纳答案by Abhijit Sarkar
@membersound answer is just breaking up the hardcoded path in 2 parts, not dynamically resolving the property. I can tell you how to achieve what you're looking for, but you need to understand is that there is NOproject.basedirwhen you're running the application as a jar or war. Outside the local workspace, the source code structure doesn't exist.
@membersound 答案只是将硬编码路径分为两部分，而不是动态解析属性。我可以告诉你如何实现你要找的东西，但你要明白的是，有NOproject.basedir，当你运行的应用程序作为一个罐子或War。在本地工作区之外，源代码结构不存在。
If you still want to do this for testing, that's feasible and what you need is to manipulate the PropertySources. Your simplest option is as follows:
如果您仍然想这样做以进行测试，那是可行的，您需要的是操作PropertySources。您最简单的选择如下：
Define an ApplicationContextInitializer, and set the property there. Something like the following:
定义一个ApplicationContextInitializer, 并在那里设置属性。类似于以下内容：
public class MyApplicationContextInitializer implements ApplicationContextInitializer<ConfigurableApplicationContext> {
@Override
public void initialize(ConfigurableApplicationContext appCtx) {
try {
// should be /<path-to-projectBasedir>/build/classes/main/
File pwd = new File(getClass().getResource("/").toURI());
String projectDir = pwd.getParentFile().getParentFile().getParent();
String conf = new File(projectDir, "db/init").getAbsolutePath();
Map<String, Object> props = new HashMap<>();
props.put("spring.datasource.url", conf);
MapPropertySource mapPropertySource = new MapPropertySource("db-props", props);
appCtx.getEnvironment().getPropertySources().addFirst(mapPropertySource);
} catch (URISyntaxException e) {
throw new RuntimeException(e);
}
}}
Looks like you're using Boot, so you can just declare context.initializer.classes=com.example.MyApplicationContextInitializerin your application.propertiesand Boot will run this class at startup.
看起来你正在使用 Boot，所以你可以context.initializer.classes=com.example.MyApplicationContextInitializer在你的中声明application.properties，Boot 将在启动时运行这个类。
Words of cautionagain:
再次警告：
This will not work outside the local workspace as it depends on the source code structure.
I've assumed a Gradle project structure here
/build/classes/main. If necessary, adjust according to your build tool.If
MyApplicationContextInitializeris in thesrc/test/java,pwdwill be<projectBasedir>/build/classes/test/, not<projectBasedir>/build/classes/main/.
这在本地工作区之外不起作用，因为它取决于源代码结构。
我在这里假设了一个 Gradle 项目结构
/build/classes/main。如有必要，请根据您的构建工具进行调整。如果
MyApplicationContextInitializer是在src/test/java，pwd将会是<projectBasedir>/build/classes/test/，不是<projectBasedir>/build/classes/main/。
回答by chendu
I'm using spring boot to build a upload sample, and meet the same problem, I only want to get the project root path. (e.g. /sring-boot-upload)
我在用spring boot构建一个上传示例，遇到同样的问题，我只想获取项目根路径。（例如 /sring-boot-upload）
I find out that below code works:
我发现下面的代码有效：
upload.dir.location=${user.dir}\uploadFolder
回答by membersound
your.basedir=${project.basedir}/db/init
spring.datasource.url=jdbc:hsqldb:file:${your.basedir}
@Value("${your.basedir}")
private String file;
new ClassPathResource(file).getURI().toString()
