Java 如何以编程方式获取资源目录路径
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How to get resources directory path programmatically
提问by whyem
I have the following directory layout:
我有以下目录布局：
- src
- main
- java
- resources
- sql (scripts for database)
- spring (configuration)
- webapp
- 源文件
- 主要的
- 爪哇
- 资源
- sql（数据库脚本）
- 弹簧（配置）
- 网络应用程序
Within a ServletContextListener class, I want to access the files under the SQL directory and list them. Basically my problem is with the path, because I know that listing files under a directory in a nutshell is:
在 ServletContextListener 类中，我想访问 SQL 目录下的文件并列出它们。基本上我的问题出在路径上，因为我知道在一个目录下列出文件简而言之是：
File folder = new File(path);
File[] listOfFiles = folder.listFiles();
Maybe I could use the ServletContextEventObject to try and build a path to resources/sql
也许我可以使用ServletContextEvent对象来尝试构建一条路径resources/sql
public void contextInitialized(ServletContextEvent event) {
event.getServletContext(); //(getRealPath etc.)
}
Does something exist to set that path in a relative, non-hardcoded way?
Something like new File("classpath:sql")(preferably spring if possible) or what should I do with the servletContext to point at resources/sql?
是否存在以相对的非硬编码方式设置该路径的方法？像new File("classpath:sql")（如果可能的话最好是弹簧）或者我应该如何处理 servletContext 指向resources/sql？
采纳答案by Dev
I'm assuming the contents of src/main/resources/is copied to WEB-INF/classes/inside your .war at build time. If that is the case you can just do (substituting real values for the classname and the path being loaded).
我假设在构建时将内容src/main/resources/复制到WEB-INF/classes/您的 .war 中。如果是这种情况，您可以这样做（用实际值替换类名和正在加载的路径）。
URL sqlScriptUrl = MyServletContextListener.class
.getClassLoader().getResource("sql/script.sql");
回答by samlewis
import org.springframework.core.io.ClassPathResource;
...
File folder = new ClassPathResource("sql").getFile();
File[] listOfFiles = folder.listFiles();
It is worth noting that this will limit your deployment options, ClassPathResource.getFile()only works if the container has exploded (unzipped) your war file.
值得注意的是，这将限制您的部署选项，ClassPathResource.getFile()仅当容器已分解（解压缩）您的 war 文件时才有效。
回答by whyem
Finally, this is what I did:
最后，这就是我所做的：
private File getFileFromURL() {
URL url = this.getClass().getClassLoader().getResource("/sql");
File file = null;
try {
file = new File(url.toURI());
} catch (URISyntaxException e) {
file = new File(url.getPath());
} finally {
return file;
}
}
...
...
File folder = getFileFromURL();
File[] listOfFiles = folder.listFiles();
回答by grep
Just use com.google.common.io.Resourcesclass. Example:
只需使用com.google.common.io.Resources类。例子：
URL url = Resources.getResource("file name")
After that you have methods like: .getContent(), .getFile(), .getPath() etc
之后你有像这样的方法：.getContent(), .getFile(), .getPath() 等
