我有一个小型的maven项目。
项目结构如下:
src
|-main
|-java
|-com.my.group
S3FileUploader
|-resources
s3.properties
pom文件如下所示:
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.my.group</groupId>
<artifactId>try-s3</artifactId>
<version>1.0-SNAPSHOT</version>
<packaging>jar</packaging>
<dependencies>
<dependency>
<groupId>com.amazonaws</groupId>
<artifactId>aws-java-sdk</artifactId>
<version>1.11.141</version>
</dependency>
</dependencies>
<build>
<resources>
<resource>
<directory>src/main/resources</directory>
<includes>
<include>*</include>
</includes>
</resource>
</resources>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-assembly-plugin</artifactId>
<executions>
<execution>
<phase>package</phase>
<goals>
<goal>single</goal>
</goals>
<configuration>
<archive>
<manifest>
<mainClass>
com.my.group.S3FileUploader
</mainClass>
</manifest>
</archive>
<descriptorRefs>
<descriptorRef>jar-with-dependencies</descriptorRef>
</descriptorRefs>
</configuration>
</execution>
</executions>
</plugin>
</plugins>
</build>
</project>
我的目标是: 1.将/ src / main / resources目录下的文件包含在jar包中。 2.将S3FileUploader作为主类。 / src / main / resources目录下有一个s3.properties文件。
在我的代码中,我尝试使用以下方法读取属性文件:
String propertiesFile = "s3.properties";
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
File file = new File(classLoader.getResource(propertiesFile).getFile());
InputStream input = new FileInputStream(file);
但是运行java -jar命令会出现此错误:
java.io.FileNotFoundException: file:/try-s3-1.0-SNAPSHOT-jar-with-dependencies.jar!/s3.properties (No such file or directory)
我的问题是: 我是否以错误的方式包装罐子? 2.如果没有,在代码中如何处理以读取s3.properties文件?
提前致谢:)
答案 0 :(得分:0)
尝试以下方法:
URL url = this.getClass().getResource("file.ext")
File file = new File(url.toURI());