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Java Spring框架-如何将目录作为资源注入?_Java_Spring_Io_Path - Fatal编程技术网

Java Spring框架-如何将目录作为资源注入?

Java Spring框架-如何将目录作为资源注入?,java,spring,io,path,Java,Spring,Io,Path,这是我的当前配置,工作正常: <bean id="foo" class="foo.Foo"> <constructor-arg> <list value-type="org.springframework.core.io.Resource"> <value>classpath:bar/01.lookup</value> <value>clas

这是我的当前配置,工作正常:

<bean id="foo"
      class="foo.Foo">
    <constructor-arg>
        <list value-type="org.springframework.core.io.Resource">
            <value>classpath:bar/01.lookup</value>
            <value>classpath:bar/02.lookup</value>
            <value>classpath:bar/03.lookup</value>
        </list>
    </constructor-arg>
</bean>

我希望调用
listFiles
并循环查找所有
.lookup
文件,但这似乎不起作用,而且我得到了一个NullPointerException,因为传递的路径未解析为目录。

您需要使用
路径匹配源模式解析程序
,请参阅文档:

代码示例:

ResourcePatternResolver resolver = new PathMatchingResourcePatternResolver();
Resource[] resources = resolver.getResources("classpath*:bar/*.lookup") ;
for (Resource resource: resources){
    ... process resource here ...
}

您需要使用
PathMatchingResourcePatternResolver
,请参阅文档:

代码示例:

ResourcePatternResolver resolver = new PathMatchingResourcePatternResolver();
Resource[] resources = resolver.getResources("classpath*:bar/*.lookup") ;
for (Resource resource: resources){
    ... process resource here ...
}