Java Spring框架-如何将目录作为资源注入?
这是我的当前配置,工作正常: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
<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 ...
}