如何在运行时获取属性文件的路径并将其传递给Bean [英] How to get the path to a properties file and pass it to a bean at runtime

问题描述

我有一个由Spring创建的bean.实际的类与Spring驻留在不同的JAR中.该bean被传递了一个路径作为构造函数参数.但是，我在检索文件句柄时遇到困难.该文件位于WEB-INF/classes/中.我已经尝试过基于WEB-INF的相对路径，但是显然那是行不通的.

I have a bean that is created by Spring. The actual class resides in a different JAR than Spring. This bean is passed a path as a constructor argument. However, I am having difficulty retrieving a handle to the file. The file is in WEB-INF/classes/. I've tried relative pathing based on WEB-INF, but obviously that didn't work.

XML:

 <bean id="configurationManager" class="package.ConfigurationManager" 
      scope="singleton">           
      <property name="configurationMapping">
            <bean class="package.PropertiesFileConfigurationMapper">
                <constructor-arg type="java.lang.String">
                    <value>/path/to/file</value>
                </constructor-arg>
            </bean> 
      </property>                     
</bean> 

Bean:

public class ConfigurationMapper {

    public ConfigurationMapper(String resource) {
            _map = new HashMap<String, String>();
        String property = null;
        BufferedReader reader = null;
        try {
            FileReader file = new FileReader(resourcePath);
            reader = new BufferedReader(file);
            while ((property = reader.readLine()) != null) {
                if (property.matches("(.+)=(.+)")) {
                    String[] temp = property.split("(.+)=(.+)");
                    _map.put(temp[0], temp[1]);
                }
            }
        } catch (Exception ex){
            ex.printStackTrace();
        } finally {
            if (reader != null)
                reader.close();
        }
    }

    //other methods to manipulate settings
}

如何获取rm.properties文件的正确路径，并在运行时将其传递给Bean?

How can I get the proper path to the rm.properties file and pass it to the bean at runtime?

添加了构造函数代码.

我明白了.我将构造函数参数更改为不再采用路径.现在需要一个资源，因此Spring已经找到了我想要加载的资源.

I got it. I changed the constructor argument to no longer take a path. It now takes a Resource, so Spring has found the resource that I wanted loaded.

推荐答案

java.io.File和FileReader仅适用于实际文件.打包在JAR文件中的资源本身不是文件.

java.io.File and FileReader only work for actual files. A resource packed inside a JAR file isn't itself a file.

最简单的加载方法是作为类路径资源:

The easiest way to load it is as a classpath resource:

替换此:

FileReader file = new FileReader(resourcePath);
reader = new BufferedReader(file);

具有这样的内容:

InputStream inputStream = getClass().getResourceAsStream();
reader = new BufferedReader(new InputStreamReader(inputStream));

更好的是，通过将构造函数参数声明为org.springframework.core.io.Resource:来使用Spring的Resource抽象:

Better yet, use Spring's Resource abstraction, by declaring the constructor parameter as org.springframework.core.io.Resource:

public ConfigurationMapper(Resource resource) {
   ...
   InputStream inputStream = resource.getInputStream();
   reader = new BufferedReader(new InputStreamReader(inputStream));

然后提供路径:

<constructor-arg value="classpath:/path/to/file"/>

Spring将为该路径自动创建一个ClasspathResource(使用类路径)，并将其传递给您的构造函数.

Spring will automatically create a ClasspathResource for that path (using a classpath) , and pass it to your constructor.

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