I am using CXF to create a java web service.
i have a file path, e.g "C:\ftproot", ne开发者_开发问答ed to be configurable so i want to put this path into a properties file e.g application.properties
but how can i read the properties file in my java code?
can anyone help?
Thanks
You need to put the properties file in your resources
or WEB-INF/classes
Then
Properties properties = new Properties();
try {
properties.load(new FileInputStream("classpath:application.properties"));
} catch (IOException e) {
}
See Also
- Properties
Create a PropertyPlaceholderConfigurer
for Spring (Refer to the API for options).
Example:
<bean id="propertyConfigurer" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="searchSystemEnvironment" value="true"/>
<property name="systemPropertiesModeName" value="SYSTEM_PROPERTIES_MODE_OVERRIDE"/>
<property name="ignoreUnresolvablePlaceholders" value="true"/>
<property name="locations">
<list>
<value>classpath:build.properties</value>
<value>classpath:other.properties</value>
</list>
</property>
</bean>
Assuming you have a property file.path
in the property file and you are using component scanning you can then use:
@Value("file.path") private String filePath;
That will then be populated with the value of file.path
in the property file (if the bean is created by Spring)
Or if you are creating your beans in XML:
<bean class="yourClassName">
<property name="filePath" value="${file.path} />
</bean>
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