java 属性文件作为 web.xml 中的 init-param
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Properties file as init-param in web.xml
提问by TV Nath
I am migrating a Jsp-servlet based Java project that was hosted in websphere to tomcat. Following init-param is in web.xml inside a filter definition. I moved the properties file src folder which is classpath. How to change the following in the web.xml. Can I define properties file as init-param because most of the answers I saw has used context-param to define properties file. I dont think its an option to me as the existing application needs the properties file to be init-param.
我正在将托管在 websphere 中的基于 Jsp-servlet 的 Java 项目迁移到 tomcat。init-param 位于过滤器定义内的 web.xml 中。我移动了类路径的属性文件 src 文件夹。如何更改 web.xml 中的以下内容。我可以将属性文件定义为 init-param 吗，因为我看到的大多数答案都使用 context-param 来定义属性文件。我不认为这对我来说是一个选择，因为现有的应用程序需要属性文件是 init-param。
<init-param>
<param-name>configPath</param-name>
<param-value>/pws/WebSphere/AppServer/properties/fyp/filterConfig/filter.properties</param-value>
</init-param>
I tried
我试过
<init-param>
<param-name>configPath</param-name>
<param-value>classpath:filter.properties</param-value>
</init-param>
It did not work.Thank you in advance,
它没有用。提前谢谢你，
回答by Harry.Chen
Check your servlet implementation,you will find something like the following:
检查您的 servlet 实现，您会发现如下内容：
- get the context root path from ServletContext;
- append the property file path get from init-param;
- do some file operation
- 从 ServletContext 获取上下文根路径；
- 追加从 init-param 获取的属性文件路径；
- 做一些文件操作
As you asked,you can config the servlet as :
正如您所问，您可以将 servlet 配置为：
<init-param>
<param-name>configPath</param-name>
<param-value>filter.properties</param-value>
</init-param>
then change your code to
然后将您的代码更改为
- get the file name from init-param
- open the stream this.getClass().getClassLoader().getResourceAsStream("fileName");
- do some file operation
- 从 init-param 获取文件名
- 打开流 this.getClass().getClassLoader().getResourceAsStream("fileName");
- 做一些文件操作
