如何在Struts 2中获得项目路径? [英] How do you get the project path in Struts 2?

问题描述

例如:

<constant name="struts.multipart.saveDir" value="temp" />

将保存到:

\ Apache Tomcat 7.0.14 \ bin \ temp \ up ____ 781d3178_13831f5e207__8000_00000003.tmp

\Apache Tomcat 7.0.14\bin\temp\upload__781d3178_13831f5e207__8000_00000003.tmp

所以，相反，我该如何去做:

So, instead, how do I make it go to:

C:\ my_project \ build \ web \ temp \

C:\my_project\build\web\temp\

不使用绝对文件路径，因为我不想每次项目移动时都重新配置它.我只想要基本的相对路径.

Without using the absolute file path because I don't want to reconfigure it each time the project moves. I just want the relative path basically.

我认为这并不重要，但我使用的是Struts 2 2.1.8.1版

I don't think it matters, but I'm using Struts 2 version 2.1.8.1

我正在使用Apache Tomcat.那就是我要将项目部署到的地方.没有办法在struts.xml中引用Tomcat的项目部署位置吗?

I'm using Apache Tomcat. That's where I'm deploying the project to. Is there no way to make a reference to Tomcat's project deployment location in the struts.xml?

类似的东西:

<constant name="struts.multipart.saveDir" value="..\webapps\project\build\web" />

推荐答案

有几种方法可以做到这一点.这是最简单的方法:

there are several ways to do this. here is the most easy way:

<constant name="struts.multipart.saveDir" value="${catalina.home}/webapps/project/web/temp"/>

您必须设置 CATALINA_HOME

更新

也许是这样，在您的web.xml中:

maybe this way, in your web.xml:

<context-param>
  <param-name>catalina.home</param-name>
  <param-value>path/to/your/tomcat</param-value>
</context-param>

更新

我还找到了一个链接，它用于Log4j，但它是相同的. 链接到解决方案

I also found a link, it is for Log4j, but it is the same. Link-To-Solution

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