如何用 Maven 属性替换 web.xml 中的值? [英] How to replace a value in web.xml with a Maven property?

问题描述

我有一个 Maven 项目，它将一些测试文件下载到它的构建目录 ./target/files.然后这些文件应该可供 servlet 使用，我可以通过将完整路径硬编码为 servlet 的 <init-param> 来轻松实现:

I have a Maven project that downloads some test files into its build directory ./target/files. These files should then be available to a servlet, which I can easily achieve by hardcoding the full path as an <init-param> of the servlet:

<servlet>
    <servlet-name>TestServlet</servlet-name>
    <servlet-class>my.package.TestServlet</servlet-class>
    <init-param>
        <param-name>filepath</param-name>
        <param-value>/home/user/testproject/target/files</param-value>
    </init-param>
</servlet>

如何避免对完整路径进行硬编码并改用动态参数替换?我尝试了以下方法，但没有奏效:

How can I avoid hardcoding the full path and use a dynamic parameter replacement instead? I tried the following, but it did not work:

<param-value>${project.build.directory}/files</param-value>

推荐答案

添加到您的 pom 部分:

Add to your pom section:

<plugin>
    <groupId>org.apache.maven.plugins</groupId>
    <artifactId>maven-war-plugin</artifactId>
    <configuration>
        <webResources>
            <resource>
                <filtering>true</filtering>
                <directory>src/main/webapp</directory>
                <includes>
                    <include>**/web.xml</include>
                </includes>
            </resource>
        </webResources>
        <warSourceDirectory>src/main/webapp</warSourceDirectory>
        <webXml>src/main/webapp/WEB-INF/web.xml</webXml>
    </configuration>
</plugin>

有关详细信息，请参阅Maven:自定义 Web 应用项目的 web.xml

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