小编典典

Maven:将生成的资源放入tomcat-maven-plugin的位置?

tomcat

src/main/webapp我的项目目录中有CSS和JavaScript文件。我想将这些资源的合并和缩小版本添加到我的WAR文件以及将其tomcat- maven-plugin拾取的位置。

我使用yuicompressor-maven-
plugin创建文件并将其放入${project.build.directory}/${project.build.finalName}。它对maven包非常有用,那些资源进入了WAR文件,但是某种程度上tomcat- maven-plugin根本看不到。我应该使用其他目录吗?

我的pom:

    <plugins>
        <plugin>
            <groupId>org.codehaus.mojo</groupId>
            <artifactId>tomcat-maven-plugin</artifactId>
            <version>1.1</version>
            <configuration>
                <path>/MyApp</path>
                <warDirectory>${project.build.directory}/${project.build.finalName}</warDirectory>
            </configuration>
        </plugin>
        <plugin>
            <version>2.5.1</version>
            <inherited>true</inherited>
            <groupId>org.apache.maven.plugins</groupId>
            <artifactId>maven-compiler-plugin</artifactId>
            <configuration>
                <source>1.6</source>
                <target>1.6</target>
                <optimize>true</optimize>
                <debug>true</debug>
            </configuration>
        </plugin>
        <plugin>
            <groupId>org.apache.maven.plugins</groupId>
            <artifactId>maven-resources-plugin</artifactId>
            <version>2.5</version>
            <configuration>
                <encoding>UTF-8</encoding>
            </configuration>
        </plugin>
        <plugin>
            <groupId>org.apache.maven.plugins</groupId>
            <artifactId>maven-war-plugin</artifactId>
            <version>2.2</version>
            <configuration>
                <webResources>
                    <resource>
                        <directory>${basedir}/src/main/resources/META-INF</directory>
                        <filtering>true</filtering>
                        <targetPath>META-INF</targetPath>
                        <includes>
                            <include>context.xml</include>
                        </includes>
                    </resource>
                </webResources>
                <archive>
                    <addMavenDescriptor>false</addMavenDescriptor>
                    <manifest>
                        <addDefaultImplementationEntries>true</addDefaultImplementationEntries>
                        <addClasspath>true</addClasspath>
                    </manifest>
                </archive>
            </configuration>
        </plugin>
        <plugin>
            <groupId>net.alchim31.maven</groupId>
            <artifactId>yuicompressor-maven-plugin</artifactId>
            <version>1.3.0</version>
            <executions>
                <execution>
                    <phase>process-resources</phase>
                    <configuration>
                        <excludes>
                            <exclude>**/*</exclude>
                        </excludes>
                        <aggregations>
                            <aggregation>
                                <output>${project.build.directory}/${project.build.finalName}/js/commons-pack.js</output>
                                <includes>
                                    <include>${project.build.sourceDirectory}/../webapp/js1.js</include>
                                    <include>${project.build.sourceDirectory}/../webapp/js2.js</include>
                     ...

我还应该怎么做才能 mvn tomcat:run拾取生成的文件?


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2020-06-16

共1个答案

小编典典

用途warSourceDirectory

<warSourceDirectory>${project.build.directory}/${project.build.finalName}</warSourceDirectory>

代替此配置属性(warDirectorytomcat-maven-plugin

<warDirectory>${project.build.directory}/${project.build.finalName}</warDirectory>

根据tomcat-maven-plugin文档warSourceDirectoryWeb资源是从此处获取的,其默认值为${basedir}/src/main/webapp。这意味着,如果您未设置该属性,则需要在下生成统一/最小化的JavaScript文件${basedir}/src/main/webapp

如果设置warSourceDirectory为输出文件夹,则意味着您需要在启动Tomcat之前生成此文件。

2020-06-16