Question

我正在使用maven-compile插件来编译类。现在我想在当前的类路径中添加一个jar文件。该文件保留在另一个位置（假设c：/jars/abc.jar。我更喜欢将此文件保留在此处）。我怎么能这样做？

如果我在参数中使用classpath：

<configuration>
 <compilerArguments>
  <classpath>c:/jars/abc.jar</classpath>
 </compilerArguments>
</configuration>

它将无效，因为它将覆盖当前的类路径（包括所有依赖项）

请帮帮我。

Answer 1

之前可能会问过这个问题。见Can I add jars to maven 2 build classpath without installing them?

简而言之：将jar包含为系统范围的依赖项。这需要指定jar的绝对路径。

另见http://maven.apache.org/guides/introduction/introduction-to-dependency-mechanism.html

Answer 2

从docs和example来看，不清楚是否允许类路径操作。

<configuration>
 <compilerArgs>
  <arg>classpath=${basedir}/lib/bad.jar</arg>
 </compilerArgs>
</configuration>

但请参阅Java docs（也https://www.cis.upenn.edu/~bcpierce/courses/629/jdkdocs/tooldocs/solaris/javac.html）

-classpath path指定javac用于查找运行javac或被其他类引用所需的类的路径编译。覆盖缺省值或CLASSPATH环境变量如果设置了。

也许有可能获得当前的类路径并扩展它，
见in maven, how output the classpath being used?

    <properties>
      <cpfile>cp.txt</cpfile>
    </properties>

  <plugin>
    <groupId>org.apache.maven.plugins</groupId>
    <artifactId>maven-dependency-plugin</artifactId>
    <version>2.9</version>
    <executions>
      <execution>
        <id>build-classpath</id>
        <phase>generate-sources</phase>
        <goals>
          <goal>build-classpath</goal>
        </goals>
        <configuration>
          <outputFile>${cpfile}</outputFile>
        </configuration>
      </execution>
    </executions>
  </plugin>

读取文件（Read a file into a Maven property）

<plugin>
  <groupId>org.codehaus.gmaven</groupId>
  <artifactId>gmaven-plugin</artifactId>
  <version>1.4</version>
  <executions>
    <execution>
      <phase>generate-resources</phase>
      <goals>
        <goal>execute</goal>
      </goals>
      <configuration>
        <source>
          def file = new File(project.properties.cpfile)
          project.properties.cp = file.getText()
        </source>
      </configuration>
    </execution>
  </executions>
</plugin>

最后

  <plugin>
    <groupId>org.apache.maven.plugins</groupId>
    <artifactId>maven-compiler-plugin</artifactId>
    <version>3.6.1</version>
    <configuration>
      <compilerArgs>
         <arg>classpath=${cp}:${basedir}/lib/bad.jar</arg>
      </compilerArgs>
    </configuration>
   </plugin>

Answer 3

编译器插件的classpath设置是两个args。像这样更改它并为我工作：

<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<version>3.6.1</version>
<configuration>
  <compilerArgs>
     <arg>-cp</arg>
     <arg>${cp}:${basedir}/lib/bad.jar</arg>
  </compilerArgs>
</configuration>

我使用gmavenplus-plugin读取路径并创建属性'cp'：

      <plugin>
    <!--
      Use Groovy to read classpath and store into
      file named value of property <cpfile>

      In second step use Groovy to read the contents of
      the file into a new property named <cp>

      In the compiler plugin this is used to create a
      valid classpath
    -->
    <groupId>org.codehaus.gmavenplus</groupId>
    <artifactId>gmavenplus-plugin</artifactId>
    <version>1.12.0</version>
    <dependencies>
      <dependency>
        <groupId>org.codehaus.groovy</groupId>
        <artifactId>groovy-all</artifactId>
        <!-- any version of Groovy \>= 1.5.0 should work here -->
        <version>3.0.6</version>
        <type>pom</type>
        <scope>runtime</scope>
      </dependency>
    </dependencies>
    <executions>
      <execution>
        <id>read-classpath</id>
        <phase>validate</phase>
        <goals>
          <goal>execute</goal>
        </goals>
      </execution>

    </executions>
    <configuration>
      <scripts>
        <script><![CDATA[
                def file = new File(project.properties.cpfile)
                /* create a new property named 'cp'*/
                project.properties.cp = file.getText()
                println '<<< Retrieving classpath into new property named <cp> >>>'
                println 'cp = ' + project.properties.cp
              ]]></script>
      </scripts>
    </configuration>
  </plugin>

Maven：将文件夹或jar文件添加到当前类路径中

3 个答案: