我知道launch4j并没有将JRE捆绑在.exe
中,而是必须将它放在它旁边。我的问题是,我该怎么做?maven是否有一种方法可以自动定位和复制用于构建应用程序并将其复制到给定目录的JDK的JRE?
我试过这样做:
<plugin>
<artifactId>maven-resources-plugin</artifactId>
<version>2.6</version>
<executions>
<execution>
<id>copy-resources</id>
<!-- here the phase you need -->
<phase>package</phase>
<goals>
<goal>copy-resources</goal>
</goals>
<configuration>
<outputDirectory>${basedir}/target/windows/jre</outputDirectory>
<resources>
<resource>
<directory>${java.home}</directory>
</resource>
</resources>
</configuration>
</execution>
</executions>
</plugin>
但节目还没开始。它显示了一个立即消失的小对话框(它似乎是空白的,但它消失得太快了,我没有注意到)。
发布于 2018-04-06 13:21:36
更新:删除我以前的答案,用经过测试的工作示例替换
更新2:这个pom.xml现在下载JRE并解压缩它,launch4j exe使用它并工作。我添加了一些评论来解释它是如何工作的。
我建议只使用32位exe和JRE。使用64位JRE的唯一原因是如果您的程序需要使用超过4GB的RAM。
当然,现在您需要一个安装程序将所有这些都安装到Program中。我以前也用过NSIS。NSIS有一条学习曲线,但也不算太糟。
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.akathist.encc</groupId>
<artifactId>mavenproject1</artifactId>
<version>1.0-SNAPSHOT</version>
<packaging>jar</packaging>
<properties>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
<maven.compiler.source>1.8</maven.compiler.source>
<maven.compiler.target>1.8</maven.compiler.target>
</properties>
<dependencies>
<!-- This is the win32 JRE tgz hosted by alfresco - https://mvnrepository.com/artifact/com.oracle.java/jre -->
<dependency>
<groupId>com.oracle.java</groupId>
<artifactId>jre</artifactId>
<classifier>win32</classifier>
<type>tgz</type>
<version>1.8.0_131</version>
</dependency>
</dependencies>
<repositories>
<repository>
<!-- this repository has the JRE tgz -->
<id>alfresco</id>
<url>https://artifacts.alfresco.com/nexus/content/repositories/public/</url>
</repository>
</repositories>
<build>
<plugins>
<plugin>
<!-- this is to extract the JRE tgz file we downloaded -->
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-dependency-plugin</artifactId>
<version>2.5.1</version>
<executions>
<execution>
<phase>generate-resources</phase>
<goals>
<goal>unpack-dependencies</goal>
</goals>
<configuration>
<includeGroupIds>com.oracle.java</includeGroupIds>
<includeTypes>tgz</includeTypes>
<includeArtifactIds>jre</includeArtifactIds>
<includeClassifiers>win32</includeClassifiers>
<outputDirectory>target/win32</outputDirectory>
</configuration>
</execution>
</executions>
</plugin>
<plugin>
<!-- This calls launch4j to create the program EXE -->
<groupId>com.akathist.maven.plugins.launch4j</groupId>
<artifactId>launch4j-maven-plugin</artifactId>
<executions>
<execution>
<id>l4j-clui</id>
<phase>package</phase>
<goals>
<goal>launch4j</goal>
</goals>
<configuration>
<headerType>console</headerType>
<outfile>target/encc.exe</outfile>
<jar>target/mavenproject1-1.0-SNAPSHOT.jar</jar>
<errTitle>encc</errTitle>
<classPath>
<mainClass>com.akathist.encc.Clui</mainClass>
<addDependencies>false</addDependencies>
<preCp>anything</preCp>
</classPath>
<jre>
<path>./win32/java</path>
</jre>
<versionInfo>
<fileVersion>1.2.3.4</fileVersion>
<txtFileVersion>txt file version?</txtFileVersion>
<fileDescription>a description</fileDescription>
<copyright>my copyright</copyright>
<productVersion>4.3.2.1</productVersion>
<txtProductVersion>txt product version</txtProductVersion>
<productName>E-N-C-C</productName>
<internalName>ccne</internalName>
<originalFilename>original.exe</originalFilename>
</versionInfo>
</configuration>
</execution>
</executions>
</plugin>
</plugins>
</build>
</project>
更新3:可悲的事实是,没有官方的,甚至是最新的maven回购与你想要的JRE。您可以宿主您自己的具有所需JRE的maven repo。当完成新版本时,您将不得不更新它。在发布之前使用新版本进行测试也是个好主意。每个月的第三个星期二是完成新Java发行版的时候。您可以设置一个提醒,以检查是否发布了一个新版本并下载。自动化这是一个痛苦,因为许可证协议检查。这篇文章可能会有所帮助,但您可能无法以这样的方式下载JRE的tar.gz版本:Java以编程方式检查最新版本
如果您想支持多个平台,那么托管您自己的maven回购是一个很好的方法。每次发布时,您都可以宿主自己的tar.gz并使用新的JRE https://stackoverflow.com/a/29261502/35264更新它:https://stackoverflow.com/a/29261502/35264
最简单的选择是完成您已经设定的目标,只使用您正在构建的JRE。这将允许您支持Windows 32和64,只要使用32位JRE构建即可。您可以偶尔更新这一点,因为您有时间用新版本进行测试。下面是一个工作的pom.xml,它可以这样做:
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.akathist.encc</groupId>
<artifactId>mavenproject1</artifactId>
<version>1.0-SNAPSHOT</version>
<packaging>jar</packaging>
<properties>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
<maven.compiler.source>1.8</maven.compiler.source>
<maven.compiler.target>1.8</maven.compiler.target>
</properties>
<build>
<plugins>
<plugin>
<!-- This copies the JRE used to do the build from java.home - should be 32 bit Windows JRE -->
<artifactId>maven-resources-plugin</artifactId>
<version>2.6</version>
<executions>
<execution>
<id>copy-resources</id>
<!-- here the phase you need -->
<phase>package</phase>
<goals>
<goal>copy-resources</goal>
</goals>
<configuration>
<outputDirectory>${basedir}/target/win32/java</outputDirectory>
<resources>
<resource>
<directory>${java.home}</directory>
</resource>
</resources>
</configuration>
</execution>
</executions>
</plugin>
<plugin>
<!-- This calls launch4j to create the program EXE -->
<groupId>com.akathist.maven.plugins.launch4j</groupId>
<artifactId>launch4j-maven-plugin</artifactId>
<executions>
<execution>
<id>l4j-clui</id>
<phase>package</phase>
<goals>
<goal>launch4j</goal>
</goals>
<configuration>
<headerType>console</headerType>
<outfile>target/encc.exe</outfile>
<jar>target/mavenproject1-1.0-SNAPSHOT.jar</jar>
<errTitle>encc</errTitle>
<classPath>
<mainClass>com.akathist.encc.Clui</mainClass>
<addDependencies>false</addDependencies>
<preCp>anything</preCp>
</classPath>
<jre>
<path>./win32/java</path>
</jre>
<versionInfo>
<fileVersion>1.2.3.4</fileVersion>
<txtFileVersion>txt file version?</txtFileVersion>
<fileDescription>a description</fileDescription>
<copyright>my copyright</copyright>
<productVersion>4.3.2.1</productVersion>
<txtProductVersion>txt product version</txtProductVersion>
<productName>E-N-C-C</productName>
<internalName>ccne</internalName>
<originalFilename>original.exe</originalFilename>
</versionInfo>
</configuration>
</execution>
</executions>
</plugin>
</plugins>
</build>
</project>
发布于 2018-04-13 08:02:46
我相信这就是你想要的:
<plugin>
<groupId>com.akathist.maven.plugins.launch4j</groupId>
<artifactId>launch4j-maven-plugin</artifactId>
<executions>
<execution>
...
<configuration>
...
<jre>
<path>${java.home}</path> <!-- SEE THIS -->
</jre>
...
</configuration>
</execution>
</executions>
</plugin>
Java允许您使用java.home
system属性访问JRE路径。你也可以访问pom中的java系统属性。而且你有一个插件可以包装launch4j。把所有这些放在一起,你就有了解决办法。
https://stackoverflow.com/questions/49651902
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