相当于Gradle的Maven配置文件 [英] Maven profiles equivalent of Gradle

问题描述

我试图在我的spring引导项目构建中实现一个简单的场景：包括/排除依赖关系，并根据环境打包war或jar。例如，对于环境 dev 包含devtools和包jar，对于生成的
$ b

package war等。

我知道这不是基于XML的配置，我基本上可以在我的build.gradle中编写if语句，但是在那里一个推荐的方式来实现这一点？

我可以声明一些常见的依赖关系并将它们引用到一个文件中，而不是创建多个构建文件吗？

是否有基于构建目标环境更改构建配置的最佳实践？

解决方案

  ext {
 devDependencies = ['org.foo:dep1:1.0'，'org.foo:dep2:1.0'] 
 prodDependencies = ['org.foo:dep3:1.0 '，'org.foo:dep4:1.0'] 
 isProd = System.properties ['env'] =='prod'
 isDev = System.properties ['env'] =='dev' 
} 
 
 apply plugin：'java'
 
 dependencies {
 compile'org.foo： common：1.0'
 if（isProd）{
 compile prodDependencies 
} 
 if（isDev）{
 compile devDependencies 
} 
} 
 
 if（isDev）tasks.withType（War）.all {it.enabled = false} 
  

I'm trying to achieve a simple scenario in my spring boot project build: including / excluding dependencies and packaging war or jar depending on the environment.

So for example, for the environment dev include devtools and package jar, for prod package war etc.

I know it is not XML based configuration anymore and I can basically write if statements in my build.gradle but is there a recommended way of achieving this?

Can I declare some common dependencies and refer them in a single file instead of creating multiple build files?

Is there a best practice changing build configuration based on the build target environment?

解决方案

ext {
    devDependencies = ['org.foo:dep1:1.0', 'org.foo:dep2:1.0']
    prodDependencies = ['org.foo:dep3:1.0', 'org.foo:dep4:1.0']
    isProd = System.properties['env'] == 'prod'
    isDev = System.properties['env'] == 'dev'
}

apply plugin: 'java'

dependencies {
    compile 'org.foo:common:1.0'
    if (isProd) {
       compile prodDependencies
    }
    if (isDev) {
       compile devDependencies
    }
}

if (isDev) tasks.withType(War).all { it.enabled = false }

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