Azure DevOps Build Pipeline - Build Spring Boot based on specific application.profile

Question

我有一个 Spring 引导应用程序，我利用 Azure DevOps 自动 build/test 项目并部署工件。

我的问题如下：如何根据特定的 application.profile 进行构建？现在它刚刚构建，但在我的项目中我有以下 3 个文件：

application.properties
应用-test.properties
应用-acceptance.properties

例如，我如何基于 "application-test.properties" 构建？

目前我的代码如下所示：

stages: 
 - stage: Build
displayName: Build stage
  jobs:
 - job: MavenPackageAndPublishArtifacts
displayName: Maven Package and Publish Artifacts
pool:vmImage: $(vmImageName)

steps:
- task: Maven@3
  displayName: 'Maven Package'
  inputs:
    mavenPomFile: 'pom.xml'
    javaHomeOption: 'JDKVersion'
    jdkVersionOption: '1.11'
    jdkArchitectureOption: 'x64'

- task: CopyFiles@2
  displayName: 'Copy Files to artifact staging directory'
  inputs:
    SourceFolder: '$(System.DefaultWorkingDirectory)'
    Contents: '**/target/*.?(war|jar)'
    TargetFolder: $(Build.ArtifactStagingDirectory)

- upload: $(Build.ArtifactStagingDirectory)
  artifact: drop

Answer 1

根据

Setting the SPRING_PROFILES_ACTIVE environment variable will activate the intended profile. for example, in order to activate the qa profile, set the SPRING_PROFILES_ACTIVE environment variable to qa. Read more about profile specific properties here in Spring Boot Documentation.

关于如何使用环境，你也可以看看这个博客：Using Environment Variables in Azure DevOps Pipelines

Azure DevOps Build Pipeline - Build Spring Boot based on specific application.profile

Azure DevOps Build Pipeline - Build Spring Boot based on specific application.profile

environment

spring-boot

azure-devops

application.properties