Question

如何创建动态设置App Engine项目ID并运行appengineDeploy任务的任务？

在此示例中，当我运行deployStaging appengineDeploy与项目'a'一起执行时，如何重写此代码以使其与项目'b'一起运行？

buildscript {
    dependencies {
        classpath("com.google.cloud.tools:appengine-gradle-plugin:2.0.1")
    }
}
apply plugin: 'com.google.cloud.tools.appengine'

def gcpProject = 'a'

appengine {
    deploy {
        projectId = gcpProject
    }
}

task deployStaging() {
    doLast {
        gcpProject = 'b'
    }
}
deployStaging.finalizedBy appengineDeploy

Answer 1

如何使用https://symfony.com/doc/master/bundles/FOSJsRoutingBundle/usage.html代替呢？可以通过-P命令行参数传递项目属性：

buildscript {
    dependencies {
        classpath("com.google.cloud.tools:appengine-gradle-plugin:2.0.1")
    }
}
apply plugin: 'com.google.cloud.tools.appengine'

def gcpProject = project.findProperty('stageName') ?: 'a'

appengine {
    deploy {
        projectId = gcpProject
    }
}

现在，如果您仅调用./gradlew appengineDeploy，则gcpProject变量将具有值'a'。如果您调用./gradlew appengineDeploy -PstageName=b，则gcpProject变量的值将为'b'。

Answer 2

gradle插件伙计们提供的答案：

appengine {
  deploy {
     version = "123"
     // do not define projectId here
  }
}

task deployStaging {
  dependsOn appengineDeploy
}
task deployProduction {
  dependsOn appengineDeploy
}

// here's the weird gradle logic, use at your own risk
if (project.gradle.startParameter.taskNames.contains("deployStaging")) {
  appengine.deploy.projectId = "potato-stage"
}
else if(project.gradle.startParameter.taskNames.contains("deployProduction")) {
  appengine.deploy.projectId = "tomato-prod"
}

如何动态设置AppEngine Gradle属性

2 个答案: