我有项目(“后端”)配置: dependencies { compile project(':model') //I would like to pass some params into this being build. ... 和其他项目(“前端”)配置非常相似: dependencies { compile project(':model') //but fro
dependencies { compile project(':model') //I would like to pass some params into this being build. ...
和其他项目(“前端”)配置非常相似:
dependencies { compile project(':model') //but from this project I will not pass params. ...
当我构建依赖项目模型时 – 我需要将参数传递给它的构建参数并在条件上执行某些操作.因此,对于需要子项目的所有项目,它将根据他们的需求进行构建.
更具体一点:当我构建“后退” – “模型”的项目应该运行一些任务,但是当我构建“前端”时我不这样做.
是否有可能强制清理子项目?
您可以轻松地从命令行传递参数,也可以通过gradle.properties或ext中定义的属性传递参数.这是一个小项目:/build.gradle
ext { var = "alpha" } task hello { println "Name: $project.name" println "Var: $var" println "Param: $param" println "Prop: $prop" }
/gradle.properties
prop=gama
/settings.gradle
project.name = 'foo' include 'bar'
/bar/build.gradle
task hello { println "Name: $project.name" println "Var: $var" println "Param: $param" println "Prop: $prop" }
如果你运行gradle hello -Pparam = beta它会打印:
Name: foo Var: alpha Param: beta Prop: gama Name: bar Var: alpha Param: beta Prop: gama :hello UP-TO-DATE :bar:hello UP-TO-DATE