本文介绍了创建一个在gradle中的所有其他任务之前运行的任务的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我需要创建一个初始化任务,该任务在执行时将在所有其他任务之前运行.
I need to create an initialize task that will run before all other task when I execute it.
task A { println "Task A" } task initializer { println "initialized" }如果我执行gradle -q A,输出将是:
If I execute gradle -q A, the output will be:
>initialized >Task A现在,如果我要添加:
task B { println "Task B" }执行gradle -q B,我得到:
>initialized >Task B所以我执行哪个任务都没有关系,它总是首先被初始化".
So it doesn't matter which task I execute, it always get "initialized" first.
推荐答案您可以使每个名称不是"initializer"的Task都依赖于"initializer"任务.例如:
You can make every Task who's name is NOT 'initializer' depend on the 'initializer' task. Eg:
task initializer { doLast { println "initializer" } } task task1() { doLast { println "task1" } } // make every other task depend on 'initializer' // matching() and all() are "live" so any tasks declared after this line will also depend on 'initializer' tasks.matching { it.name != 'initializer' }.all { Task task -> task.dependsOn initializer } task task2() { doLast { println "task2" } }或者您可以添加 BuildListener (或使用一个便利方法,例如: Gradle.buildStarted(...))
Or you could add a BuildListener (or use one of the convenience methods eg: Gradle.buildStarted(...))
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创建一个在gradle中的所有其他任务之前运行的任务
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