gradle

gradle由简入深

邮差的信 提交于 2020-02-26 00:19:31
众所周知android编译有两种方式分别是,make和gradle Gradle 附 gradle文档 gradle这种方式包含project、task、action、closures,他们关系如下图: Project 一个project与一个build.gradle文件是一对一的关系 project与project的关系是树形结构的,通过getParent()获得父project,通过getSubprojects()获得所有子project 提出一个问题,build.gradle是一个project对象吗? 答案:不是的,根据官方文档介绍可以知道每一个build.gradle脚本与关联的project对象有一个委托关系。所以脚本使用的任何属性或方法都通过关联的项目对象进行委托。 举个例子 我们在build.gradle文件中常用的Project属性,实际代理的是 Project getProject(); Project Properties 一个project的properties可以有5种范围 project自身的,比如任何 getters and setters修饰的属性 extra 比如在ext{}里定义的 由插件添加的Convention 由插件添加的 extensions task的名字,比如task:compile,它的名字就是一个属性

使用public.xml来自定义资源id,支持最新gradle 5.4.1

天大地大妈咪最大 提交于 2020-02-25 23:10:10
来源: https://blog.csdn.net/m0_37165550/article/details/102716420 总结: 了解 aapt 是Android Asset Packaging Tool的缩写,是编译和打包资源的工具。而aapt2是在aapt上做了优化 。 熟悉aapt的运行过程,可以在中间实现自己想要的功能 新建: App\src\main\res\values\public.xml 内容类似如下: <resources> <public type="dimen" name="status_bar_height" id="0x7f0600e9" /> </resources> 在project根部放置脚本public-xml.gradle import org.gradle.util.GFileUtils apply plugin: PublicPlugin class PublicPlugin implements Plugin<Project> { void apply(Project project) { project.afterEvaluate { if (project.plugins.hasPlugin("com.android.application")) { def android = project.extensions

Gradle 入门--只此一篇

╄→гoц情女王★ 提交于 2020-02-25 19:37:49
是什么? 在语法上是基于Groovy语言的( Groovy 是一种基于JVM的敏捷开发语言,可以简单的理解为强类型语言java的弱类型版本),在项目管理上是基于Ant和Maven概念的项目自动化建构工具。 基础知识准备 Java基础,命令行使用基础 官方文档 : https://docs.gradle.org/current/dsl/ Gradle使用指南: https://gradle.org/docs/current/userguide/userguide Android插件文档 : https://github.com/google/android-gradle... AndroidGradle使用文档 : http://tools.android.com/tech-docs/new-build-system/user-guide Groovy基础: http://attis-wong-163-com.iteye.com/blog/1239819 Groovy闭包的Delegate机制 : https://www.cnblogs.com/zqlxtt/p/5741297.html 搭建Gradle运行环境 Gradle 运行依赖JVM,也就是java运行的环境。所以要安装jdk和jre,好像目前的Gradle的运行环境要求jdk的版本在1.6以上,应该的,现在jdk都到1

Exception in thread “main” java.lang.NoClassDefFoundError: kotlin/KotlinPackage & Caused by: java.lang.ClassNotFoundException: kotlin.KotlinPackage

守給你的承諾、 提交于 2020-02-25 16:59:12
问题 When I try to add IntArrays in my codes, the Android Studio gives me these errors: "C:\Program Files\Android\Android Studio\jre\bin\java.exe" "-javaagent:C:\Program Files\Android\Android Studio\lib\idea_rt.jar=50719:C:\Program Files\Android\Android Studio\bin" -Dfile.encoding=UTF-8 -classpath C:\Users\bosun\AppData\Local\Android\Sdk\platforms\android-29\android.jar;C:\Users\bosun\AppData\Local\Android\Sdk\platforms\android-29\data\res;C:\Users\bosun\AndroidStudioProjects\Test2\app\build

两行配置完全解放gradle编译慢问题

大憨熊 提交于 2020-02-25 16:05:21
Android Studio编译经常出现gradle编译缓慢甚至超时问题,抛开电脑硬件配置不说,主要问题还是国内网络环境的因素影响,可以通过修改项目根目录下的build.gradle文件如下: buildscript { ext.kotlin_version = '1.3.50' repositories { google() //jcenter() maven { url "https://jitpack.io" } maven { url "http://maven.aliyun.com/nexus/content/groups/public/" } } dependencies { classpath 'com.android.tools.build:gradle:3.5.2' classpath "org.jetbrains.kotlin:kotlin-gradle-plugin:$kotlin_version" } } allprojects { repositories { google() //jcenter() maven { url "https://jitpack.io" } maven { url "http://maven.aliyun.com/nexus/content/groups/public/" } } } 但是上面这种方式

Gradle Module Build Order

倖福魔咒の 提交于 2020-02-25 04:22:06
问题 I have a Gradle root project P with two subprojects P:foo and P:bar . Naturally, Gradle builds them in alphabetical order: bar , foo . But I need foo to be built first when I say gradle build in the P root directory. This is because bar depends on the AAR (Android library) artifact that foo publishes to the local Maven repository. Both bar and foo are such Android-library projects. This looks like an easy problem, but I can't figure it out. I read about evaluationDependsOn , so in bar/build

Python Script to out put dependencies of build.Gradle File

你。 提交于 2020-02-25 04:11:09
问题 My Python Script import re filename = "build.gradle" pattern = re.compile(r"dependencies", re.IGNORECASE) with open(filename, "rt") as in_file: for line in in_file: if pattern.search(line) != None: #for line in in_file: #print(line) print(line, end='') My Build.Gradle File buildscript { ext { springBootVersion = '2.1.0.RELEASE' } repositories { mavenCentral() } dependencies { classpath("org.springframework.boot:spring-boot-gradle-plugin:${springBootVersion}") } } apply plugin: 'java' apply

Python Script to out put dependencies of build.Gradle File

六眼飞鱼酱① 提交于 2020-02-25 04:10:25
问题 My Python Script import re filename = "build.gradle" pattern = re.compile(r"dependencies", re.IGNORECASE) with open(filename, "rt") as in_file: for line in in_file: if pattern.search(line) != None: #for line in in_file: #print(line) print(line, end='') My Build.Gradle File buildscript { ext { springBootVersion = '2.1.0.RELEASE' } repositories { mavenCentral() } dependencies { classpath("org.springframework.boot:spring-boot-gradle-plugin:${springBootVersion}") } } apply plugin: 'java' apply

Python Script to out put dependencies of build.Gradle File

孤街浪徒 提交于 2020-02-25 04:10:11
问题 My Python Script import re filename = "build.gradle" pattern = re.compile(r"dependencies", re.IGNORECASE) with open(filename, "rt") as in_file: for line in in_file: if pattern.search(line) != None: #for line in in_file: #print(line) print(line, end='') My Build.Gradle File buildscript { ext { springBootVersion = '2.1.0.RELEASE' } repositories { mavenCentral() } dependencies { classpath("org.springframework.boot:spring-boot-gradle-plugin:${springBootVersion}") } } apply plugin: 'java' apply

Supporting a Dagger 2 dependency in a Dagger 1 project

折月煮酒 提交于 2020-02-25 02:13:10
问题 This is a follow-up to my question about Play Services Cast Framework, where the solution seems to be updating from v17.0.0 to v18.0.0. The new version unfortunately has a dependency on Dagger 2, whilst the (large & complex) project is Dagger 1. We are currently using: api "com.google.android.gms:play-services-cast:17.0.0" api "com.google.android.gms:play-services-cast-framework:17.0.0" Updating these to 18.0.0 results in: > 1 exception was raised by workers: java.lang.RuntimeException: