android - 如何使用模块创建库？

Question

我想创建几个库(核心和模块)，将所有这些库添加到应用程序并从核心库中的模块调用方法。所有模块都应该是可选的。例如，在我看来:

申请 build.gradle :

compile('core-library:1.0@aar')
compile('module1-library:1.0@aar')
compile('module2-library:1.0@aar')
compile('module3-library:1.0@aar')

在每个模块中定义具有相同名称的类和方法:

public class ModuleClass {
    public int moduleMethod1() {
        // Do something and return result
        return 1;
    }

    public String moduleMethod2() {
        return "Some String";
    }
}

在核心库中:

for(ModuleClass c : getAllModules()) {
    Log.d("tag", "Result: " + c.moduleMethod1() + " / " + c.moduleMethod2();
}

当然这只是伪代码。如何实现它或类似的东西？

更新 :
核心模块 build.gradle :

apply plugin: 'com.android.library'

android {
    compileSdkVersion 24
    buildToolsVersion "24.0.3"

    defaultConfig {
        minSdkVersion 9
        targetSdkVersion 24
        versionCode 1
        versionName "1.0"

        testInstrumentationRunner "android.support.test.runner.AndroidJUnitRunner"

        buildConfigField "String[]", "KNOWN_MODULES", "{" +
                "\"module1\"," +
                "\"module2\"" +
                "}"
    }
    buildTypes {
        release {
            minifyEnabled false
            proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
        }
    }
}

dependencies {
    compile fileTree(include: ['*.jar'], dir: 'libs')
    compileOnly project(':module1')
    compileOnly project(':module2')
    testCompile 'junit:junit:4.12'
    androidTestCompile('com.android.support.test.espresso:espresso-core:2.2.2', {
        exclude group: 'com.android.support', module: 'support-annotations'
    })
}

Answer 1

您可以尝试添加 List<Class>在 core-library .当您尝试加载列表时，您必须非常小心，但您应该能够做到。

首先你的moduleN-library需要compileOnly所以它不会与 core-library bundle 在一起.

然后在你的core-library保持一个安全地列出所有可能的模块类的单例。

package com.jbirdvegas.test;

import java.util.HashMap;
import java.util.Map;

public class ModuleClassManager {
    private static final Object LOCK = new Object();
    private static ModuleClassManager instance;

    private Map<Class, Methods> moduleClasses = new HashMap<>();

    private ModuleClassManager() {
        try {
            moduleClasses.put(MayNotExist1.class, new MayNotExist1());
        } catch (Exception ignored) {
        }
        try {
            moduleClasses.put(MayNotExist2.class, new MayNotExist2());
        } catch (Exception ignored) {
        }
        try {
            moduleClasses.put(MayNotExist3.class, new MayNotExist3());
        } catch (Exception ignored) {
        }
    }

    public static ModuleClassManager getInstance() {
        if (instance == null) {
            synchronized (LOCK) {
                instance = new ModuleClassManager();
            }
        }
        return instance;
    }

    public void callMethodOnAll() {
        moduleClasses.forEach((aClass, methods) -> methods.methodOne());
    }

    // the below interface in the real world would be located
    // in the `core-library` and the classes that implement the
    // interface, `Methods` in this example, would exist in the
    // module's they represent.
    interface Methods {
        void methodOne();
    }

    class MayNotExist1 implements Methods {
        @Override
        public void methodOne() {
        }
    }

    class MayNotExist2 implements Methods {
        @Override
        public void methodOne() {
        }
    }

    class MayNotExist3 implements Methods {
        @Override
        public void methodOne() {
        }
    }
}

或者...使用 OSGi。 OSGi 设置起来更复杂，但是它处理这个用例非常好。