Question

如何让@UIApplicationMain class AppDelegate: UIResponder, UIApplicationDelegate { var window: UIWindow? var data: JSON = [] func application(application: UIApplication, didFinishLaunchingWithOptions launchOptions: [NSObject: AnyObject]?) -> Bool { loadData() return true } func loadData () { Alamofire.request(.GET, "...", parameters: nil) .responseJSON { response in data = JSON(response.result.value!) } } }的Gradle @Input工作？我尝试添加List<CustomClass>方法，这有所帮助，但它仍然以奇怪的方式失败。什么是使其可序列化的正确方法？这是Gradle版本2.4。

失败：toString()

List<CustomClass>

此操作失败并显示以下消息：

@Input
List<CustomClass> getMyInput() {
    List<CustomClass> simpleList = new ArrayList<>()
    simpleList.add(new CustomClass())
    return simpleList
}

static class CustomClass {
    String str
}

成功："Unable to store task input properties. Property 'myInput' with value '[null]' cannot be serialized."

List<String>

创建例外的Gradle源代码参考： https://github.com/gradle/gradle/blob/master/subprojects/core/src/main/groovy/org/gradle/api/internal/changedetection/state/InputPropertiesSerializer.java#L42

Answer 1

该类需要实现'Serializable'接口。这是为了允许Gradle完全序列化对象，以便可以在构建之间比较二进制格式以查看是否有任何更改。一旦CustomClass可序列化，那么List<CustomClass>也可以序列化，只要它使用ArrayList等标准实现。

以下是一个例子：

@EqualsAndHashCode
static class CustomClass implements Serializable {
    // Increment this when the serialization output changes
    private static final long serialVersionUID = 1L;

    String projectName

    @SuppressWarnings('unused')
    private static void writeObject(ObjectOutputStream s) throws IOException {
        s.defaultWriteObject();
    }

    // Gradle only needs to serialize objects, so this isn't strictly needed
    @SuppressWarnings('unused')
    private static void readObject(ObjectInputStream s) throws IOException {
        s.defaultReadObject();
    }
}

Gradle：无法存储任务输入属性。值'[null]'的属性'myInput'无法序列化

1 个答案: