Question

我有一个关于初始化与groovy / grails的问题。当我有以下类时，sInstance不会进入SService初始化。

class A {

    String sInstance
    String app
    String dbInstance

    SService s = new SService(sInstance:sInstance, app:app)
}

SService类：

class SService {

    String sInstance
    String app

public getSInstance{
    return sInstance
    }
}

返回null，其中

class A {

    String sInstance
    String app
    String dbInstance

public initializeSService{
    SService s = new SService(sInstance:sInstance, app:app)
    }
}

从SService类返回sInstance变量。

为什么这样，我如何使用A类构造函数初始化SService对象？

Answer 1

你不能这样做：

class A {

    String sInstance
    String app
    String dbInstance

    SService s = new SService(sInstance:sInstance, app:app)
}

问题在于，当您创建SService的实例时，尚未初始化sInstance。如果要将sInstance传递给A类中某个其他类的构造函数，则必须在为sInstance赋值之后执行此操作，就像在完全构造A之后调用的方法一样。

编辑：

试图澄清以下评论中的内容：

class A {
    String sInstance
    String app
    String dbInstance

    void anyMethod() {
        // this will work as long as you have initialized sInstance
        SService s = new SService(sInstance:sInstance, app:app)
    }
}

根据你真正想做的事情，可能会朝这个方向发展：

class A {
    String sInstance
    String app
    String dbInstance
    SService s

    void initializeS() {
        if(s == null) {
            // this will work as long as you have initialized sInstance
            s = new SService(sInstance:sInstance, app:app)
        }
    }
}

或者：

class A {
    String sInstance
    String app
    String dbInstance
    SService theService

    SService getTheService() {
        if(theService == null) {
            // this will work as long as you have initialized sInstance
            theService = new SService(sInstance:sInstance, app:app)
        }
        theService
    }

    def someMethodWhichUsesTheService() {
        getTheService().doSomethingToIt()
    }
}

Grails初始化类

1 个答案: