是否有可能使用两次ifPresent每一个检查不同的返回或分叉图

问题描述

上下文：我链接了由wsdl2java生成的对象。有一个已经创建的方法调用soap web服务，接收xml并在级联对象中解组它。

Context: I have chained objects generated by wsdl2java. There is an already created method that call the soap web services, receives the xml and unmarshalling it in cascaded objects.

欲望：我想得到两个不同的对象，它们位于不同层次的节点。从映射的角度来看，它们彼此远，但从商业角度看，它们是情侣。

Desire: I want to get two different objects which lies in different hierarchy nodes. From mapping perspective they are hold "far" from one another but from business perspective they are a "couple".

目前正在运作：

Optional.ofNullable(onderneming.getOndernemingOfVestiging()).map(OndernemingOfVestigingType::getCode) //same return
        .map(CodeOndernemingOfVestigingType::getValue) // different object return
        .ifPresent(o -> myMethod("onderneming_type", o.getValue()));

Optional.ofNullable(onderneming.getOndernemingOfVestiging()).map(OndernemingOfVestigingType::getCode) //same return
        .map(CodeOndernemingOfVestigingType::getBeschrijving) // different object return
        .ifPresent(o -> myMethod("onderneming_type_omschrijving", o.getValue()));

我正在尝试的是（伪解释）：

What I am trying is something like (pseudo-explanation):

Optional.ofNullable(onderneming.getOndernemingOfVestiging()).map(OndernemingOfVestigingType::getCode) //same return
        .map(CodeOndernemingOfVestigingType::getValue) // different object return
        .ifPresent(o -> myMethod("onderneming_type", o.getValue()))
        .map(CodeOndernemingOfVestigingType::getBeschrijving) // different object return
        .ifPresent(o -> myMethod("onderneming_type_omschrijving", o.getValue()));

我知道这不起作用，因为我只有一个终止clausure而ifPresent出现两次。

I understand that this doesn't work because I can have only one terminate clausure and ifPresent appears twice.

从虚拟角度来看，我的问题也可能是：是否可以在同一个lambda表达式中以两个不同的方向分叉/分叉/划分地图？

From dummy viewpoint, my question could be also: is it possible to fork/bifurcate/divide map in two different direction in same lambda expression?

***基于Eugene的建议我成功了

*** Based on Eugene suggestion I did successfully

Optional.ofNullable(onderneming.getOndernemingOfVestiging()).map(OndernemingOfVestigingType::getCode)
        .ifPresent(o -> {
            System.out.println(o.getValue().getValue());
            System.out.println(o.getBeschrijving().getValue());
        }); 

我从myMethod更改为System.out.println所以没有造成混淆（我不在乎）关于myMethod在这个上下文中。它可以是任何实例化对象的任何其他方法。）

I changed from myMethod to System.out.println so no confusion is caused (I don't care about myMethod in this context. It could be anyother method of any instanced object).

虽然它起作用，但与原始的相比，它听起来是一个很好的进步（而不是两个lambda）句子只有一个），我看到一些问题：o.getValue（）。getValue（）和o.getBeschrijving（）。getValue（）是空的不安全，不是吗？

Althought it works and it sounds a good progress compared to original one (instead of two lambda sentences there are only one), I see some problem: "o.getValue().getValue()" and "o.getBeschrijving().getValue()" are null unsafety, aren't they?

推荐答案

你可以抓住你的结果：

Optional.ofNullable(onderneming.getOndernemingOfVestiging()).map(OndernemingOfVestigingType::getCode)
    .map(CodeOndernemingOfVestigingType::getValue) 
    .map(o -> {
        myMethod("onderneming_type", o.getValue());
        return o.getBeschrijving();
    })
    .ifPresent(o -> myMethod("onderneming_type_omschrijving", o.getValue()));

请注意，多行lambda是坏的，难以阅读，所以你可以把它提取到私人方法。

Take a notice, that multiline lambdas are bad and hard to read, so you can extract it into private method.

编辑
您想要分支您的可选。是不可能的，您应该这样做：

EDITYou want to branch your Optional. Is not possible and you should do:

Optional<Code> op = Optional.ofNullable(onderneming.getOndernemingOfVestiging())
    .map(OndernemingOfVestigingType::getCode) 

op.map(CodeOndernemingOfVestigingType::getValue)
    .ifPresent(o -> myMethod("onderneming_type", o.getValue()));

op.map(CodeOndernemingOfVestigingType::getBeschrijving) 
    .ifPresent(o -> myMethod("onderneming_type_omschrijving", o.getValue()));

它应该比一些黑客更具可读性。

It should be more readable than do some hacks.

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