可选的获取价值(如果存在) [英] Optional get value if present

问题描述

作为一个例子，我有一个像这样的可选内容:

As an example I have an optional like this:

Optional<Desktop> opt = Optional.ofNullable(status).map(Status::getDesktop);

我想拥有桌面并在lambda表达式之外使用它.我是这样的:

I want to have the desktop and work with it outside of lambda expressions. I do it like this:

if (opt.isPresent()){
    Desktop desktop = opt.get();
    ...
}

是否有更好的解决方案来获得台式机?像这样的东西?

Is there a better solution to get the desktop; something like this ?

Desktop desktop = Optional.ofNullable(status).map(Status::getDesktop).ifPresent(get());

OrElse是我正在寻找的方法:)

OrElse was the method I was looking for :)

推荐答案

如果您的Desktop具有默认值，则可以尝试使用Optional.orElse:

If you have a default value for your Desktop, you could try with Optional.orElse:

Desktop defaultDesktop = ...;

Desktop desktop = Optional.ofNullable(status)
    .map(Status::getDesktop)
    .orElse(defaultDesktop);

但是，您不一定必须在Optional.ifPresent的lambda表达式内工作.您可以完美地使用接收Desktop实例的方法，该方法将充当Optional.ifPresent的Consumer参数:

However, you don't have to necessarily work inside a lambda expression with Optional.ifPresent. You could perfectly use a method that receives a Desktop instance, which would act as the Consumer argument of Optional.ifPresent:

Desktop desktop = Optional.ofNullable(status)
    .map(Status::getDesktop)
    .ifPresent(this::workWithDesktop);

然后:

void workWithDesktop(Desktop desktop) {
    // do whatever you need to do with your desktop
}

如果您需要其他参数(除了桌面本身)，则可以使用lambda表达式来调用该方法:

If you need additional arguments (apart from the desktop itself), you could use a lambda expression that invokes the method instead:

String arg1 = "hello";
int arg2 = 10;

Desktop desktop = Optional.ofNullable(status)
    .map(Status::getDesktop)
    .ifPresent(desktop -> this.workWithDesktop(desktop, arg1, arg2));

然后:

void workWithDesktop(Desktop desktop, String arg1, int arg2) {
    // do whatever you need to do with your desktop, arg1 and arg2
}

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