JS解构.如何处理空值或未定义的值 [英] JS destructuring. How to deal with null or undefined values

问题描述

如何在 es6 .这样的东西，在每个地方:

How to implement Nullable feature in es6? I need to support the source code of my previous co-worker, who used too much destructuring feature of es6. Something like this, every where:

dispatch(
    loadImports(response.items.map(({ importRecord: { ['import']: importId } }) => importId))
)

在此示例中，可能会出现TypeError: Cannot read property 'import' of null错误.

In this example, it's possible that I may get TypeError: Cannot read property 'import' of null error.

我不想将整个结构重写为常规条件.或者，如果没有，如何在不重写的情况下处理它们?

I don't want to rewrite the whole destructures to regular conditions. Or if there isn't, how to deal with them, without rewriting?

UPD:

同事的预期版本: https://jsbin.com/fesewiy/edit?js,console

当前版本: https://jsbin.com/qirixil/edit?js,console

推荐答案

由于使用的是地图，因此需要为每个项目获取一些结果. 更改最少的最佳解决方案是仅使用一个级别的解构，然后对数组进行过滤以保留不虚假的值数组.

Because you are using map, you need to get some result for each item. The best solution with minimum changes will be to use only one level of destructuring and then filter the array to stay with array of values that are not falsy.

dispatch(
    loadImports(response.items
        .map(({ importRecord }) => importRecord && importRecord.import)
        .filter(v => v)
))

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