如何像forEach一样遍历不可变列表?

本文介绍了如何像forEach一样遍历不可变列表?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我想遍历Immutable List，我用List.map来做到这一点，它可以工作，但效果不好.有没有更好的办法?因为我只检查数组中的每个元素，所以如果该元素符合我的规则，我会做某些事情，就像Array.forEach一样，我不想返回任何Array.map这样的东西.

I would like to loop through Immutable List, I used List.map to do it, it can be worked, but not good. is there are a better way? Because I just check each element in array, if the element match my rule, I do something, just like Array.forEach, I don't want to return anything like Array.map.

例如，这是我现在的工作:

for example, it is my work now:

let currentTheme = ''; let selectLayout = 'Layout1'; let layouts = List([{ name: 'Layout1', currentTheme: 'theme1' },{ name: 'Layout2', currentTheme: 'theme2' }]) layouts.map((layout) => { if(layout.get('name') === selectLayout){ currentTheme = layout.get('currentTheme'); } });

推荐答案

Immutable.js列表中存在方法List.forEach.

The method List.forEach exists for Immutable.js lists.

但是，一种更实用的方法将使用方法List.find如下:

However, a more functional approach would be using the method List.find as follows:

let selectLayoutName = 'Layout1'; let layouts = List([Map({ name: 'Layout1', currentTheme: 'theme1' }),Map({ name: 'Layout2', currentTheme: 'theme2' })]) selectLayout = layouts.find(layout => layout.get('name') === selectLayoutName); currentTheme = selectLayout.get('currentTheme')

然后您的代码没有副作用.

Then your code doesn't have side-effects.