别再使用 .reverse().find()：认识 findLast()

If you’ve ever needed to search an array from the end, you’ve probably reached for a combination of .slice().reverse().find(...), or looped backward manually. It works, but it’s not an elegant solution.

如果你曾经需要从数组末尾开始搜索，你可能会使用 .slice().reverse().find(...) 的组合，或者手动反向循环。虽然可行，但不够优雅。

We now have a better way: Array.prototype.findLast() and Array.prototype.findLastIndex(). These features let you search arrays from the end without reversing them, which gives you cleaner and more intuitive code.

我们现在有了更好的方式：Array.prototype.findLast() 和 Array.prototype.findLastIndex()。这些特性让你无需反转数组即可从末尾开始搜索，带来更简洁、更直观的代码。

The problem with reversing arrays

反转数组的问题所在

Working with lists in front-end apps (e.g., chat messages, activity logs, form history, etc.) often involves finding the last matching item. Prior to ES2023, you’d typically do something like:

在前端应用中处理列表（例如聊天消息、活动日志、表单历史等）时，通常需要找到 最后一个匹配项。在 ES2023 之前，你通常会这样做：

const lastError = [...logs].reverse().find(log => log.type === 'error');

Reverses a cloned array, which is inefficient
反转克隆后的数组，效率较低
Risky if you forget to clone (.slice())
如果忘记克隆（.slice()）会有风险
Less readable and more error-prone
可读性更差，更容易出错

Now, you can write:

现在，你可以这样写：

const lastError = logs.findLast(log => log.type === 'error');

No copying or reversing
禁止复制或反转
Cleaner and clearer
更简洁、更清晰
Safer and more performant, especially with large datasets
更安全且性能更高，尤其是在处理大型数据集时

`findLast(predicate)`

This method works just like .find(), but starts at the end of the array and moves backward. It takes a predicate, a function that returns true or false, for each item.

该方法与 .find() 类似，但从数组末尾开始向前遍历。它接收一个 predicate，即一个为每个元素返回 true 或 false 的函数。

const messages = [ { id: 1, text: 'Hello', read: true }, { id: 2, text: 'Hi', read: false }, { id: 3, text: 'Hey', read: true },
]; const lastUnread = messages.findLast(msg => !msg.read); console.log(lastUnread);

Like .find(), the predicate receives three arguments: (element, index, array):

与 .find() 类似，谓词函数会接收三个参数：(element, index, array)：

array.findLast((element, index, array) => { })...