Question

如何使用ES6简写将console.log放入过滤器，映射，缩小或拒绝？

words.filter(word => word.length > 6);

ES6版本

＆＃13;

const words = ["spray", "limit", "elite", "exuberant", "destruction", "present", "happy"];

let longWords = words.filter(word => word.length > 6); //NEED AN EXAMPLE FOR HERE

// Filtered array longWords is ["exuberant", "destruction", "present"]

console.log(longWords);

＆＃13;

可能在这里

＆＃13;

const words = ["spray", "limit", "elite", "exuberant", "destruction", "present", "happy"];

let longWords = words.filter(function(word) {
  console.log(word);
  return word.length > 6;
});

＆＃13;

Answer 1

您可以使用或者因为控制台不会返回任何内容

word => console.log(word) || word.length > 6

我个人只是使用括号和返回，如果我需要调试

Answer 2

如果您正在寻找一个不会遮挡关键操作的可重复使用的代码段，您可以先定义帮助程序：

const firstDo = f => g => (...args) => (f(...args), g(...args));

这需要两个函数并返回一个新函数。新函数使用相同的参数调用f和g，但返回 g(...args)的结果。

这样，您就可以点按map，reduce，filter，every等，并在调试完成后轻松删除日志记录。

对于filter示例，调试记录器可以是：

const logLength = str => console.log(`Length of "${str}": ${str.length}`);

现在，您可以将谓词包装在：

const tapLog = firstDo(logLength);

const longWords = words.filter(tapLog(x => x > 6));

在一个工作示例中，包含reduce示例：

const words = ["spray", "limit", "elite", "exuberant", "destruction", "present", "happy"];

const firstDo = f => g => (...args) => (f(...args), g(...args));

const logLength = str => console.log(`Length of "${str}": ${str.length}`);



const firstLogLength = firstDo(logLength);

const pred = x => x.length > 6;

// in a filter
console.log(
  words.filter(firstLogLength(pred))
);

// In a reduce
const sum = (a, b) => a + b;
const logSumProg = firstDo((a, b) => console.log(`${a} + ${b}`));

console.log(
  [1,2,3,4,5].reduce(logSumProg(sum))
)

Answer 3

根据您要调试的内容，可以更轻松地设置条件断点。

let longWords = words.filter(word => word.length > 6);

在Firefox调试器中，您可以右键单击要调试的代码行。然后选择＆＃34;添加条件断点＆＃34;，例如添加typeof word !== 'undefined' && word == 'some word'。现在，当您运行代码时，如果满足这些条件，代码的执行将暂停，您可以使用常用的调试器函数。

您应该尽可能避免更改代码以进行调试。

在ES6中使用简写符号进行调试 - 过滤，映射，拒绝或减少

3 个答案: