相关疑难解决方法(0)

$('button').click(function () {
   [1, 2, 3, 4, 5].forEach(function (n) {
      if (n == 3) {
         // it should break out here and doesn't alert anything after
         return false
      }
      alert(n)      
   })
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<button>Click me</button>

我的问题:虽然我打电话,为什么它仍会提醒下一个号码return？就像:忽略下面的代码并继续下一个元素

我知道有很多这样的话题.我知道基础知识:.forEach()在原始数组和.map()新数组上运行.

就我而言:

function practice (i){
    return i+1;
};

var a = [ -1, 0, 1, 2, 3, 4, 5 ];
var b = [ 0 ];
var c = [ 0 ];
console.log(a);
b = a.forEach(practice);
console.log("=====");
console.log(a);
console.log(b);
c = a.map(practice);
console.log("=====");
console.log(a);
console.log(c);

这是输出:

[ -1, 0, 1, 2, 3, 4, 5 ]
=====
[ -1, 0, 1, 2, 3, 4, 5 ]
undefined
=====
[ -1, 0, 1, 2, 3, 4, 5 ] …