var arr = [
[{a: "apple", b: "ball"}, {c: "cat", d: "dog"}],
[{e: "elephent", f: "flight"}, {g: "god", h: "hen"}]
];
预期结果是——
var exp = [
{a: "apple", b: "ball"},
{c: "cat", d: "dog"},
{e: "elephent", f: "flight"},
{g: "god", h: "hen"}
]
您可以使用flat()
:
var arr = [[{a: "apple", b: "ball"}, {c: "cat", d: "dog"}],[{e: "elephent", f: "flight"}, {g: "god", h: "hen"}]];
console.log(arr.flat());
对于 Microsoft Edge 和 IE,请使用[].concat(...arr)
尝试使用以下代码:
[].concat(...arr)
您可以访问数组的索引0
并将其分配给存储数组的现有变量
类似的东西
arr = arr[0]
*由于问题被误读,此答案需要更新
let newArr = []
newArr = arr.map((item) {
return item[0]
})
这使arr
保持不变。如果您宁愿修改arr
请使用forEach
而不是map