在同一索引值中推送多个值 JavaScript
push multiple value in same index value JavaScript
我是 javascript
的新手
问题
['name',1,2,3,4,5]
我需要:-
['name','12345']
代码:-
var abc = [];
text = 'name';
abc.push(text);
var def = [1,2,3,4,5]
$.each(def, function(index, item) {
abc.push(item);
});
你可以试试这个
var abc = [];
text = 'name';
abc.push(text);
var def = [1,2,3,4,5]
abc.push(def.join(''));
let data = ['name',1,2,3,4,5]
let result = [data[0],data.slice(1).join('')]
console.log(result)
您可以从末尾加入,直到获得所需的数组长度。
const array = ['name', 1, 2, 3, 4, 5];
while (array.length > 2) array.push(array.splice(array.length - 2, 2).join(''));
console.log(array);
您可以使用 destruction assignment(...
):
const array = ['name', 1, 2, 3, 4, 5];
const [key, ...value] = array;
const result = [key, value.join('')];
console.log(result);
我是 javascript
的新手问题
['name',1,2,3,4,5]
我需要:-
['name','12345']
代码:-
var abc = [];
text = 'name';
abc.push(text);
var def = [1,2,3,4,5]
$.each(def, function(index, item) {
abc.push(item);
});
你可以试试这个
var abc = [];
text = 'name';
abc.push(text);
var def = [1,2,3,4,5]
abc.push(def.join(''));
let data = ['name',1,2,3,4,5]
let result = [data[0],data.slice(1).join('')]
console.log(result)
您可以从末尾加入,直到获得所需的数组长度。
const array = ['name', 1, 2, 3, 4, 5];
while (array.length > 2) array.push(array.splice(array.length - 2, 2).join(''));
console.log(array);
您可以使用 destruction assignment(...
):
const array = ['name', 1, 2, 3, 4, 5];
const [key, ...value] = array;
const result = [key, value.join('')];
console.log(result);