Question

我有像我的var bar一样的数组，我想在数组中的每个第二位添加var test。

var test = "foo";

var bar = ["Wilson 1", "Wilson 2"];

var new = ["Wilson 1", "foo" "Wilson 2", "foo"];

我尝试使用concat但没有结果：

var new = bar.concat(test);

Answer 1

这没有标准功能，所以你必须自己做。你会如何解决它？我会

创建新阵列
循环遍历旧数组（bar）
对于bar的每次迭代，我都会在新数组中添加一个项目+你想要的字符串。

修改

所有其他答案都很糟糕，我觉得有必要添加一个完整的例子：

const test = "foo"; const bar = ["Wilson 1", "Wilson 2"]; const newA = []; for (const item of bar) { newA.push(item, test); }

Answer 2

这可以通过简单的for循环来实现

var test = "foo";
var bar = ["Wilson 1", "Wilson 2"];
var newA = [];

for (var i = 0; i < bar.lenght; i ++) {
    newA.push(bar[i]);
    newA.push(test);
}

Answer 3

您可以使用ES6的spread运算符，array.prototype.concat和array.prototype.map：

var test = "foo";

var bar = ["Wilson 1", "Wilson 2"];

var result = [].concat(...bar.map(e => [e, test]));

console.log(result);

Answer 4

您可以采用while循环并将值拼接为每个第二项。

＆＃13;

var item = "foo",
    array = ["Wilson 1", "Wilson 2"],
    i = 1;

while (i <= array.length) {
    array.splice(i, 0, item);
    i += 2;
}

console.log(array);

＆＃13;

或使用Array#reduce

＆＃13;

var item = "foo",
    array = ["Wilson 1", "Wilson 2"],
    result = array.reduce((r, a) => r.concat(a, item), []);

console.log(result);

＆＃13;

Answer 5

我解决了：

var new = bar.map((m, i) => (m).concat(new Array(test[i])));

Answer 6

您可以使用.forEach()：

var test = "foo";

var bar = ["Wilson 1", "Wilson 2"];

var newArr = [];

bar.forEach(el => newArr.push(el, test))

console.log(newArr)

Answer 7

您可以使用array#reduce

var test = "foo",
    bar = ["Wilson 1", "Wilson 2"],
    result = bar.reduce((r,v) => {
      r.push(v, test);
      return r;
    },[]);
console.log(result);

Answer 8

使用.reduce并在其中添加current element和test

＆＃13;

var test = "foo";
var bar = ["Wilson 1", "Wilson 2"];

var newArr = bar.reduce((acc, el) => acc.concat([el, test]) , []);

console.log(newArr);

＆＃13;

每隔一个地方连接两个数组

8 个答案: