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Javascript 如何将带有另一个数组的数组添加到新数组中?

[英]Javascript How to add array with another array into a new array?

So my challenge right now is unable to add array with another array into another new array.所以我现在的挑战是无法将带有另一个数组的数组添加到另一个新数组中。

my code:我的代码:

var disabletimerange = { basedonservicetype: "", basedonservicer: "" }
var arr = [{"StartTime":"09:00:00","EndTime":"09:10:00"},{"StartTime":"10:00:00","EndTime":"15:00:00"}]
for (var xx = 0; xx < arr.length; xx++) {
    var bookedStartTime, bookedEndTime
    bookedStartTime = arr[xx].StartTime.slice(0, -3)
    bookedEndTime = arr[xx].EndTime.slice(0, -3)

    if (disabletimerange.basedonservicer === "") {
        var disabletimealreadybookedtiming = "[" + '"' + bookedStartTime + '"' + "," + '"' + bookedEndTime + '"' + "]"
        disabletimerange.basedonservicer = JSON.parse(disabletimealreadybookedtiming)
    } else {
        var disabletimealreadybookedtiming = "[" + '"' + bookedStartTime + '"' + "," + '"' + bookedEndTime + '"' + "]"
        disabletimerange.basedonservicer = "[" + "[" + disabletimerange.basedonservicer + "]" + "," + disabletimealreadybookedtiming + "]"
        disabletimerange.basedonservicer = JSON.parse(disabletimerange.basedonservicer)
}

Uncaught SyntaxError: Unexpected number in JSON at position 3 --> at the last line it cannot parse. Uncaught SyntaxError: Unexpected number in JSON at position 3 --> 在最后一行它无法解析。

In short, what i want is basically 2 array stick tgt instead of adding into it简而言之,我想要的基本上是 2 个数组棒 tgt 而不是添加到其中

var y = ["hi", "123"]
var x = ["yo", "312"]

this
[["hi", "123"], ["yo", "312"]] instead of ["hi", "123", "yo", "312"]

and it must be array.它必须是数组。 not text form不是文本形式

You can do:你可以做:

 var y = ["hi", "123"] var x = ["yo", "312"] var result = [x, y] console.log(result)

As you described正如你所描述的

How to add array with another array into a new array?如何将带有另一个数组的数组添加到新数组中?

  • create a new array创建一个新数组

  • push your arrays into this new array one by one将您的 arrays 一个一个推入这个新阵列

 let x = [1,2]; let y = [3.4]; let z = []; z.push(x) z.push(y); console.log(z);

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