javascript - 将变量设置为以 Null 返回的字符串

Question

我现在有这些变量

var rsCash = 0;
var rsCashString = rsCash.toString();
var rsCashLength = rsCashString.length;

我在 Switch 语句中使用 rsCash 变量，我需要数字的长度，但我收到错误消息“Uncaught TypeError: Cannot set property 'innerHTML' of null

这是 Switch 语句

switch (rsCashLength){
case 1:
    document.getElementById("cash3").innerHTML = rsCash + "Gp";
    break;
case 2:
    document.getElementById("cash3").innerHTML = rsCash + "Gp";
    break;
case 3:
    document.getElementById("cash3").innerHTML = rsCash + "Gp";
    break;
case 4:
    document.getElementById("cash3").innerHTML = rsCash.substring(0,cut1) + "k";
    break;
case 5:
    document.getElementById("cash3").innerHTML = rsCash.substring(0,cut2) + "k";
    break;
case 6:
    document.getElementById("cash3").innerHTML = rsCash.substring(0,cut3) + "k";
    break;
case 7:
    document.getElementById("cash3").innerHTML = rsCash.substring(0,cut1) + "M";
    break;
case 8:
    document.getElementById("cash3").innerHTML = rsCash.substring(0,cut2) + "M";
    break;
case 9:
    document.getElementById("cash3").innerHTML = rsCash.substring(0,cut3) + "M";
    break;
case 10:
    document.getElementById("cash3").innerHTML = rsCash.substring(0,cut4) + "M";
    break;
case 11:
    document.getElementById("cash3").innerHTML = rsCash.substring(0,cut5) + "M";
    break;

我试过了

Switch (rsCash.toString().length){

我还有一个 Switch 语句，它像这样更改 rsCash 变量

switch (RewardNumber){
case 1:
    rsCash = rsCash + 200000;
    document.getElementById("Log").innerHTML = "$200k was found!";
    break;
case 2:
    rsCash = rsCash + 25000;
    document.getElementById("Log").innerHTML = "$15 was found!";
    break;
case 3:
    rsCash = rsCash + 5000;
    document.getElementById("Log").innerHTML = "5k was found!";
    break;
case 4:
    rsCash = rsCash + 500000;
    document.getElementById("Log").innerHTML = "500k was found!";
    break;
case 5:
    rsCash = rsCash + 75000;
    document.getElementById("Log").innerHTML = "75k was found!";
    break;
case 6:
    rsCash = rsCash + 5000000;
    document.getElementById("Log").innerHTML = "5m was found!";
    break;
case 7:
    rsCash = rsCash + 275000;
    document.getElementById("Log").innerHTML = "275k was found!";
    break;
case 8:
    rsCash = rsCash + 5000;
    document.getElementById("Log").innerHTML = "5k was found!";
    break;
case 9:
    rsCash = rsCash + 15000;
    document.getElementById("Log").innerHTML = "15k was found!";
    break;
case 10:
    rsCash = rsCash + 8000;
    document.getElementById("Log").innerHTML = "8k was found!";
    break;
case 11:
    rsCash = rsCash + 35000;
    document.getElementById("Log").innerHTML = "35k was found!";
    break;
case 12:
    rsCash = rsCash + 125000;
    document.getElementById("Log").innerHTML = "125k was found!";
    break;
case 13:
    rsCash = rsCash + 80000;
    document.getElementById("Log").innerHTML = "80k was found!";
    break;
case 14:
    rsCash = rsCash + 600000;
    document.getElementById("Log").innerHTML = "600k was found!";
    break;
case 15:
    rsCash = rsCash + 12000;
    document.getElementById("Log").innerHTML = "12k was found!";
    break;

}

我不明白为什么变量返回 null？

Answer 1

错误将您指向此代码(重复多次):

document.getElementById("cash3").innerHTML = rsCash + "Gp";

document.getElementById("cash3") 调用在这里返回 null。您需要正确找到您的元素。

你可能只需要这样做:

var cashElement = document.getElementById("correctID");

然后在整个过程中重用该变量。

正如评论者所说，这很可能是因为您在加载 DOM 之前执行代码。 jQuery 有简单的方法来处理这个问题，方法是将代码放在 $(document).ready 中，或者可以用 vanilla javascript 完成:

document.addEventListener("DOMContentLoaded", function() {
  // code…
});

这将允许您的代码在加载 DOM 后运行，从而使您的代码能够查看和访问 HTML。