繁体   English   中英

访问不使用变量的对象

[英]Accessing an object not working with a variable

这似乎很简单,但我撞墙了。 我的代码获取权重,访问对象数组以获取一个值,然后使用该值*权重来计算结果。 但是访问对象不适用于变量。

 function calc(){
var gender;
if(document.getElementById("male").checked){
    gender = "mensList";
} else if (document.getElementById("female").checked){
    gender = "womensList";
} else {
    alert("Please select a gender");
    return false;
}
var kg = parseInt(document.getElementById("bwKg").value);
var grams = parseFloat(document.getElementById("bwGrams").value);
var bw = parseFloat(kg + grams);
var lifted = parseFloat(document.getElementById("liftWeight").value);

var theValue = womensList[bw]; // This works
var theValue = mensList[bw];   // This also works
var theValue = gender[bw];     // This doesn't work
var theValue = gender + "[\"" + bw + "\"]" // Nor this

var result = theValue * lifted;
document.getElementById("result").textContent = result;
 }

 var womensList = {
  40.0: "1.4936",
  40.1: "1.4915",
  40.2: "1.4894",
  40.3: "1.4872",
  40.4: "1.4851",
  // ......... etc
  150.7: "0.7691",
  150.8: "0.7691",
 150.9: "0.7691"
};

var mensList = {
  40.0: "1.3354",
  40.1: "1.3311",
  40.2: "1.3268",
  40.3: "1.3225",
  40.9: "1.2975",
  // ......... etc
  205.7: "0.5318",
  205.8: "0.5318",
  205.9: "0.5318"
 };

这是你的问题:

if(document.getElementById("male").checked){
    gender = "mensList";                         //gender now contains a string only..
} else if 

由于性别仅包含字符串,因此将无法使用:

var theValue = gender[bw];     // This doesn't work

您应该做的是:

if(document.getElementById("male").checked){
    gender = mensList;                         //now gender contains an array provided mensList is defined beforehand ..
} 

在您的代码中,您将gender设置为string不变。

if(document.getElementById("male").checked){
    gender = "mensList";
} else if (document.getElementById("female").checked){
    gender = "womensList";
}

应该

if(document.getElementById("male").checked){
    gender = mensList;
} else if (document.getElementById("female").checked){
    gender = womensList;
}

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM