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我刚刚开始学习 JavaScript，所以这对我来说可能是一个简单的错误，但我遇到的问题是我应该为每个数组中的每个字符串打印不同的指令到控制台，并且当我运行我的现在，它每次都重复相同的元素，并在每次迭代时将下一个字符串添加到它们，而不是为新数组中的每个单独的字符串提供指令。

这是我的代码：

const coffees = [ 
    "light colombian roast", "hawaiian dark roast", "guatemalan blend medium roast",
    "madagascar espresso roast", "jamaican dark blue", "jamaican medium roast",
    "salvador robusto light", "bali espresso roast"]

let lightRoast = []
let medRoast = []
let darkOrEspresso = []

for (const roast of coffees) 
{
    if (roast.includes("light") )
    {
        lightRoast.push(roast)
        console.log(`I'll have the ${lightRoast} and drink it black`)
    }
    else if (roast.includes("medium"))
    {
        medRoast.push(roast)
        console.log(`I'll have the ${medRoast} and add cream only`)
    }
    else if (roast.includes("dark") || roast.includes("espresso"))
    {
        darkOrEspresso.push(roast)
        console.log(`I'll have the ${darkOrEspresso} and add cream and sugar`)
    }
}

我试图让控制台为新数组中的每个单独的字符串打印这些指令，这取决于它是在 light、med 还是 dark 数组中：

"I'll have the xxx and drink it black" 
"I'll have the xxx and add cream only"
"I'll have the xxx and add cream and sugar"

但相反，我得到这样的结果，其中字符串位于正确的数组中，但被添加到彼此之上，而不是每个都有自己的行：

I'll have the light colombian roast,salvador robusto light and drink it black
I'll have the guatemalan blend medium roast,jamaican medium roast and add cream only
I'll have the hawaiian dark roast,madagascar espresso roast,jamaican dark blue,bali espresso roast and add cream and sugar

有人介意看看我做错了什么吗？希望我的指导有意义。

4

2 回答 2

2

尝试只打印烤，而不是整个阵列：

const coffees = [ 
    "light colombian roast", "hawaiian dark roast", "guatemalan blend medium roast",
    "madagascar espresso roast", "jamaican dark blue", "jamaican medium roast",
    "salvador robusto light", "bali espresso roast"]

let lightRoast = []
let medRoast = []
let darkOrEspresso = []

for (const roast of coffees) 
{
    if (roast.includes("light") )
    {
        lightRoast.push(roast)
        console.log(`I'll have the ${roast} and drink it black`)
    }
    else if (roast.includes("medium"))
    {
        medRoast.push(roast)
        console.log(`I'll have the ${roast} and add cream only`)
    }
    else if (roast.includes("dark") || roast.includes("espresso"))
    {
        darkOrEspresso.push(roast)
        console.log(`I'll have the ${roast} and add cream and sugar`)
    }
}
console.log(lightRoast, medRoast, darkOrEspresso);

于 2020-09-27T16:43:23.383 回答
2

精确的变量名称有助于防止错误。在这里，您的lightRoast变量是迄今为止已被确定为轻的咖啡名称的数组。考虑重命名为lightRoastNames（复数），然后重命名roast为roastName（单数）。然后，问题应该很明显 - 你正在使用

console.log(`I'll have the ${lightRoastNames} and drink it black`)

当你应该使用

console.log(`I'll have the ${roastName} and drink it black`)


const coffees = [
  "light colombian roast", "hawaiian dark roast", "guatemalan blend medium roast",
  "madagascar espresso roast", "jamaican dark blue", "jamaican medium roast",
  "salvador robusto light", "bali espresso roast"
]

const lightRoastNames = [];
const medRoastNames = [];
const darkOrExpressoRoastNames = [];

for (const roastName of coffees) {
  if (roastName.includes("light")) {
    lightRoastNames.push(roastName)
    console.log(`I'll have the ${roastName} and drink it black`)
  } else if (roastName.includes("medium")) {
    medRoastNames.push(roastName)
    console.log(`I'll have the ${roastName} and add cream only`)
  } else if (roastName.includes("dark") || roastName.includes("espresso")) {
    darkOrExpressoRoastNames.push(roastName)
    console.log(`I'll have the ${roastName} and add cream and sugar`)
  }
}

不过，由于您似乎没有在任何地方使用数组，您可以完全删除它们并使代码更简单：

const coffees = [
  "light colombian roast", "hawaiian dark roast", "guatemalan blend medium roast",
  "madagascar espresso roast", "jamaican dark blue", "jamaican medium roast",
  "salvador robusto light", "bali espresso roast"
]

for (const roastName of coffees) {
  if (roastName.includes("light")) {
    console.log(`I'll have the ${roastName} and drink it black`)
  } else if (roastName.includes("medium")) {
    console.log(`I'll have the ${roastName} and add cream only`)
  } else if (roastName.includes("dark") || roastName.includes("espresso")) {
    console.log(`I'll have the ${roastName} and add cream and sugar`)
  }
}

于 2020-09-27T16:43:57.807 回答