Javascript“声明但从未使用过的变量”......即使我正在使用它？

Question

我肯定错过了什么。

在下面的代码中，我清楚地声明loopingAdjustment，然后在它下面我在fromCharCode函数中调用它。所以，我正在使用它，对吧？我应该可以调用它，因为它在同一个范围内，对吧？

为什么 VSCode 说它“从未使用过”，为什么我的终端说它“未定义”？

谢谢你。

const caesar = function(startingString, shiftAmount) {

    

    let itemizedString = startingString.split('');

    const mappedLetters = itemizedString.map(stringLetter => {

        

        //turn each letter in array into their respective character code

        let thisLetter = stringLetter.charCodeAt(stringLetter);

        // checking if character is alphabetic and converting its charcode back to a letter

        if (thisLetter < 65 || (thisLetter > 90 && thisLetter < 97) || thisLetter > 122) {

            return;

        } else {

            shiftedLetter = thisLetter + shiftAmount;

        }

        

        // making sure the shifted letters loop to beginning, or end, of alphabet

        if (thisLetter > 96 && shiftedLetter > 122) {

            let loopingAdjustment = shiftedLetter - 26;

        } else if (thisLetter > 96 && shiftedLetter < 96) {

            let loopingAdjustment = shiftedLetter + 26;

        } else if (thisLetter < 91 && shiftedLetter > 90) {

            let loopingAdjustment = shiftedLetter - 26;

        } else if (thisLetter < 91 && shiftedLetter < 65) {

            let loopingAdjustment = shiftedLetter + 26;

        } else {

            let loopingAdjustment = shiftedLetter;

        }

        let finalString = String.fromCharCode(loopingAdjustment);

        return finalString;

    });

    console.log(mappedLetters);

    return mappedLetters.join('');

}

module.exports = caesar

Answer 1

这是因为范围而发生的，因为您仅在 if 语句中声明和定义了 loopingAdjustment 变量，因此范围仅限于 if 语句。要解决此问题，您可以在 if 之外声明 loopingAdjustment 并在 if 语句中分配它。您可以从下面的代码中获取参考，您可以修改 acc. 根据您的需要。const caesar = function(startingString, shiftAmount) {&nbsp; &nbsp;&nbsp;&nbsp; &nbsp; let itemizedString = startingString.split('');&nbsp; &nbsp; const mappedLetters = itemizedString.map(stringLetter => {&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; //turn each letter in array into their respective character code&nbsp; &nbsp; &nbsp; &nbsp; let thisLetter = stringLetter.charCodeAt(stringLetter);&nbsp; &nbsp; &nbsp; &nbsp; // checking if character is alphabetic and converting its charcode back to a letter&nbsp; &nbsp; &nbsp; &nbsp; if (thisLetter < 65 || (thisLetter > 90 && thisLetter < 97) || thisLetter > 122) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return;&nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; shiftedLetter = thisLetter + shiftAmount;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; // making sure the shifted letters loop to beginning, or end, of alphabet&nbsp; &nbsp; &nbsp; &nbsp; let loopingAdjustment = 0;&nbsp; &nbsp; &nbsp; &nbsp; if (thisLetter > 96 && shiftedLetter > 122) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; loopingAdjustment = shiftedLetter - 26;&nbsp; &nbsp; &nbsp; &nbsp; } else if (thisLetter > 96 && shiftedLetter < 96) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; loopingAdjustment = shiftedLetter + 26;&nbsp; &nbsp; &nbsp; &nbsp; } else if (thisLetter < 91 && shiftedLetter > 90) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; loopingAdjustment = shiftedLetter - 26;&nbsp; &nbsp; &nbsp; &nbsp; } else if (thisLetter < 91 && shiftedLetter < 65) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;loopingAdjustment = shiftedLetter + 26;&nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;loopingAdjustment = shiftedLetter;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; let finalString = String.fromCharCode(loopingAdjustment);&nbsp; &nbsp; &nbsp; &nbsp; return finalString;&nbsp; &nbsp; });&nbsp; &nbsp; console.log(mappedLetters);&nbsp; &nbsp; return mappedLetters.join('');}module.exports = caesar

Answer 2

您在 let 有自己的范围的条件中声明 loopingAdjustment 。在条件句之外声明它，只在条件句中更改它的值一次。