将数字等级转换为字母等级-带有显示等级的问题

Question

我在运行下面的if语句时遇到了问题。我正在上一个在线课程，我唯一的资源就是课本。感谢您的帮助。我能够让它返回除提示、错误消息之外的任何内容，然后退出程序。它正确地运行了一个连续循环。

var numberGrade;
var grade = [];
do {
    numberGrade = prompt("Enter number grade from 0 through 100\nOr enter -1 to end entries\n");
    numberGrade = parseInt(numberGrade);
    if (numberGrade >= 0 && numberGrade <= 100 ) {
        grade[grade.length] = numberGrade;
    } else if (numberGrade != -1) {
        alert("Entry not valid!\n Entry must be a valid number from 0 through 100 \n Or enter -1 to end entries");
    } 
    }
    while (numberGrade != -1);

for (var num = 0; num <= grade; num++) {
    if (grade >= 0 && grade <= 59) {
        alert ("Number grade =" + grade + "\nLetter Grade = F");
    }
    if (grade >= 60 && grade <= 62) {
             alert ("Number grade =" + numberGrade + "\nLetter Grade = D-");
        }
    if (grade >= 63 && grade <= 66) {
        alert ("Number grade =" + numberGrade + "\nLetter Grade = D");
    }
    if (numberGrade >= 67 && numberGrade <= 69) {
        alert ("Number grade =" + numberGrade + "\nLetter Grade = D+");
    }
    if (numberGrade >= 70 && numberGrade <= 76){
        alert ("Number grade =" + numberGrade + "\nLetter Grade = C");
    }
    if (numberGrade >= 77 && numberGrade <= 79) {
        alert ("Number grade =" + numberGrade + "\nLetter Grade = C+");
    }
    if (numberGrade >= 80 && numberGrade <= 82) {
        alert ("Number grade =" + numberGrade + "\nLetter Grade = B-");
    }
    if (numberGrade >= 83 && numberGrade <= 86) {
        alert ("Number grade =" + numberGrade + "\nLetter Grade = B");
    }
    else if (numberGrade >= 87 && numberGrade <= 89) {
        alert ("Number grade =" + numberGrade + "\nLetter Grade = B+");
    }
    if (numberGrade >= 90 && numberGrade <= 92) {
        alert ("Number grade =" + numberGrade + "\nLetter Grade = A-");
    }
    if (numberGrade >= 93 && numberGrade <= 100) {
        alert ("Number grade =" + numberGrade + "\nLetter Grade = A");
    }
}

Answer 1

在循环初始化中，因为grade是一个数组而不是数字，所以语句num <= grade是没有意义的。将您的循环更改为for (var num = 0; num < grade.length; num++)。然后，您的if语句应该针对grade数组的元素进行测试：

if (grade[num] >= 0 && grade[num] <= 59) {
    alert ("Number grade =" + grade[num] + "\nLetter Grade = F");
}

Answer 2

我稍微修改了一下你的代码，以便正常工作。它可以用更好的方式来完成，但我选择了保持简单，这样你仍然可以理解它。

我创建了一个将数字等级转换为字母的函数，返回要输出的字符串。

var numberGrade;
do {
    numberGrade = prompt("Enter number grade from 0 through 100\nOr enter -1 to end entries\n");
    numberGrade = parseInt(numberGrade);

    if (numberGrade >= 0 && numberGrade <= 100 ) {
        // If the input is valid, calls the function and creates an alert with the return
        alert(convertGrade(numberGrade));
    } else if (numberGrade != -1) {
        alert("Entry not valid!\n Entry must be a valid number from 0 through 100 \n Or enter -1 to end entries");
    } 
} while (numberGrade != -1);

// Function that converts the grade, returning the message
function convertGrade(grade) {
    if (grade >= 0 && grade <= 59) {
        return ("Number grade =" + grade + "\nLetter Grade = F");
    }
    if (grade >= 60 && grade <= 62) {
        return ("Number grade =" + grade + "\nLetter Grade = D-");
    }
    if (grade >= 63 && grade <= 66) {
        return ("Number grade =" + grade + "\nLetter Grade = D");
    }
    if (grade>= 67 && grade <= 69) {
        return ("Number grade =" + grade + "\nLetter Grade = D+");
    }
    if (grade>= 70 && grade <= 76){
        return ("Number grade =" + grade + "\nLetter Grade = C");
    }
    if (grade>= 77 && grade <= 79) {
        return ("Number grade =" + grade + "\nLetter Grade = C+");
    }
    if (grade>= 80 && grade <= 82) {
        return ("Number grade =" + grade + "\nLetter Grade = B-");
    }
    if (grade>= 83 && grade <= 86) {
        return ("Number grade =" + grade + "\nLetter Grade = B");
    }
    if (grade>= 87 && grade <= 89) {
        return ("Number grade =" + grade + "\nLetter Grade = B+");
    }
    if (grade>= 90 && grade <= 92) {
        return ("Number grade =" + grade + "\nLetter Grade = A-");
    }
    if (grade>= 93 && grade <= 100) {
        return ("Number grade =" + grade + "\nLetter Grade = A");
    }
}