Eric 很好地解释了如何去做你似乎想要完成的事情，但让我澄清一下你哪里出错了。
switch/case 结构将给定变量（switch 参数）与可能的值（case 参数）进行比较，然后在匹配的 case 语句和下一个 break 语句之间执行代码（或者，如果语言不支持 fall -through，在下一个 case 语句之前）。
您要做的不是将变量与常量表达式进行比较，而是将变量与条件进行比较。 if/elseif 结构可能是一种更简洁的表达方式：
if (this.yourNumberGrade >= 0.90) {
this.yourLetterGrade = 'A';
} else if (this.yourNumberGrade >= 0.80) {
this.yourLetterGrade = 'B';
} else if (this.yourNumberGrade >= 0.70) {
this.yourLetterGrade = 'C';
} else if (this.yourNumberGrade >= 0.60) {
this.yourLetterGrade = 'D';
} else { // you left the default out, but I assume this will be an F for Failed
this.yourLetterGrade = 'F';
}
如果您希望它更短，您可以尝试像这样尝试使用三元运算符：
this.yourLetterGrade = (
this.yourNumberGrade >= 0.90 ? 'A' : (
this.yourNumberGrade >= 0.80 ? 'B' : (
this.yourNumberGrade >= 0.70 ? 'C' : (
this.yourNumberGrade >= 0.60 ? 'D' : 'F'
)
)
)
)
如您所见，这会降低您的可读性，因此 if/else 可能是最简洁的方法。
Eric 试图向您展示的是这样的结构：
switch (true) { // We compare the boolean constant "true" to the case arguments
case this.yourNumberGrade >= 0.90:
// this is a boolean expression and evaluates either
// to "true" (matches the switch argument) or
// to "false" (does not match the switch argument)
this.yourLetterGrade = 'A';
break;
case this.yourNumberGrade >= 0.80:
this.yourLetterGrade = 'B';
break;
case this.yourNumberGrade >= 0.70:
this.yourLetterGrade = 'C';
break;
case this.yourNumberGrade >= 0.90:
this.yourLetterGrade = 'D';
break;
default:
// This is executed if none of the case arguments evaluate
// to the value of the switch argument.
this.yourLetterGrade = 'F';
// No break needed, because the end of the switch structure follows:
}
我希望这可以为您解决问题。您可能必须更加注意您尝试使用的结构的确切语义。这些结构在大多数语言中都非常相似。
对于踢腿和咯咯笑，你甚至可以用数组来做：
// Our letter grades in ascending order (from bad to good).
String letterGrades[] = {'F','D','C','B','A'};
// Our number grade is in the range [0.0;1.0]. As floating point numbers are
// too precise for indexes, we want to round them down to the cut-off
// (0.9, 0.8, etc) and turn them into integer values we can use as array indices.
int gradeIndex = (int) Math.floor(this.yourNumberGrade*10);
// The lowest cut-off is 0.6, so we can treat everything lower than that the same
gradeindex = gradeindex - 5;
gradeIndex = Math.max(gradeIndex, 0);
// With Math.max we have ensured that no index can be lower than 0, now we need
// to make sure that no index is larger than the largest index in our array
// (which by definition is equal to the array's length (i.e. number of elements)
// minus 1 (because the lowest index is 0, an array of e.g. size 4 has the
// indices 0,1,2,3, but lacks an index 4 -- better get used to it, that's how
// programmers count, too).
gradeIndex = Math.min(gradeIndex, letterGrades.length-1);
// Now that our index is clean and guaranteed to be within range, we can use it
// to look up the letter grade:
this.yourLetterGrade = letterGrades[gradeIndex];
没有 cmets 和一些速记，这甚至更短：
// Grades are as follows: A: 90%+, B: 80%+, C: 70%+, D: 60%+, F: <60%
String letterGrades[] = {'F','D','C','B','A'};
int gradeIndex = Math.min(
Math.max(0, (int) Math.floor(this.yourNumberGrade*10) - 5),
letterGrades.length-1
);
this.yourLetterGrade = letterGrades[gradeIndex];
请注意，这使得字母等级的确切截止点在哪里变得不太清楚，这就是它需要 cmets 的原因。此外，如果临界值因任何原因发生变化（例如 A：85%+ 或 F：Math.floor(this.yourNumberGrade*10)-5 部分），但这将使其更难遵循，并且如果成绩不仅仅是渐进式的，也无济于事。但是，对于传统系统，这是一种快速简便的方法。