这里的目的是让用户输入一个数字来确定它是否是偶数,或者输入'q'来退出程序。
var readlineSync = require('readline-sync');
var i = 0;
while (i <= 3) {
var num = readlineSync.question("Enter q to quit, or enter an integer to continue?");
if (num === 'q') {
console.log("You have quit the application. Thanks for using.");
break;
}
else if (num % 2) {
console.log("You have entered an odd number");
}
else if (num !== 'q') {
console.log("You have not entered a valid character. Please try again.");
break;
}
else {
console.log("You have entered an even number.");
break;
}
}
按q启动适当的响应并退出程序。输入奇数也会生成适当的响应。但是,如果输入的是偶数,则程序不会生成适当的响应,而是读取您没有输入有效字符。请再试一次。我忽略了什么?
这是因为您的第三个条件(num !== 'q')。当输入偶数时,它的计算结果为true。