javascript - 为什么我的 javascript 函数被打印为未定义?
问题描述
我是 JavaScript 新手,目前正在学习函数。每当我尝试将其打印到控制台时,它都会以未定义的形式返回。我确定它非常简单,但我错过了什么?这是我的代码:
let getSleepHours = day => {
switch (day) {
case 'monday':
return 8;
break;
case 'tuesday':
return 7;
break;
case 'wednesday':
return 6;
break;
case 'thursday':
return 9;
break;
case 'friday':
return 10;
break;
case 'saturday':
return 8;
break;
case 'sunday':
return 11;
break;
default:
return 'error';
}
}
const getActualSleepHours = () => {
return
getSleepHours('monday') +
getSleepHours('tuesday') +
getSleepHours('wednesday') +
getSleepHours('thursday') +
getSleepHours('friday') +
getSleepHours('saturday') +
getSleepHours('sunday');
};
console.log(getActualSleepHours());
解决方案
由于这一行,它返回 undefined
const getActualSleepHours=()=>{ return
这里 return 关键字后没有返回任何内容,因此 Javascript 将认为其未定义。
你应该做这个:
const getActualSleepHours =() =>{
return getSleepHours('monday')+
getSleepHours('tuesday')+
getSleepHours('wednesday')+
getSleepHours('thursday')+
getSleepHours('friday')+
getSleepHours('saturday')+
getSleepHours('sunday');
};
完整片段:
let getSleepHours = day => {
switch (day) {
case 'monday':
return 8;
break;
case 'tuesday':
return 7;
break;
case 'wednesday':
return 6;
break;
case 'thursday':
return 9;
break;
case 'friday':
return 10;
break;
case 'saturday':
return 8;
break;
case 'sunday':
return 11;
break;
default:
return 'error';
}
}
const getActualSleepHours =() =>{
return getSleepHours('monday')+
getSleepHours('tuesday')+
getSleepHours('wednesday')+
getSleepHours('thursday')+
getSleepHours('friday')+
getSleepHours('saturday')+
getSleepHours('sunday');
};
console.log(getActualSleepHours());
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