Question

在尝试获得班级中的成绩总数时出现NAN错误。在学生[i] .grade上尝试了parseInt（）和.value。我是全新的。

var students= [{name:"David", grade:80}, 
               {name:"Vinoth", grade:77}, 
               {name:"Divya", grade:88}, 
               {name:"Ishitha", grade:95},
               {name:"Thomas", grade:68}];


var average;

for(i = 0; i < students.length; i++){
    console.log(students[i].grade);
    average += Number(students[i].grade.value);
    console.log(average);
}

控制台日志的输出是：

80 为NaN 77 为NaN 88 为NaN 95 为NaN 68 为NaN

Answer 1

1）您需要初始化平均值：var average = 0;

2）您也不需要.value，因为价值不是有效的财产。

average += Number(students[i].grade);

3）如果您确定所有访问过的属性都已经是数字类型，则不需要将数字值冗余地转换为Number()的数字。

这样做会很好：

＆＃13;

var students= [{name:"David", grade:80}, 
           {name:"Vinoth", grade:77}, 
           {name:"Divya", grade:88}, 
           {name:"Ishitha", grade:95},
           {name:"Thomas", grade:68}];

var average = 0;

for(i = 0; i < students.length; i++){
    console.log(students[i].grade);
    average += Number(students[i].grade);
    console.log(average);
}

console.log("average: ", average/students.length);

＆＃13;

试图从数组中获取数字得到NaN错误

1 个答案: