js如何高效更改对象中属性名

Question

如图中的一个json对象：

[
    {
        "Id":"3972679ef2c04151972b376dd88e6413",
        "T_CourseId":"7a4494aae1804d3e94094583249750fe",
        "CourseName":"英语",
        "Code":"english"
    },
    {
        "Id":"5665d803e7994b26a56c6287d12c2090",
        "T_CourseId":"75761ad2ce23498c9f9db134ab844aec",
        "CourseName":"药物化学",
        "Code":"ywhx"
    }
]

怎么使用js高效的将其中的CourseName更改为title，得到结果如下：

[
    {
        "Id":"3972679ef2c04151972b376dd88e6413",
        "T_CourseId":"7a4494aae1804d3e94094583249750fe",
        "title":"英语",
        "Code":"english"
    },
    {
        "Id":"5665d803e7994b26a56c6287d12c2090",
        "T_CourseId":"75761ad2ce23498c9f9db134ab844aec",
        "title":"药物化学",
        "Code":"ywhx"
    }
]

Answer 1

      var aaa = [
    {
        "Id":"3972679ef2c04151972b376dd88e6413",
        "T_CourseId":"7a4494aae1804d3e94094583249750fe",
        "CourseName":"英语",
        "Code":"english"
    },
    {
        "Id":"5665d803e7994b26a56c6287d12c2090",
        "T_CourseId":"75761ad2ce23498c9f9db134ab844aec",
        "CourseName":"药物化学",
        "Code":"ywhx"
    }
]
   var bbb= JSON.parse(JSON.stringify(aaa).replace(/CourseName/g,"title"));
    console.log(bbb)

Answer 2

arr = JSON.parse(jsonObj);
arr.map(item => {
    item.title = item.CourseName;
    delete item.CourseName;
    return item;
})

Answer 3

var arr = [
    {
        "Id":"3972679ef2c04151972b376dd88e6413",
        "T_CourseId":"7a4494aae1804d3e94094583249750fe",
        "CourseName":"英语",
        "Code":"english"
    },
    {
        "Id":"5665d803e7994b26a56c6287d12c2090",
        "T_CourseId":"75761ad2ce23498c9f9db134ab844aec",
        "CourseName":"药物化学",
        "Code":"ywhx"
    }
];
arr.forEach(function(item){
    item.title = item.CourseName;
    delete item.CourseName;
})


或者对每一项用JSON.stringify()转换成字符串，用replace替换，再用JSON.parse()转回来

Answer 4

如果只是json的话。。那就简单多了
replace一下就行了。

Answer 5

目前觉得还是 转换字符串然后replace，如果是遇到多重对象嵌套也可以用

Answer 6

用foreach的方式，不要用replace的方式，万一你的属性里面包含了该字符串就被替换掉了，当然这个可能性很小。

Answer 7

var aaa = [
    {
        "Id":"3972679ef2c04151972b376dd88e6413",
        "T_CourseId":"7a4494aae1804d3e94094583249750fe",
        "CourseName":"英语",
        "Code":"english"
    },
    {
        "Id":"5665d803e7994b26a56c6287d12c2090",
        "T_CourseId":"75761ad2ce23498c9f9db134ab844aec",
        "CourseName":"药物化学",
        "Code":"ywhx"
    }
]
var copyTrans = function(obj, typeArr) {
    let result;
    let toString = Object.prototype.toString;
    if(toString.call(obj) === '[object Array]'){
        result = [];
        for(let i = 0; i < obj.length; i++){
            result[i] = copyTrans(obj[i], arguments[1])
        }
    }else if(toString.call(obj) === '[object Object]'){
        result = {};
        for (let _key in obj) {
            if (obj.hasOwnProperty(_key)) {
                let flag = 0, _value = null;
                for (let j = 0; j < arguments[1].length; j ++) {
                    if (arguments[1][j].key === _key) {
                        flag = 1;
                        _value = arguments[1][j].value
                    }
                }
                if (flag)
                    result[_value] = copyTrans(obj[_key], arguments[1])
                else
                    result[_key] = copyTrans(obj[_key], arguments[1])
            }
        }
    }else{
        return obj
    }
    return result
};
console.log(copyTrans(aaa, [{ key: "CourseName", value: "title" }]));