我有一个这样定义的枚举:

export enum GoalProgressMeasurements {
    Percentage = 1,
    Numeric_Target = 2,
    Completed_Tasks = 3,
    Average_Milestone_Progress = 4,
    Not_Measured = 5
}

然而,我希望它被表示为一个对象数组/列表从我们的API如下:

[{id: 1, name: 'Percentage'}, 
 {id: 2, name: 'Numeric Target'},
 {id: 3, name: 'Completed Tasks'},
 {id: 4, name: 'Average Milestone Progress'},
 {id: 5, name: 'Not Measured'}]

是否有简单和本地的方法来做到这一点,或者我必须构建一个函数,将枚举转换为int和字符串,并将对象构建为数组?


当前回答

我使用

Object.entries(GoalProgressMeasurement).filter(e => !isNaN(e[0]as any)).map(e => ({ name: e[1], id: e[0] }));

一条简单的线就可以做到这一点。

它只需要3个简单的步骤 -使用Object.entries加载键和值的组合。 -过滤掉非数字(因为typescript会为反向查找生成值)。 -然后我们将它映射到我们喜欢的数组对象。

其他回答

如果你使用的是ES8

只有在这种情况下,它才会工作得很好。它会给你给定枚举的值数组。

enum Colors {
  WHITE = 0,
  BLACK = 1,
  BLUE = 3
}

const colorValueArray = Object.values(Colors); //[ 'WHITE', 'BLACK', 'BLUE', 0, 1, 3 ]

你会得到这样的colorValueArray ['WHITE', 'BLACK', 'BLUE', 0,1,3]。所有的键都在数组的前半部分,所有的值都在数组的后半部分。

甚至这种枚举也可以正常工作

enum Operation {
    READ,
    WRITE,
    EXECUTE
}

但是这种解决方案不适用于这样的异构枚举

enum BooleanLikeHeterogeneousEnum {
  No = 0,
  Yes = "YES",
}

我不认为顺序是可以保证的,否则就很容易切片Object的后半部分。条目的结果和映射来自那里。

上述答案的唯一(非常小的)问题是

字符串和数字之间有很多不必要的类型转换。 当一次迭代同样干净有效时,条目将迭代两次。

type StandardEnum = { [id: string]: number | string; [nu: number]: string;}

function enumToList<T extends StandardEnum> (enm: T) : { id: number; description: string }[] {
    return Object.entries(enm).reduce((accum, kv) => {
        if (typeof kv[1] === 'number') {
            accum.push({ id: kv[1], description: kv[0] })
        }
        return accum
    }, []) // if enum is huge, perhaps pre-allocate with new Array(entries.length / 2), however then push won't work, so tracking an index would also be required
}

另一种方法是

export const GoalNames = {
    [GoalProgressMeasurements.Percentage] = 'Percentage',
    [GoalProgressMeasurements.Numeric_Target] = 'Numeric Target',
    [GoalProgressMeasurements.Completed_Tasks] = 'Completed Tasks',
    [GoalProgressMeasurements.Average_Milestone_Progress] = 'Average Milestone Progress',
    [GoalProgressMeasurements.Not_Measured] = 'Not Measured'
}

你可以调用:

const name = GoalNames[goalEnumVal];
class EnumHelpers {

    static getNamesAndValues<T extends number>(e: any) {
        return EnumHelpers.getNames(e).map(n => ({ name: n, value: e[n] as T }));
    }

    static getNames(e: any) {
        return EnumHelpers.getObjValues(e).filter(v => typeof v === 'string') as string[];
    }

    static getValues<T extends number>(e: any) {
        return EnumHelpers.getObjValues(e).filter(v => typeof v === 'number') as T[];
    }

    static getSelectList<T extends number, U>(e: any, stringConverter: (arg: U) => string) {
        const selectList = new Map<T, string>();
        this.getValues(e).forEach(val => selectList.set(val as T, stringConverter(val as unknown as U)));
        return selectList;
    }

    static getSelectListAsArray<T extends number, U>(e: any, stringConverter: (arg: U) => string) {
        return Array.from(this.getSelectList(e, stringConverter), value => ({ value: value[0] as T, presentation: value[1] }));
    }

    private static getObjValues(e: any): (number | string)[] {
        return Object.keys(e).map(k => e[k]);
    }
}

首先,我们得到这个枚举的键数组。然后,使用map()函数将数据转换为所需的格式。Id从键获取,name从enum通过相同的键获取。

const converted = Object.keys(GoalProgressMeasurements).map(key => {
        return {
            id: GoalProgressMeasurements[key],
            name: key,
        };
    });