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fixed grammer
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I know typescript from just a few months, and theThe solution below worked for me. Hope it may help someone as well -

export enum ScheduleType {
  Basic = <any>'B',
  Consolidated = <any>'C',
}

scheduleTypes = Object.keys(ScheduleType)
.filter((k, i) => i % 2)
.map((key: any) => {
  return {
    systemValue: key,
    displayValue: ScheduleType[key],
  };
});

It gave the following result - [{displayValue: "Basic", systemValue: "B"}, {displayValue: "Consolidated", systemValue: "C"}]

I know typescript from just a few months, and the solution below worked for me. Hope it may help someone as well -

export enum ScheduleType {
  Basic = <any>'B',
  Consolidated = <any>'C',
}

scheduleTypes = Object.keys(ScheduleType)
.filter((k, i) => i % 2)
.map((key: any) => {
  return {
    systemValue: key,
    displayValue: ScheduleType[key],
  };
});

It gave the following result - [{displayValue: "Basic", systemValue: "B"}, {displayValue: "Consolidated", systemValue: "C"}]

The solution below worked for me. Hope it may help someone as well -

export enum ScheduleType {
  Basic = <any>'B',
  Consolidated = <any>'C',
}

scheduleTypes = Object.keys(ScheduleType)
.filter((k, i) => i % 2)
.map((key: any) => {
  return {
    systemValue: key,
    displayValue: ScheduleType[key],
  };
});

It gave the following result - [{displayValue: "Basic", systemValue: "B"}, {displayValue: "Consolidated", systemValue: "C"}]

Source Link

I know typescript from just a few months, and the solution below worked for me. Hope it may help someone as well -

export enum ScheduleType {
  Basic = <any>'B',
  Consolidated = <any>'C',
}

scheduleTypes = Object.keys(ScheduleType)
.filter((k, i) => i % 2)
.map((key: any) => {
  return {
    systemValue: key,
    displayValue: ScheduleType[key],
  };
});

It gave the following result - [{displayValue: "Basic", systemValue: "B"}, {displayValue: "Consolidated", systemValue: "C"}]