如何使用擴展語法es6進行簡化

Question

如何簡化代碼，我嘗試生成對象並為每個change方法調用，並取決於arguments（newValues）我更改了某些字段，似乎我有重復的代碼，如何避免這種情況？ 我應該使用一些更困難的方法來避免使用擴展語法嗎

const formName = {
    fieldRange: 'miConfiguration.fieldRange',
    defaultTimeout: 'miConfiguration.doorConfiguration.defaultTimeout',
    standAlone: 'miConfiguration.doorConfiguration.standAlone',
    overrideTimeout: 'miConfiguration.doorConfiguration.overrideTimeout',
    inputMode: 'miConfiguration.doorConfiguration.inputMode',
    stopMi: 'miConfiguration.doorConfiguration.stopMi',
    activeLow: 'miConfiguration.doorConfiguration.activeLow',
    enableDualTechnology: 'miConfiguration.enableDualTechnology',
    passageName: 'miConfiguration.passageName',
}

let {fieldRange, defaultTimeout, standAlone, overrideTimeout, inputMode, stopMi, activeLow, enableDualTechnology, passageName} = formName

let configurationsMi = {
    [passageName]:  null,

    [fieldRange]: null,
    [activeLow]: false,
    [standAlone]: null,
    [defaultTimeout]: null,
    [overrideTimeout]: null,
    [inputMode]: null,
    [stopMi]: null,
    [enableDualTechnology]: false,
}

const defaultValues = {
    [MiConfigurationTypes.AccessPointOnly]: {
        ...configurationsMi,
        [fieldRange]:  MiFieldRanges.Disabled,
    },

    [MiConfigurationTypes.WanderingDetection]: {
        ...configurationsMi,
        [fieldRange]:  MiFieldRanges.Small,
    },
    [MiConfigurationTypes.MuteWanderingDetection]: {
        ...configurationsMi,
        [fieldRange]:  MiFieldRanges.Small,
    },
    [MiConfigurationTypes.LockedWanderingControl]: {
        ...configurationsMi,
        [fieldRange]:  MiFieldRanges.Small,
        [standAlone]:  DoorStates.Locked,
        [defaultTimeout]: '00:00:03',
        [overrideTimeout]: '00:00:30',
        [inputMode]: InputModes.NotUsed,
        [stopMi]: false,
    },
    [MiConfigurationTypes.OpenWanderingControl]: {
        ...configurationsMi,
        [fieldRange]:  MiFieldRanges.Small,
        [standAlone]:  DoorStates.Locked,
        [defaultTimeout]: '00:00:03',
        [overrideTimeout]: '00:00:30',
        [inputMode]: InputModes.NotUsed,
        [stopMi]: false,
    },
}

 onChange={(e, newValue) => {
                console.log(defaultValues)
                Object.keys(defaultValues[newValue]).forEach(key => change(key, defaultValues[newValue][key]))
            }}

Answer 1

您可以將重復的部分放在一個額外的對象文字中，就像您已經對configurationsMi所做的一樣，然后引用它。

const configurationsWanderingControl = {
    [fieldRange]:  MiFieldRanges.Small,
    [standAlone]:  DoorStates.Locked,
    [defaultTimeout]: '00:00:03',
    [overrideTimeout]: '00:00:30',
    [inputMode]: InputModes.NotUsed,
    [stopMi]: false,
};
const defaultValues = {
    …
    [MiConfigurationTypes.LockedWanderingControl]: {
        ...configurationsMi,
        ...configurationsWanderingControl,
    },
    [MiConfigurationTypes.OpenWanderingControl]: {
        ...configurationsMi,
        ...configurationsWanderingControl,
    },
};

當然，還有避免重復的默認方法：將重復的代碼放在一個函數中（可能帶有參數以表示不同的細節），然后從多個位置進行調用。