如何在Javascript中基于对象键从对象数组添加对象值

我想根据另一个对象键从对象数组中添加一些对象值。我想知道如何通过普通Javascript或使用像lodash这样的助手库来实现这一点

我已经尝试过使用lodash的

.groupBy

和

Array.prototype.reduce（）

，但还没有成功。以下是数据的外观：

{
  "2019-01-04": [
    {
        "payments": [
            {
                "sum": "12",
                "currency": "€"
            }
        ]
    }
  ],

  "2019-01-06": [
    {
        "payments": [
            {
                "sum": "50",
                "currency": "€"
            },
            {
                "sum": "30",
                "currency": "£"
            },
            {
                "sum": "40",
                "currency": "Lek"
            },
            {
                "sum": "2",
                "currency": "£"
            },
            {
                "sum": "60",
                "currency": "£"
            }
        ]
    }
  ]
}

我希望sum属性从该日期起具有相同类型的所有货币的总和：

{
  "2019-01-04": [
    {
        "payments": [
            {
                "sum": "12",
                "currency": "€"
            }
        ]
    }
  ],

  "2019-01-06": [
    {
        "payments": [
            {
                "sum": "50",
                "currency": "€"
            },
            {
                "sum": "92",
                "currency": "£"
            },
            {
                "sum": "40",
                "currency": "Lek"
            }
        ]
    }
  ]
}

---

给定您提供的数据结构，使用以下方法可获得所需的输出：

function sumByCurrency(history) {
    _.forOwn(history, value => {
        const newPayments = [];

        _.forEach(value["0"].payments, v => {
            let existingPayment = _.find(
                newPayments,
                newPayment => newPayment && newPayment.currency === v.currency
            );

            if (existingPayment) {
                let existingSum = +existingPayment.sum;
                let incomingSum = +v.sum;

                existingSum += incomingSum ? incomingSum : 0;

                existingPayment.sum = "" + existingSum;
            } else {
                newPayments.push({
                    currency: v.currency,
                    sum: v.sum ? v.sum : 0
                });
            }
        });

        value["0"].payments = newPayments;
    });

    return history;
}

使用它传递您的对象，假设您将其称为

paymentHistory

to

sumByCurrency

函数，如下所示：

sumByCurrency(paymentHistory);

---

请注意：如果

值[“0”]，您可能需要进行一些回退/确保它不会中断。付款
不可用。
给定您提供的数据结构，使用以下方法可提供所需的输出：

function sumByCurrency(history) {
    _.forOwn(history, value => {
        const newPayments = [];

        _.forEach(value["0"].payments, v => {
            let existingPayment = _.find(
                newPayments,
                newPayment => newPayment && newPayment.currency === v.currency
            );

            if (existingPayment) {
                let existingSum = +existingPayment.sum;
                let incomingSum = +v.sum;

                existingSum += incomingSum ? incomingSum : 0;

                existingPayment.sum = "" + existingSum;
            } else {
                newPayments.push({
                    currency: v.currency,
                    sum: v.sum ? v.sum : 0
                });
            }
        });

        value["0"].payments = newPayments;
    });

    return history;
}

const result = {};
Object.keys(input).forEach(date => {
    result[date] = [{ }];
    result[date][0].payments = input[date][0].payments.reduce((payments, c) => {
        const grp = payments.find(p => p.currency === c.currency);
        grp ? grp.sum = +grp.sum + +c.sum : payments.push(c);
        return payments;
    }, []);
});

使用它传递您的对象，假设您将其称为
paymentHistory
to
sumByCurrency
函数，如下所示：

sumByCurrency(paymentHistory);

请注意：如果
value[“0”]，您可能需要进行一些回退/确保它不会中断.payments
不可用。
请将您的尝试添加到问题中…重述：如果某一货币在某一日期存在多次，则获取金额？您知道正手操作的货币是什么吗？请将您的尝试添加到问题中…重述：如果某一货币在某一日期存在多次，则获取金额？您知道正手操作的货币是什么吗你有什么货币？我喜欢你的解决方案，因为它比我的更干净：）谢谢！我喜欢你的解决方案，因为它比我的更干净：）谢谢！
const result = {};
Object.keys(input).forEach(date => {
    result[date] = [{ }];
    result[date][0].payments = input[date][0].payments.reduce((payments, c) => {
        const grp = payments.find(p => p.currency === c.currency);
        grp ? grp.sum = +grp.sum + +c.sum : payments.push(c);
        return payments;
    }, []);
});