按 id 合并数组中的多个对象 - javascript

Question

这是我的第一个问题，所以请温和:)

我知道我的问题与许多其他问题类似，我尝试了很多解决方案，但没有得到我需要的结果。

我有一个对象数组，这些对象可以有重复的 ID。 id 'THOM01':

有 3 个对象

[
    { id: 'MARS03', name: 'Employee 3', T1: 9.23, A6: '', SAT: '', SUN: '', PHW: '', SAN: '', COMM:'', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' },
    { id: 'THOM01', name: 'Employee 4', T1: '', A6: '', SAT: 4.12, SUN: '', PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' },
    { id: 'THOM01', name: 'Employee 4', T1: '', A6: '', SAT: '', SUN: 6.12, PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' },
    { id: 'THOM01', name: 'Employee 4', T1: 15, A6: '', SAT: '', SUN: '', PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' }
]

需要合并具有相同 ID 的对象（每个员工一个对象）。所以结果看起来像这样：

[
    { id: 'MARS03', name: 'Employee 3', T1: 9.23, A6: '', SAT: '', SUN: '', PHW: '', SAN: '', COMM:'', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' },
    { id: 'THOM01', name: 'Employee 4', T1: 15, A6: 4.12, SAT: 4.12, SUN: '', PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' }
]

当我输出到 csv 以上传到工资单系统时，我需要维护 key/value 对（包括空值对）的 structure/placement，但是对象无关紧要。

我尝试过的其他解决方案最终总是只有一个支付率值（'T1' 到 'other'）。它们似乎被覆盖了，因为每个对象都有所有字段，即使它们是空的。

我希望这些信息足够了。谢谢！

戴夫

Answer 1

您可以获取一组键并通过获取一个对象进行分组来减少数据。

const
    data = [{ id: 'MARS03', name: 'Employee 3', T1: 9.23, A6: '', SAT: '', SUN: '', PHW: '', SAN: '', COMM:'', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' }, { id: 'THOM01', name: 'Employee 4', T1: '', A6: '', SAT: 4.12, SUN: '', PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' }, { id: 'THOM01', name: 'Employee 4', T1: '', A6: '', SAT: '', SUN: 6.12, PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' }, { id: 'THOM01', name: 'Employee 4', T1: 15, A6: '', SAT: '', SUN: '', PHW: '', SAN: '', COMM: '', BON: '', other: '', quantity: '', leaveDays: '', from: '', to: '', reason: '', costCentre: '', message: '' }],
    keys = ['T1', 'A6', 'SAT', 'SUN', 'PHW', 'SAN', 'COMM', 'BON', 'other'],
    result = Object.values(data.reduce((r, o) => {
        if (r[o.id]) keys.forEach(k => { if (o[k] !== '') r[o.id][k] = (r[o.id][k] || 0) + o[k]; });
        else r[o.id] = { ...o };
        return r;
    }, {}));

console.log(result);

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