按 id 合并数组中的多个对象 - javascript

Question

这是我的第一个问题，所以请温柔:)

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

我有一个可以有重复 id 的对象数组。 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）。所以结果应该是这样的：

[
    { 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 以上传到工资系统时，我需要维护键/值对（包括空值对）的结构/位置，但对象的顺序并不重要。

我尝试过的其他解决方案总是最终只有一个工资率值（“T1”到“其他”）。它们似乎被覆盖，因为每个对象都有所有字段，即使它们是空的。

我希望这是足够的信息。谢谢！

戴夫

Answer 1

您可以获取一个键数组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);

.as-console-wrapper { max-height: 100% !important; top: 0; }