javascript - 减少node.js应用程序中的函数，忽略json json数据，并且太麻烦了

我在客户变量中有此json：

{
    "customers": [
        {
            "id": 6,
            "name": "Joe Bloggs",
            "email": "[email protected]",
            "supplier_id": 1,
            "title": "First Supplier",
            "code": "firstsupplier"
        },
        {
            "id": 7,
            "name": "Paul",
            "email": "[email protected]",
            "supplier_id": 1,
            "title": "First Supplier",
            "code": "firstsupplier"
        },
        {
            "id": 7,
            "name": "Paul",
            "email": "[email protected]",
            "supplier_id": 3,
            "title": "New Supplier",
            "code": "new"
        },
        {
            "id": 7,
            "name": "Paul",
            "email": "[email protected]",
            "supplier_id": 4,
            "title": "PC supplier 3",
            "code": "PC03"
        },
        {
            "id": 8,
            "name": "John Doe",
            "email": "[email protected]",
            "supplier_id": 1,
            "title": "First Supplier",
            "code": "firstsupplier"
        }
    ]
}

我想要更改它，所以它看起来像这样：

{
    "customers": [
        {
            "id": 6,
            "name": "Joe Bloggs",
            "email": "[email protected]",
            "suppliers": [
              {
                "supplier_id": 1,
                "title": "First Supplier",
                "code": "firstsupplier"
              }
            ]
        },
        {
            "id": 7,
            "name": "Paul",
            "email": "[email protected]",
            "suppliers": [
              {
                "supplier_id": 1,
                "title": "First Supplier",
                "code": "firstsupplier"
              },
              {
                "supplier_id": 3,
                "title": "New Supplier",
                "code": "new"
              },
              {
                "supplier_id": 4,
                "title": "PC supplier 3",
                "code": "PC03"
              }
            ]
        },
        {
            "id": 8,
            "name": "John Doe",
            "email": "[email protected]",
            "suppliers": [
              {
                "supplier_id": 1,
                "title": "First Supplier",
                "code": "firstsupplier"
              }
            ]
        }
    ]
}

我设法让其中的一部分可以单独工作，但无法在单个函数中全部工作。

例如。我可以创建suppliers数组，但是对于客户ID为7的情况，该数组中只有一个供应商。

let customers = await query.offset(offset).limit(limit);

customers = customers.reduce((accumulator, item) => {
  if (accumulator[item.id]) {
    const group = accumulator[item.id];
    group.suppliers = [];
    group.suppliers.push({
      id: item.supplier_id,
      code: item.code,
      title: item.title
    });
  } else {
    accumulator[item.id] = item;
    accumulator[item.id].suppliers = [];
    accumulator[item.id].suppliers.push({
      id: item.supplier_id,
      code: item.code,
      title: item.title
    });
    delete accumulator[item.id].supplier_id;
    delete accumulator[item.id].code;
    delete accumulator[item.id].title;
  }
  return accumulator;
}, {});

我所拥有的也看起来很麻烦。如何在更清洁的功能中获得所需的结果？

最佳答案

每次累加器包含一个具有与初始数组中相同的ID的元素时，您只是在重置suppliers属性。正确的代码是：

let customers = await query.offset(offset).limit(limit);

customers = customers.reduce((accumulator, item) => {
  if (accumulator[item.id]) {
    const group = accumulator[item.id];
    //group.suppliers = [];
    group.suppliers.push({
      id: item.supplier_id,
      code: item.code,
      title: item.title
    });
  } else {
    accumulator[item.id] = item;
    accumulator[item.id].suppliers = [];
    accumulator[item.id].suppliers.push({
      id: item.supplier_id,
      code: item.code,
      title: item.title
    });
    delete accumulator[item.id].supplier_id;
    delete accumulator[item.id].code;
    delete accumulator[item.id].title;
  }
  return accumulator;
}, {});