javascript - 在javascript中合并多个数组？

Question

我有这段代码应该基于匹配的 siteid 合并 3 个数组.
这在一定程度上起作用。基本上，它只找到 1 个站点标识并忽略相同的站点标识!
这是代码:
如果你运行下面的代码，你会看到它只添加了一个带有 的表单。 siteid 为 77 到合并的数组!但它应该将所有具有相同 siteid(本例中为 77)的表单添加到合并数组中。

var siteS1 = [{
    Date: '2021-02-02 11:19',
    siteid: 77,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 76,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 66,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 96,
  }
];


var siteS2 = [{
    Date: '2021-02-02 11:19',
    siteid: 77,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 76,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 66,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 96,
  }
];

var form = [{
    data: {},
    siteid: 77,
  },
  {
    data: {},
    siteid: 77,
  },
  {
    data: {},
    siteid: 77,
  },
  {
    data: {},
    siteid: 96,
  }
];

const merge = (...arrayList) => {
  const siteids = arrayList
    .map(array => array.map(item => item.siteid))
    .flat()
  return Array.from(new Set(siteids))
}

const find = (siteid, array) => {
  const found = array.find(item => item.siteid === siteid)
  if (!found) return undefined // returns undefined if item was not found
  const {
    siteid: _,
    ...restFoundData
  } = found // omit `siteid` from found item
  return restFoundData
}

const populate = (siteids) => siteids.map(siteid => {
  const foundSiteS1 = find(siteid, siteS1)
  const foundSiteS2 = find(siteid, siteS2)
  const foundForm = find(siteid, form)
  return {
    siteS1: foundSiteS1,
    siteS2: foundSiteS2,
    form: foundForm ? [foundForm.data] : [], // empty array if no items matched
  }
})

const result = populate(merge(siteS1, siteS2, form))
console.log(result)

我在这里想念什么？
编辑:
这是示例代码:

var siteS1 = [{
Date: '2021-02-02 11:19',
siteid: 77,
}];


var siteS2 = [{
Date: '2021-02-02 11:19',
siteid: 77,
}];

var form = [{
data: {},
siteid: 77,
},
{
data: {},
siteid: 77,
},
{
data: {},
siteid: 77,
},
{
data: {},
siteid: 77,
},
{
data: {},
siteid: 78,
}];

预期的结果是这样的:

[{
    "siteS1": {
        "Date": "2021-02-02 11:19"
    },
    "siteS2": {
        "Date": "2021-02-02 11:19"
    },
    "form": [{},{},{},{}]
}]

但是当前代码的结果是这样的:

[{
    "siteS1": {
        "Date": "2021-02-02 11:19"
    },
    "siteS2": {
        "Date": "2021-02-02 11:19"
    },
    "form": [{}]
}]

如您所见，它只找到 1 个 siteid 为 77 的表单，而忽略其余的表单。

Answer 1

Array.find仅找到第一个匹配元素，我使用另一个函数来查找表单的所有匹配数据

const findAll = (siteid, array) => {
    return array.filter(item => item.siteid === siteid).map(({siteid, ...data}) => data);
}

var siteS1 = [{
    Date: '2021-02-02 11:19',
    siteid: 77,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 76,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 66,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 96,
  }
];


var siteS2 = [{
    Date: '2021-02-02 11:19',
    siteid: 77,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 76,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 66,
  },
  {
    Date: '2021-02-02 11:19',
    siteid: 96,
  }
];

var form = [{
    data: {},
    siteid: 77,
  },
  {
    data: {},
    siteid: 77,
  },
  {
    data: {},
    siteid: 77,
  },
  {
    data: {},
    siteid: 96,
  }
];

const merge = (...arrayList) => {
  const siteids = arrayList
    .map(array => array.map(item => item.siteid))
    .flat()
  return Array.from(new Set(siteids))
}

const find = (siteid, array) => {
  const found = array.find(item => item.siteid === siteid)
  if (!found) return undefined // returns undefined if item was not found
  const {
    siteid: _,
    ...restFoundData
  } = found // omit `siteid` from found item
  return restFoundData
}

const findAll = (siteid, array) => {
  return array.filter(item => item.siteid === siteid).map(({siteid, ...data}) => data);
}

const populate = (siteids) => siteids.map(siteid => {
  const foundSiteS1 = find(siteid, siteS1)
  const foundSiteS2 = find(siteid, siteS2)
  const foundForm = findAll(siteid, form)
  return {
    siteS1: foundSiteS1,
    siteS2: foundSiteS2,
    form: foundForm, // empty array if no items matched
  }
})

const result = populate(merge(siteS1, siteS2, form))
console.log(result)