如何映射数组并使用索引以消除空值？答案

【问题标题】：How to map over an array and use index such that empty values are eliminated?如何映射数组并使用索引以消除空值？
【发布时间】：2020-06-24 16:46:25
【问题描述】：

我有两个数组 keyVals 和 rows 定义如下：

    keyVals = 
    [
      "Status",
      "ErrorHeading",
      "ErrorDetail"
    ]
    
rows =
[
  {
    "Hostname": "ABC",
    "name": "Check1",
    "Status": "PASS",
    "ErrorHeading": "",
    "ErrorDetail": ""
  },
  {
    "Hostname": "ABC",
    "name": "Check2",
    "Status": "PASS",
    "ErrorHeading": "",
    "ErrorDetail": ""
  },
  {
    "Hostname": "ABC",
    "name": "Check3",
    "Status": "FAIL",
    "ErrorHeading": "Heading3",
    "ErrorDetail": "ErrorDetails3"
  }
]

我编写了以下代码来生成Res：

function dArray(rows, keyVals) {
    const response = Object.values(rows.reduce((res, {Hostname, name, ...o}) => {
        res[Hostname] = res[Hostname] || {Hostname};
        keyVals.forEach(kv => res[Hostname][name + kv] = o[kv]);
        return res;
    }, {}));
    return response;
}

const Res = dArray(rows, keyVals);

Res 值如下所示：

Res = 
[
  {
    "Hostname": "ABC",
    "Check1kStatus": "PASS",
    "Check1ErrorHeading": "",
    "Check1ErrorDetail": "",
    "Check2Status": "PASS",
    "Check2ErrorHeading": "",
    "Check2ErrorDetail": "",
    "Check3Status": "FAIL",
    "Check3ErrorHeading": "Heading3",
    "Check3ErrorDetail": "ErrorDetails3"
  }
]

我想修改function dArray(rows, keyVals) 使Res 排除所有空值，如下所示：

Res = 
[
  {
    "Hostname": "ABC",
    "Check1kStatus": "PASS",
    "Check2Status": "PASS",
    "Check3Status": "FAIL",
    "Check3ErrorHeading": "Heading3",
    "Check3ErrorDetail": "ErrorDetails3"
  }
]

【问题讨论】：

There's no such thing as a "JSON Object"
keyVals 和 rows 是数组
问题是什么？ if (o[kv]) { ... }

标签： javascript node.js arrays dictionary node-modules

【解决方案1】：

function dArray(rows, keyVals) {
    const response = Object.values(rows.reduce((res, {Hostname, name, ...o}) => {
        res[Hostname] = res[Hostname] || {Hostname};
        keyVals.forEach(kv => (o[kv] && (res[Hostname][name + kv] = o[kv])));
        return res;
    }, {}));
    return response;
}

【讨论】：

【解决方案2】：

您可以使用特殊的filterObject 函数来实现。

const keyVals = ["Status", "ErrorHeading", "ErrorDetail"]
    
const rows = [
  { "Hostname": "ABC", "name": "Check1", "Status": "PASS", "ErrorHeading": "", "ErrorDetail": "" },
  { "Hostname": "ABC", "name": "Check2", "Status": "PASS", "ErrorHeading": "", "ErrorDetail": "" },
  { "Hostname": "ABC", "name": "Check3", "Status": "FAIL", "ErrorHeading": "Heading3", "ErrorDetail": "ErrorDetails3" },
]

const filterObject = (fn, x) => {
  const y = {}
  for (const k in x) {
    const ok = fn(x[k])
    if (ok) { y[k] = x[k] }
  }
  return y
}

function dArray(rows, keyVals) {
    const response = Object.values(rows.reduce((res, {Hostname, name, ...o}) => {
        res[Hostname] = res[Hostname] || {Hostname};
        keyVals.forEach(kv => res[Hostname][name + kv] = o[kv]);
        return res;
    }, {}));
    return response.map(obj => filterObject(x => x !== '', obj))
}

const Res = dArray(rows, keyVals);

console.log(Res)

filterObject 改编自我的函数式编程库 rubico 中的多态过滤器。过滤器的文档here，如果您有兴趣。

【讨论】：

【解决方案3】：

function dArray(rows, keyVals) {
  const response = Object.values(rows.reduce((res, {Hostname, name}) => {
      res[Hostname] = res[Hostname] || {Hostname};
      keyVals.forEach(kv => res[Hostname][name + kv] = o[kv]);
      return res;
  }, {}));

  let keys=Object.keys(response[0]);
  for(let i=0,l=keys.length;i<l;i++){
    if(!response[0][keys[i]]){
      delete response[0][keys[i]];
    }  
  }

  return response;
}

【讨论】：

有没有办法可以更新function dArray(rows, keyVals) 而不是创建另一个函数？

【解决方案4】：

在返回响应之前，使用map 和Object.entries 过滤对象中的空值。

function dArray(rows, keyVals) {
  const response = Object.values(
    rows.reduce((res, { Hostname, name, ...o }) => {
      res[Hostname] = res[Hostname] || { Hostname };
      keyVals.forEach((kv) => (res[Hostname][name + kv] = o[kv]));
      return res;
    }, {})
  );
  return response.map((obj) =>
    Object.fromEntries(Object.entries(obj).filter(([, val]) => val))
  );
}

keyVals = ["Status", "ErrorHeading", "ErrorDetail"];

rows = [
  {
    Hostname: "ABC",
    name: "Check1",
    Status: "PASS",
    ErrorHeading: "",
    ErrorDetail: "",
  },
  {
    Hostname: "ABC",
    name: "Check2",
    Status: "PASS",
    ErrorHeading: "",
    ErrorDetail: "",
  },
  {
    Hostname: "ABC",
    name: "Check3",
    Status: "FAIL",
    ErrorHeading: "Heading3",
    ErrorDetail: "ErrorDetails3",
  },
];

console.log(dArray(rows, keyVals));

【讨论】：

【解决方案5】：

这是通过过滤Object.values() 调用的输出来实现的。

const
  keyVals = [ "Status", "ErrorHeading", "ErrorDetail" ],
  rows = getRows();

function dArray(rows, keyVals) {
  const reducedValues = Object.values(
    rows.reduce( (res, { Hostname, name, ...o }) => {
      res[Hostname] = res[Hostname] || { Hostname };
      keyVals.forEach(kv => res[Hostname][name + kv] = o[kv]);
      return res;
    }, {})
  ),
  response = [
    Object.fromEntries(
      Object.entries(reducedValues[0]).filter(en=>en[1])
    )
  ];
  return response;
}
const Res = dArray(rows, keyVals);
console.log(Res);


function getRows(){
  return [{
      "Hostname": "ABC",
      "name": "Check1",
      "Status": "PASS",
      "ErrorHeading": "",
      "ErrorDetail": ""
    },
    {
      "Hostname": "ABC",
      "name": "Check2",
      "Status": "PASS",
      "ErrorHeading": "",
      "ErrorDetail": ""
    },
    {
      "Hostname": "ABC",
      "name": "Check3",
      "Status": "FAIL",
      "ErrorHeading": "Heading3",
      "ErrorDetail": "ErrorDetails3"
    }
  ];
}

【讨论】：