Question

我想从此文本中删除所有空格，然后将结果整理成对象

var notOrganizedObj = {
    "06fe562bc26ad23ec2a717c893ccc4407297de136b2761a035e80eea75637d3c-0" : "  ENABLED 10035 mYkSVf9AcAMCScw91kcPBiKawe51LYzJRP        8.9.30.2:19662 1557196699  4242543 1557188820",
    "0282ab652d149e092051df77db70187efe5d3e61ed714a9a4efa69d0cc4452a5-0" : "  ENABLED 10035 mMni8ALvkdUHN3jxgHnhJ99S2VthctFbM4       8.9.30.12:19662 1557196829  3169507 1557194872",
}

我已经尝试过

Object.keys(masternodelist).map(function(key, index) {
 var list = masternodelist[key].split(' ', '')
  console.log(masternodelist[key])
});

所以我可以拆分键的值，然后将其拆分为对象中的自定义键

预期结果将是这样

[{
    "address": "94.177.163.40:4836",
    "tx": "06fe562bc26ad23ec2a717c893ccc4407297de136b2761a035e80eea75637d3c",
    "payee": "WNijK2poAXXoEPp87YT1paWBGsTJYaQhcL",
    "status": "ENABLED",
    "protocol": 70210,
    "daemonversion": "0.12.3.3",
    "sentinelversion": "1.1.0",
    "sentinelstate": "current",
    "lastseen": 1557198027,
    "activeseconds": 4705452,
    "lastpaidtime": 1557181730,
    "lastpaidblock": 184934
  },
  {
    "address": "95.179.229.205:4836",
    "tx": "0282ab652d149e092051df77db70187efe5d3e61ed714a9a4efa69d0cc4452a5",
    "payee": "WcpfARQfyE4SfzLacehgoRrjxQg3wK3YBY",
    "status": "ENABLED",
    "protocol": 70210,
    "daemonversion": "0.12.3.3",
    "sentinelversion": "1.1.0",
    "sentinelstate": "current",
    "lastseen": 1557198094,
    "activeseconds": 8397933,
    "lastpaidtime": 1556925620,
    "lastpaidblock": 180869
  }]

Answer 1

说实话，由于许多键/属性缺失，我无法给您一个完整的答案，而且我不知道如何获取其余的值。

但是基本上，我建议您遍历该对象，将字符串拆分为字符串数组（.filter(i => i)删除数组中的空字符串”，然后将它们映射到所需的结果中。 / p>

我建议您这样做：

const notOrganizedObj = {
  "06fe562bc26ad23ec2a717c893ccc4407297de136b2761a035e80eea75637d3c-0": "  ENABLED 10035 mYkSVf9AcAMCScw91kcPBiKawe51LYzJRP        8.9.30.2:19662 1557196699  4242543 1557188820",
  "0282ab652d149e092051df77db70187efe5d3e61ed714a9a4efa69d0cc4452a5-0": "  ENABLED 10035 mMni8ALvkdUHN3jxgHnhJ99S2VthctFbM4       8.9.30.12:19662 1557196829  3169507 1557194872",
}

const res = [];
for (let key in notOrganizedObj) {
  const wordList = notOrganizedObj[key].split(' ').filter(i => i);
  res.push({
    address: wordList[3],
    tx: key.split('-')[0],
    payee: '',
    status: wordList[1],
    protocol: '',
    daemonversion: '',
    sentinelversion: '',
    sentinelstate: '',
    lastseen: '',
    activeseconds: '',
    lastpaidtime: '',
    lastpaidblock: ''
  })
}

console.log(res);

Answer 2

我将创建一个映射数组，该数组按您希望它们在字符串中出现的顺序列出这些键。

在map回调中，您可以使用Array.reduce()来配对键和值。

const mapToKeys = (obj, keys) => {
  return Object.keys(obj).map(key => {                                 //for each key in the input object
    let values = obj[key].match(/\S+/g) || [];                         //split value by whitespace
    return values.reduce((o,v,i) => ({...o, [keys[i]]: v}), {tx:key}); //zip key-value pairs
  });
};

const input = {"06fe562bc26ad23ec2a717c893ccc4407297de136b2761a035e80eea75637d3c-0" : "  ENABLED 10035 mYkSVf9AcAMCScw91kcPBiKawe51LYzJRP        8.9.30.2:19662 1557196699  4242543 1557188820", "0282ab652d149e092051df77db70187efe5d3e61ed714a9a4efa69d0cc4452a5-0" : "  ENABLED 10035 mMni8ALvkdUHN3jxgHnhJ99S2VthctFbM4       8.9.30.12:19662 1557196829  3169507 1557194872"}
const keys = ["status", "protocol", "payee", "address", "lastseen", "activeseconds", "lastpaidtime"];
console.log( mapToKeys(input,keys) );

Answer 3

您可以forEach对象键和split值，然后push每个具有结果属性的数组项映射。

var notOrganizedObj = {
    "06fe562bc26ad23ec2a717c893ccc4407297de136b2761a035e80eea75637d3c-0" : "  ENABLED 10035 mYkSVf9AcAMCScw91kcPBiKawe51LYzJRP        8.9.30.2:19662 1557196699  4242543 1557188820",
    "0282ab652d149e092051df77db70187efe5d3e61ed714a9a4efa69d0cc4452a5-0" : "  ENABLED 10035 mMni8ALvkdUHN3jxgHnhJ99S2VthctFbM4       8.9.30.12:19662 1557196829  3169507 1557194872",
}
let result = [];
Object.keys(notOrganizedObj).forEach( c=> {
 var list = notOrganizedObj[c].split(' ').filter(item=>item != null && item != "");
 //console.log(list)
  result.push({
    "address": list[3],
    "tx": c,
    "payee": list[2],
    "status": list[0],
    "protocol": list[1],
    "daemonversion": "0.12.3.3",
    "sentinelversion": "1.1.0",
    "sentinelstate": "current",
    "lastseen": 1557198027,
    "activeseconds": 4705452,
    "lastpaidtime": list[6],
    "lastpaidblock": 184934
  })
});

console.log(result);

var notOrganizedObj = {
    "06fe562bc26ad23ec2a717c893ccc4407297de136b2761a035e80eea75637d3c-0" : "  ENABLED 10035 mYkSVf9AcAMCScw91kcPBiKawe51LYzJRP        8.9.30.2:19662 1557196699  4242543 1557188820",
    "0282ab652d149e092051df77db70187efe5d3e61ed714a9a4efa69d0cc4452a5-0" : "  ENABLED 10035 mMni8ALvkdUHN3jxgHnhJ99S2VthctFbM4       8.9.30.12:19662 1557196829  3169507 1557194872",
}
let result = [];
Object.keys(notOrganizedObj).forEach( c=> {
 var list = notOrganizedObj[c].split(' ').filter(item=>item != null && item != "");
 //console.log(list)
  result.push({
    "address": list[3],
    "tx": c,
    "payee": list[2],
    "status": list[0],
    "protocol": list[1],
    "daemonversion": "0.12.3.3",
    "sentinelversion": "1.1.0",
    "sentinelstate": "current",
    "lastseen": 1557198027,
    "activeseconds": 4705452,
    "lastpaidtime": list[6],
    "lastpaidblock": 184934
  })
});

console.log(result);

如何将包含空格的字符串转换为有组织的对象

3 个答案: