如何将包含空格的字符串转换为有组织的对象

Question

I want to remove all spaces from this text then organize the result into object

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",
}

I've tried this

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

so i can split the value of keys then some how added to custom key in my object

the expected result is to be something like this

[{
    "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

TO be honest, I can't give you a complete answer, as many of the key/properties are missing, and I do not know how to get the rest of the values.

But basically, I recommend you to iterate through the object, split the string into an array of strings ( .filter(i => i) removes the empty strings '' within the array), and then map them into your desired result.

I would recommend you to do this:

 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

You can forEach Object key and split value then push each array item map with result property.

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); 

Answer 3

I'd create a mapping array that lists the keys in the order you expect them to appear in the string.

Within your map callback you could use Array.reduce() to pair the keys and values.

 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) );