Question

我有一个网络应用程序，它将分隔的字段传递到另一个网页。它工作正常！但是......我想列出javascript对象中不存在的字段（Name）。如何实现这一目标？

JS对象：

var members = [ { "Class": "E", "Rating": "1000", "ID": "16720664", "Name": "Adeyemon, Murie", "Expires": "1000.10.10" }, 
  { "Class": "B", "Rating": "1735", "ID": "12537964", "Name": "Ahmed, Jamshed", "Expires": "2018.10.18" }, 
  { "Class": "C", "Rating": "1535", "ID": "12210580", "Name": "Attaya, James", "Expires": "2019.01.12" }, 
  { "Class": "F", "Rating": "0001", "ID": "16281977", "Name": "Auld, Thomas", "Expires": "1000.10.10" }, 
  { "Class": "B", "Rating": "1793", "ID": "10117780", "Name": "Badamo, Anthony", "Expires": "2018.09.12" }
]

JS CODE：

let dataString = "Adeyemon, Murie|Ahmed, Jamshed|Attaya, James|Badamo, Anthony|Birmingham, Gerald|"; 
let splitString = dataString.split("|"); 

for (let i = 0; i < splitString.length; i++) { 
    $temp = splitString[i - 1]; 
    if ($temp > "") { 
        members.find(x => x.Name === $temp); 
    } 
}

Answer 1

使用filter

＆＃13;

var dataString =
  'Adeyemon, Murie|Ahmed, Jamshed|Attaya, James|Badamo, Anthony|Birmingham, Gerald|'
var members = [{"Class":"E","Rating":"1000","ID":"16720664","Name":"Adeyemon, Murie","Expires":"1000.10.10"},{"Class":"B","Rating":"1735","ID":"12537964","Name":"Ahmed, Jamshed","Expires":"2018.10.18"},{"Class":"C","Rating":"1535","ID":"12210580","Name":"Attaya, James","Expires":"2019.01.12"},{"Class":"F","Rating":"0001","ID":"16281977","Name":"Auld, Thomas","Expires":"1000.10.10"},{"Class":"B","Rating":"1793","ID":"10117780","Name":"Badamo, Anthony","Expires":"2018.09.12"}]

var res = dataString.split('|').filter(
  name => !members.map(o => o.Name).find(n => n === name)
).filter(name=>name.trim()!=='')
console.log(res);

＆＃13;

Answer 2

将members数组reducing用于Set个名称。然后，您可以使用Set.prototype.has()

过滤splitString数组

const members = [{"Class":"E","Rating":"1000","ID":"16720664","Name":"Adeyemon, Murie","Expires":"1000.10.10"},{"Class":"B","Rating":"1735","ID":"12537964","Name":"Ahmed, Jamshed","Expires":"2018.10.18"},{"Class":"C","Rating":"1535","ID":"12210580","Name":"Attaya, James","Expires":"2019.01.12"},{"Class":"F","Rating":"0001","ID":"16281977","Name":"Auld, Thomas","Expires":"1000.10.10"},{"Class":"B","Rating":"1793","ID":"10117780","Name":"Badamo, Anthony","Expires":"2018.09.12"}]
const dataString = "Adeyemon, Murie|Ahmed, Jamshed|Attaya, James|Badamo, Anthony|Birmingham, Gerald|";

const names = members.reduce((c, {Name}) => c.add(Name), new Set())
const missing = dataString.split('|')
    .filter(name => name.trim() && !names.has(name))
    .join('; ') // output from your comment on another answer

console.info(missing)

我已添加name.trim()来过滤掉|中结尾dataString创建的空记录。

创建Set的原因是为了避免在members中搜索整个dataString数组中的每个名称。 Set.prototype.has()应为 O（1）

Answer 3

您可以先创建Name对象映射，然后在此对象/映射中搜索字符串中的名称，这将花费O（n）作为n个名称。

var members = [ { "Class": "E", "Rating": "1000", "ID": "16720664", "Name": "Adeyemon, Murie", "Expires": "1000.10.10" }, 
  { "Class": "B", "Rating": "1735", "ID": "12537964", "Name": "Ahmed, Jamshed", "Expires": "2018.10.18" }, 
  { "Class": "C", "Rating": "1535", "ID": "12210580", "Name": "Attaya, James", "Expires": "2019.01.12" }, 
  { "Class": "F", "Rating": "0001", "ID": "16281977", "Name": "Auld, Thomas", "Expires": "1000.10.10" }, 
  { "Class": "B", "Rating": "1793", "ID": "10117780", "Name": "Badamo, Anthony", "Expires": "2018.09.12" }
];

var nameMap = members.reduce((prev, next) => {
  prev[next.Name] = next;
  return prev;
}, {});

let dataString = "Adeyemon, Murie|Ahmed, Jamshed|Attaya, James|Badamo, Anthony|Birmingham, Gerald|"; 
let names = dataString.split("|");

let result = names.filter(name => name && !(name in nameMap));

console.log(result);

检查javascript对象中是否存在记录

3 个答案: