check if record exists in javascript object

Question

I have a web app which passes delimited fields to another web page. It works fine! But... I want to list the fields (Name) that don't exist in the javascript object. How can this be accomplished?

JS object:

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); 
    } 
}

If I could do that, I would. But at verlager.com/super-dev.php I am populating a grid that looks up class and rating. But some of the names don't exist in the members object. I want to list them so the user can cut and paste them accurately into a form. Then the grid can be redrawn. — verlager
– verlager, Commented May 16, 2018 at 1:25
besides the index error (i -1?) this code members.find(x => x.Name === $temp); ... you are finding it, but throwing away the result of .find — Jaromanda X
– Jaromanda X, Commented May 16, 2018 at 1:27
but when i == 0, $temp = splitString[-1] just use i .. and check $temp.length — Jaromanda X
– Jaromanda X, Commented May 16, 2018 at 1:28

Phil · Accepted Answer · 2018-05-16 02:13:22Z

1

use filter

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);

edited May 16, 2018 at 2:13

Phil

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answered May 16, 2018 at 1:29

xianshenglu

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3 Comments

verlager Over a year ago

Except that the output formatting is ... a little confusing. Please check localhost/www/verlager/super-dev.php the output could be Fischer, Robert; Petrosian, Tigran Any chance you could improve this?

xianshenglu Over a year ago

@verlager,I can't open the url you gave ,please use codepen or jsfiddler instead

verlager Over a year ago

I got it to format to my liking. Thanks for your work.

Zohaib Ijaz · Accepted Answer · 2018-05-16 01:33:16Z

You can first create Name to object mapping and then search name from string in this object/map which will cost O(n) for n names.

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);

Phil · Accepted Answer · 2018-05-16 02:09:22Z

Try reducing the members array to a Set of names. Then you can filter your splitString array using Set.prototype.has()

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)

I've added in the name.trim() to filter out the empty record created by the trailing | in your dataString.

The reason for creating a Set is to avoid searching the entire members array for each name in dataString. Set.prototype.has() should be O(1)

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