TypeScript flat Array to object tree

Question

I implement a UI for better Overview about our LDAP branchs. For that I want to use Angular Materials Tree. It´s great and very intuitiv to click through all branches. (https://material.angular.io/components/tree/overview)

What I have:

Array of multiple LDAP paths as a String:

groups = [
     'cn=devops,ou=smallUnit1,ou=unit1,o=group1,c=de',
     'cn=devops,ou=smallUnit1,ou=unit1,o=group2,c=de',        
     'cn=devops,ou=smallUnit2,ou=unit1,o=group1,c=de',
     'cn=dev,ou=smallUnit1,ou=unit2,o=group1,c=de',
     'cn=dev,ou=smallUnit1,ou=unit1,o=group2,c=de',
     'cn=ops,ou=smallUnit1,ou=unit1,o=group1,c=de'
]

What I already did:

I convertet this Strings to Array of standalone paths with dependencies. example for groups[0]:

dependencies = [
   {id: 'c=de',          parent: NULL,            child: 'o=group1'},
   {id: 'o=group1',      parent: 'c=de',          child: 'ou=unit1'},
   {id: 'ou=unit1',      parent: 'o=group1',      child: 'ou=smallUnit1'},
   {id: 'ou=smallUnit1', parent: 'ou=unit1',      child: 'cn=devops'},
   {id: 'cn=devops',     parent: 'ou=smallUnit1', child: NULL}

]

What I need:

I need an Object in which all keys are paths of our LDAP:

{
   c=de: {
       o=group1: {
          ou=unit1: {
             ou=smallUnit1: {
                cn=devops: {},
                cn=ops: {}
             }
             ou=smallUnit2: {
                cn=devops: {}
             }
          },
          ou=unit2: {
             ou=smallUnit1: {
                cn=dev: {}
             }
          }
       },
       o=group2: {
          ou=unit1: {
             ou=smallUnit1: {
                cn=devops: {},
                cn=dev: {}
             }
          }
       }
   }
}

I already tried to use methods like that: Build tree array from flat array in javascript But this algorithmus use push function to add the new branches to an arraykey. I need that the key is an object with more keys.

is the data sorted? id not, how do you distinuish between of ` ou=smallUnit1` from `ou=unit1` and `ou=unit2`? — Nina Scholz, Jul 25 '18 at 08:02

Nina Scholz · Accepted Answer · 2018-07-25T08:51:23.353

You could use directly groups, because all information in the reversed order are present, and dependencies is not.

Basically you need to

iterate groups
split strings of groups
take the values from the right side as key for an object and return for every step the inner object.

var groups = ['cn=devops,ou=smallUnit1,ou=unit1,o=group1,c=de', 'cn=devops,ou=smallUnit1,ou=unit1,o=group2,c=de', 'cn=devops,ou=smallUnit2,ou=unit1,o=group1,c=de', 'cn=dev,ou=smallUnit1,ou=unit2,o=group1,c=de', 'cn=dev,ou=smallUnit1,ou=unit1,o=group2,c=de', 'cn=ops,ou=smallUnit1,ou=unit1,o=group1,c=de'],
    tree = groups.reduce((object, string) => {
        string
            .split(',')
            .reduceRight((o, k) => o[k] = o[k] || {}, object);
        return object;
    }, {});

console.log(tree);

.as-console-wrapper { max-height: 100% !important; top: 0; }

score 0 · Answer 2 · answered Jul 25 '18 at 07:54

Seems to me that this should do the trick (given your current example data) :

const dependencies = [
   {id: 'c=de',          parent: null,            child: 'o=group1'},
   {id: 'o=group1',      parent: 'c=de',          child: 'ou=unit1'},
   {id: 'ou=unit1',      parent: 'o=group1',      child: 'ou=smallUnit1'},
   {id: 'ou=smallUnit1', parent: 'ou=unit1',      child: 'cn=devops'},
   {id: 'cn=devops',     parent: 'ou=smallUnit1', child: null}
];

let transformed = {};
let tracker = transformed;

dependencies.forEach(d => {
  tracker[d.id] = {};
  tracker = tracker[d.id];
});

console.log(transformed);

hi thank you for your help. it looks like i can create only one branch. not a whole tree. or am i wrong? — Nico Schuck, Jul 26 '18 at 08:42
No, it means you can create as much as you want, but the `dependencies` array you provided contains only one branch. — , Jul 26 '18 at 08:44

TypeScript flat Array to object tree

2 Answers2