how to filter from multiple objects in javascript

Question

I need to get results from firstname or lastname, current behaviour gives me results from firstname only. Or a better approach to this using filter() :) should not be case sensitive too

const people = [
      { firstName: 'Bob', lastName: 'Smith', status: 'single' },
      { firstName: 'bobby', lastName: 'Suxatcapitalizing', status: 'single' },
      { firstName: 'Jim', lastName: 'bob', status: 'complicated' },
    ]

    const searchString = 'bob'

    const found = people.filter(
      (person) => new RegExp(searchString, 'i')
        .test(
          person.firstName || person.lastName
          ))

    console.log(found)

Thomas · Accepted Answer · 2020-08-26 11:21:56Z

2

The || operator returns the first argument that is truthy (not false, null, undefined, 0 or ""), so this searches for bob in person.firstName only.

You need to do two separate searches. To avoid repeating yourself, store the regexp object in a variable:

    const searchString = 'bob'
    const regexp = new RegExp(searchString, 'i')

    const found = people.filter(
      (person) => regexp.test(person.firstName) || regexp.test(person.lastName)
    )

    console.log(found)

answered Aug 26, 2020 at 11:21

Thomas

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5 Comments

Alnitak Over a year ago

it's not just a case of "repeating yourself" - the original code would create a brand new RegExp object for every element in the array.

Thomas Over a year ago

Premature optimization is the root of all evil.

Alnitak Over a year ago

most refactoring of this nature will also lead to optimisation.

BizMAN Over a year ago

@Thomas how about if I search Bob Smith it returns nothing....how can I combine the two terms in search?

Alnitak Over a year ago

@BizMAN that's a separate question.

Isaac · Accepted Answer · 2020-08-26 11:25:56Z

0

const people = [
      { firstName: 'Bob', lastName: 'Smith', status: 'single' },
      { firstName: 'bobby', lastName: 'Suxatcapitalizing', status: 'single' },
      { firstName: 'Jim', lastName: 'bob', status: 'complicated' },
    ]

    const searchString = 'bob'

    const found = people.filter(person =>
      (person.firstName.toLowerCase().includes(searchString) ||
       person.lastName.toLowerCase().includes(searchString))
    )

    console.log(found)

Alternative solution without using Regex is by converting to lowercase using toLowerCase and using includes to search of substring existence

edited Aug 26, 2020 at 11:25

answered Aug 26, 2020 at 11:23

Isaac

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1 Comment

Thomas Over a year ago

filter functions should return true or false, not the object. It works the way you wrote it, but is a bit unexpected.

Tigran Abrahamyan · Accepted Answer · 2020-08-26 11:30:13Z

0

Or use reduce

const found = people.reduce((acc, n) => {
  if (
    n.firstName.toLowerCase() === searchString
    || n.lastName.toLowerCase() === searchString
   ) {
    acc[n] = acc.push(n);
  }
  
  return acc;
}, []);

answered Aug 26, 2020 at 11:30

Tigran Abrahamyan

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1 Comment

JacobIRR Over a year ago

How does this perform compared to regex solutions?

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