CODEWITHSUNDEEP
javascript
LesterKingsley
LesterKingsley

Reputation: 51

How does filter using or/|| works together?

I know how filter function works but this my first time encountering the use of filter in this way can someone explain to me how the isLost: true objects are the only one being printed with the use of

!filtered || guest.isLost ?

const filtered=true

const users = [
  { name: "Jack", isLost: true },
  { name: "Sawyer", isLost: true },
  { name: "Lupin", isLost: false }
];

const filteredUsers = users.filter(user => !filtered || user.isLost).map(user => user)

console.log(filteredUsers)

console:

[{"name":"Jack","isLost":true},{"name":"Sawyer","isLost":true}]

Upvotes: 0

Views: 58

Answers (3)

MaximilianK
MaximilianK

Reputation: 63

!filterd is always false. So your condition is only true if isLost is true

If you want to filter out everything that has the property isLost:true Use this condition !user.isLost

const users = [
  { name: "Jack", isLost: true },
  { name: "Sawyer", isLost: true },
  { name: "Lupin", isLost: false }
];

const filteredUsers = users.filter(user => !user.isLost)

console.log(filteredUsers)

Upvotes: 1

Namysh
Namysh

Reputation: 4667

expr1 || epxr2 returns expr1 if it can be evaluate to true, else it returns expr2

In your case, as filtered is true (then !filtered is false), the filter works with the value of user.isLost

Upvotes: 1

axiac
axiac

Reputation: 72336

filtered is initialized with true and never modified.
Consequently, ! filtered is always false and false || x is the same as x.
And that means the filter condition is user.isLost.

Because .map(user => user) does not have any effect, your code has the same effect as:

const filteredUsers = users.filter(user => user.isLost)

Upvotes: 0

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