javascript operators using AND and OR in conditional statements

Question

I'm trying to display a message if only the conditional statements meet a certain requirement but with what I'm getting it fails me. Currently, the script is like this

 if ((err.status === 401 && !request.url.includes("users") || !request.url.includes("friends")) ) {
      console.log('hello')

    }

What I want to achieve is when the status is 401 and the URL doesn't include either users or friends the console should log hello but it fails to work

when I change the code to this

   if ((err.status === 401 && !request.url.includes("users")) ) {
          console.log('hello')

        }

or this

 if ((err.status === 401 && !request.url.includes("friends")) ) {
          console.log('hello')

        }

it works, but I want it to check for both conditions

Just change || to && if you need to check that url not included 'users' AND not included 'friends' — Darth
– Darth, Commented Aug 20, 2020 at 9:29
if(err.status==401 && !(request.url.includes("users") || request.url.includes("friends"))) — iAmOren
– iAmOren, Commented Aug 20, 2020 at 9:37
a AND (not (b OR c)) === a AND (not b) AND (not c) en.wikipedia.org/wiki/De_Morgan%27s_laws — malarres
– malarres, Commented Aug 20, 2020 at 9:44

MauriceNino · Accepted Answer · 2020-08-20 09:34:47Z

2

Just swap the || for a &&

if (err.status === 401 && !request.url.includes("users") && !request.url.includes("friends")) {
    console.log('hello')
}

If you think it through in your head it would sound like the following.

The status has to be 401
AND the url should NOT include users
AND the url should NOT include friends

If you want to code it like you wrote it:

The status has to be 401
AND The url should NOT include users OR friends

You could still do that, but just a little different:

if (err.status === 401 && !(request.url.includes("users") || request.url.includes("friends"))) {
    console.log('hello')
}

Both these codes mean exactly the same, just write it how you prefer it more. But always write it like it flows in your head. Just type out what you think :)

edited Aug 20, 2020 at 9:34

answered Aug 20, 2020 at 9:29

MauriceNino

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Comments

Malik Bagwala · Accepted Answer · 2020-08-20 10:30:05Z

1

For string validations I suggest you to use Regular Expressions (It can allow more complex validations)

if(error.status === 401 && !/users|friends/.test(request.url)){
  console.log("hello")
}

edited Aug 20, 2020 at 10:30

answered Aug 20, 2020 at 9:51

Malik Bagwala

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Comments

iAmOren · Accepted Answer · 2020-08-20 09:43:29Z

0

The way that is parallel to the wording:

if(err.status==401 && !(request.url.includes("users") || request.url.includes("friends")))

Visually:

if
(
  err.status==401
  &&
  !
    (
      request.url.includes("users")
      ||
      request.url.includes("friends")
    )
)

answered Aug 20, 2020 at 9:43

iAmOren

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Comments

subhadip pahari · Accepted Answer · 2020-08-20 10:11:35Z

0

your code is almost ok, just some issue with brackets ().

the format is :

if( cond1 && !(  cond2 || cond3  ) ) {
 ----- your logic ------
}

so your example should be like :

if ( err.status === 401 && !( request.url.includes("users") || request.url.includes("friends") ) )  {
            console.log('hello')
}

edited Aug 20, 2020 at 10:11

answered Aug 20, 2020 at 9:50

subhadip pahari

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1 Comment

subhadip pahari Over a year ago

yes, you are right. I am going to change the code. thanks for pointing out the mistake.

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