How to make "OR" condition using a JavaScript method?

Question

I'm learning functional programming and trying to avoid using statements.

I have a long condition statement seen below:

if (!event.body || 
  !event.body.Id || 
  !event.body.source.accountId || 
  !event.body.sourceId || !event.body.targetId || 
  !event.body.targetName) 
{ return 0; }

I'm thinking of doing something like:

let invalid = [event.body, event.body.Id, event.body.source.accountId, event.body.sourceId, event.body.targetId, event.body.targetName]

if(invalid.includes("")){return 0;}

but I'm not sure about using "" and !, are they the same? I know ! means null or undefined, but not sure "" empty string will do here?

Another way I can think of is:

let invalid = [event.body, event.body.Id, event.body.source.accountId, event.body.sourceId, event.body.targetId, event.body.targetName]

if(invalid.every(x=>{if(x){return x}})){return 0;}

I'm not sure which method is better?

what would you like to check? is zero a valid value (for an id)? — Nina Scholz
– Nina Scholz, Commented Jul 12, 2018 at 13:48

Jonas Wilms · Accepted Answer · 2018-07-12 13:49:21Z

2

includes("") won't detect null, undefined, false, 0 and NaN, just"". Therefore using every is better here:

if(!invalid.every(x => x))

or the same with some:

if(invalid.some(x => !x))

edited Jul 12, 2018 at 13:49

answered Jul 12, 2018 at 13:48

Jonas Wilms

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2 Comments

Adelin Over a year ago

The way the array is named (invalid) it makes me believe it should be x => !x instead

Jonas Wilms Over a year ago

@adelin close, but yes I had to improve a bit :)

ronapelbaum · Accepted Answer · 2018-07-12 13:49:41Z

2

Note that !a || !b is equivalent to !(a && b).

So, consider this:

const isValid = event => event.body && event.body.Id && event.body.source.accountId ....

if(!isValid(event)){
  return 0;
}

answered Jul 12, 2018 at 13:49

ronapelbaum

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3 Comments

Ivan Over a year ago

Off-topic here, but: Why use a function instead of a boolean for isValid?

ronapelbaum Over a year ago

you said you want a more functional approach... what if you'll need to verify the event validity in other places?

Ivan Over a year ago

I'm not the OP but ok ;)

Marco · Accepted Answer · 2018-07-12 13:50:13Z

I'm thinking can I do like

let invalid = [event.body, event.body.Id, event.body.source.accountId, event.body.sourceId, event.body.targetId, event.body.targetName]

if(invalid.includes("")){return 0;}

This works, but only if the values are always strings. For instance, if event.body.Id is null or undefined (or another falsy value that isn't exactly ""), it will still go through.

Another way I can think is

let invalid = [event.body, event.body.Id, event.body.source.accountId, event.body.sourceId, event.body.targetId, event.body.targetName]

if(invalid.every(x=>{if(x){return x}})){return 0;}

This also works, but it uses an if statement, which isn't considered very functional.

My Suggestion:

Do

let invalid = [event.body, event.body.Id, event.body.source.accountId, event.body.sourceId, event.body.targetId, event.body.targetName]

return (invalid.every(x => x) ? 1 : 0)

Using invalid.every(x => x) avoids the if statement and will properly handle all falsy values. And the ternary makes it return either 1 or zero, without the use of an if.

Niladri Basu · Accepted Answer · 2018-07-12 13:53:04Z

0

You can use array.every like this:

let invalid = [event.body, event.body.Id, event.body.source.accountId, event.body.sourceId, event.body.targetId, event.body.targetName]

invalid.every(e => e)

answered Jul 12, 2018 at 13:53

Niladri Basu

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