CODEWITHSUNDEEP
javascript
byrdr
byrdr

Reputation: 5457

javascript checking multiple conditions in if statement

I have an if statement which checks multiple conditions. I would like it to skip the conditional statement if at least one thing returns false. I can see that it is currently picking up the conditional statement if at least one thing is true.

if((api != 'abosp') || (api !='sersp') || (api !='volsp') || (api !='consp') || (api !='givsp') || (api !='blosp')){
console.log("api: " + api  );
api = 'cussp'
}

What would be the correct way to implement this kind of logic?

Currently if api == 'abosp' it will still pass into the conditional statement instead of skipping it.

Upvotes: 1

Views: 2957

Answers (3)

JuniorCompressor
JuniorCompressor

Reputation: 20005

Actually your if can't be evaluated to true since api can't be all the values you check. I guess you need an and (&&) instead of an or (||):

if ((api != 'abosp') && (api !='sersp') && ...

In this case, you can use an associative array to have more succinct code:

d = {abosp: 1, sersp: 1, volsp: 1, consp: 1, givsp: 1, blosp: 1};
if (!d[api]) api = 'cussp';

Upvotes: 1

APD
APD

Reputation: 1519

I think you are looking for && rather than ||.

This will skip the conditional statement if any of the individual comparisions return false.

Upvotes: 0

juvian
juvian

Reputation: 16068

You should change your || for and operator: &&. Still, a more "clean" method would be:

banned = ["abosp", "sersp", "volsp", "consp", "givsp", "blosp"]

if(banned.indexOf(api) == -1){ // if api is not in list
   console.log("api: " + api  );
   api = 'cussp'  
}

Upvotes: 4

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