javascript checking multiple conditions in if statement

Question

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.

this concept is called "short circuit evaluation"

DLeh
– DLeh

2015-03-18 20:39:34 +00:00
Commented Mar 18, 2015 at 20:39 — DLeh
– DLeh, Commented Mar 18, 2015 at 20:39

juvian · Accepted Answer · 2015-03-18 20:39:01Z

4

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'  
}

answered Mar 18, 2015 at 20:39

juvian

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JuniorCompressor · Accepted Answer · 2015-03-18 20:47:18Z

1

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';

edited Mar 18, 2015 at 20:47

answered Mar 18, 2015 at 20:39

JuniorCompressor

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Comments

APD · Accepted Answer · 2015-03-18 20:39:11Z

0

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

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

answered Mar 18, 2015 at 20:39

APD

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