CODEWITHSUNDEEP
javascriptconditional-statements
Spedwards
Spedwards

Reputation: 4492

And Or Conditions

I currently have a rather large conditional:

a[i1][i2 + 1] == '1' || a[i1][i2 - 1] == '1' || a[i1 + 1][i2] == '1' || a[i1 - 1][i2] == '1'

This works great however I just realised this is only half of what I need. Basically I need at least two of those conditionals to be true. Is there an easy way to accomplish this or do I have to rewrite the whole thing as needed?

An easy way to showcase what I mean is by the following.

Say I have three conditionals: a, b, and c. What I need in short is (a&&b) || (a&&c) || (b&&c)

Upvotes: 1

Views: 98

Answers (4)

Amy
Amy

Reputation: 7496

Another method is to utilize an Array to store the matched conditions. This way you can still define what evaluates to true and use the Array to store which conditions matched and view it afterwards.

var matched = [];

if (a[i1][i2 + 1] == '1') matched.push({x: i1, y: i2 + 1});
if (a[i1][i2 - 1] == '1') matched.push({x: i1, y: i2 - 1});
if (a[i1 + 1][i2] == '1') matched.push({x: i1 + 1, y: i2});
if (a[i1 - 1][i2] == '1') matched.push({x: i1 - 1, y: i2});

if (matched.length >= 2) {
    console.log('These matched!', matched);
}

Upvotes: 1

James
James

Reputation: 6123

Since true+true+false === 2, you can simply say:

   ((a[i1][i2 + 1] == '1') +
    (a[i1][i2 - 1] == '1') +
    (a[i1 + 1][i2] == '1') +
    (a[i1 - 1][i2] == '1'_) >= 2

Upvotes: 1

Özgür Kaplan
Özgür Kaplan

Reputation: 2136

You can try something like this.

var numberOfTrue=Number(a[i1][i2 + 1] == '1')+ Number(a[i1][i2 - 1] == '1'); // +....
if(numberOfTrue>=2){
//TODO
}

http://jsbin.com/wasuwoxe/1/edit

Upvotes: 2

jhyap
jhyap

Reputation: 3837

what about something like this?

var o = i1 + 1;
var p = i1 - 1;
var q = i2 + 1;
var r = i2 -1;
a[i1][q] == '1' || a[i1][r] == '1' || a[o][i2] == '1' || a[p][i2] == '1'

or something like this:

var a; // you should init this array in somewhere else above

function isValid(i1, i2){

return (a[i1][i2 + 1] == '1' || a[i1][i2 - 1] == '1' || a[i1 + 1][i2] == '1' || a[i1 - 1][i2] == '1')
}

Upvotes: 0

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