CODEWITHSUNDEEP
functionjavascript
user1038065
user1038065

Reputation:

Difficulty splitting array and returning a value from within it; Javascript

I have got an array that consists of strings. I have made a function that searches the array based on the search term parameter. However, when i run the code it only ever outputs the string at index 0 of the array. I want it to return the corresponding url in the array when a search is run.

Any help would be very much appreciated. Thanks in advance.

Upvotes: 1

Views: 123

Answers (4)

hugomg
hugomg

Reputation: 69934

I think you should be doing

var terms = arrayOfURL[i].toLowerCase().split('~');

if(0 <= terms[1].indexOf(searchToLower))
           // ^    ^ 
           // |    |-- 0 <= indexOf method determines
           // |        if searchToLower is a substring of terms[1] 
           // |               
           // |-- term[1] gets the part after the first "~"

and 

return terms[0]; //terms[0] is the part before the first "~"

I would also consider returning null or the empty string "" in case of failure (instead of returning the arbritrary "Nothing Found!" message)

Upvotes: 0

xdevel2000
xdevel2000

Reputation: 21364

This is a correct version of your program:

var arrayOfURL = [
                 "http://www.google.co.uk~Google is a search engine.",
                 "http://www.yahoo.co.uk~Yahoo is another search engine.",
                 "http://bing.com~Bing is a decision engine."
                 ];

function findURL(arrayOfURL,search)
{   
    var searchtoLower = search.toLowerCase();
    for (var i = 0; i < arrayOfURL.length; i++)
    {     
        var z = arrayOfURL[i].toLowerCase().split('~')[1]; 
        if (z.indexOf(searchtoLower) != -1)
            return arrayOfURL[i];
    }

     return "Nothing Found!";
}


findURL(arrayOfURL,"decision")

I hope it can help you.

Upvotes: 0

Thanh Nguyen
Thanh Nguyen

Reputation: 1056

The problem may be in the second i param:

var z = arrayOfURL[i].toLowerCase().split('~')[i];

The string will be splitted into 2 parts (index 0, 1). Why did you select part i?

Upvotes: 1

TS-
TS-

Reputation: 4411

So you are trying to return URL based on the String after the ~?

Do the line 

arrayOfURL[i].toLowerCase().split('~')[i];

seem weird to you? Imagine as i increases, eg. i = 4

 arrayOfURL[4].toLowerCase().split('~')[4];

Does that last [4] make sense? I am guessing the reason it never got past the first element is because the code actually erroring out on that part.

I think what you want is (likewise for the return line, you'll want [0]

 arrayOfURL[i].toLowerCase().split('~')[1];

I would also take a look at 

if (z >= searchtoLower)

what are you trying to compare there?

Upvotes: 1

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