3
\$\begingroup\$

How can this be written a bit shorter?

jQuery.konami = function(fn, code) {
    // ↑ ↑ ↓ ↓ ← → ← → B A
    code = code || [38, 38, 40, 40, 37, 39, 37, 39, 66, 65];

    var kkeys = '',
        i = 0;

    $(document).keydown(function(e) {
        var char = String.fromCharCode(e.which).toLowerCase();
        if (char === code[i++]) {
            kkeys += char;

            if (kkeys === code) {
                fn();
                kkeys = '';
                i = 0;
            }
        } else if (e.which === code[kkeys++]) {
            if (kkeys === code.length) {
                fn();
                kkeys = '';
                i = 0;
            }
        } else {
            kkeys = '';
            i = 0;
        }
    });
};
\$\endgroup\$
1
\$\begingroup\$

This is almost like code golf; this could probably be shortened. You just need to keep track of i and when it's past the end of the array, you know all the keys were hit in the correct order.

jQuery.konami = function() {
    function KonamiCode(kFn, kCode) {
        var i = 0;

        $(document).keydown(function(e) {
            var char = typeof kCode === 'string' ? String.fromCharCode(e.which).toLowerCase() : e.which;
            i = char === kCode[i] ? i + 1 : 0;
            if (i === kCode.length) {
                kFn();
                i = 0;
            }
        });
    }
    return function(fn, code) {
        // ↑ ↑ ↓ ↓ ← → ← → B A
        kCode = code || [38, 38, 40, 40, 37, 39, 37, 39, 66, 65];
        new KonamiCode(fn, kCode);
    };
}();
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ nice approach, but works not as expected... \$\endgroup\$ – yckart Nov 30 '12 at 15:19
  • \$\begingroup\$ @yckart what's wrong with it? \$\endgroup\$ – Nick Larsen Nov 30 '12 at 15:20
  • \$\begingroup\$ yeah it just simply isn't working, :p jsfiddle.net/dxPdw/5 I haven't investigated as to why though \$\endgroup\$ – user400654 Nov 30 '12 at 15:22
  • \$\begingroup\$ @NickLarsen it breaks on multiple instantiations and even fails with strings (words)... jsfiddle.net/ARTsinn/dxPdw/6 \$\endgroup\$ – yckart Nov 30 '12 at 15:22
  • \$\begingroup\$ Silly logic error, i = char === code[i] ? 0 : i + 1; needed to be i = char === code[i] ? i + 1 : 0;. \$\endgroup\$ – Nick Larsen Nov 30 '12 at 15:25
0
\$\begingroup\$

Here's a slightly shortened version:

jQuery.konami = function(fn, code) {
    // ↑ ↑ ↓ ↓ ← → ← → B A
    code = code || [38, 38, 40, 40, 37, 39, 37, 39, 66, 65];

    var i = 0;

    $(document).keydown(function(e) {
        var char = $.type(code[i]) === "string" ? String.fromCharCode(e.which).toLowerCase() : e.which;
        if (char === code[i]) {
            i++;
            if (i == code.length) {
                fn();
                i = 0;
            }
        } else {
            i = 0;
        }
    });
};

the kkeys var wasn't needed, and you could move the if else into the retrieval of char, allowing you to reuse what was inside the if originally.

http://jsfiddle.net/dxPdw/12/

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.