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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;
 }
 });
};
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asked Nov 30, 2012 at 15:08
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0

2 Answers 2

1
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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);
 };
}();
answered Nov 30, 2012 at 15:17
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17
  • \$\begingroup\$ nice approach, but works not as expected... \$\endgroup\$ Commented Nov 30, 2012 at 15:19
  • \$\begingroup\$ @yckart what's wrong with it? \$\endgroup\$ Commented Nov 30, 2012 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\$ Commented Nov 30, 2012 at 15:22
  • \$\begingroup\$ @NickLarsen it breaks on multiple instantiations and even fails with strings (words)... jsfiddle.net/ARTsinn/dxPdw/6 \$\endgroup\$ Commented Nov 30, 2012 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\$ Commented Nov 30, 2012 at 15:25
0
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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/

answered Nov 30, 2012 at 15:40
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1

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