Question

就复活节而言,我很想在我所开发的场所“东Egg Hunt”略感惊讶。其中2个Easter Eggs 我的藏匿将受到关键压制。这只赢了一种“压力CTRL和TAB同时”的类型,但将是一种“UP”三倍,然后是三倍”。这将寻求一系列关键压力,而不是一劳永逸地施加压力。我放弃了这一职能,但出于某种原因,它没有像它那样做。

NOTE: The script below is looking for the following keypress series:
surprise1 - LEFT (x3), RIGHT (x3), UP (x3), DOWN (x3)
surprise2 - SHIFT (x3), TAB (x3), CTRL (x3)

$(document.body).keydown(function(e) {
            surprise1(e);
            surprise2(e);
});

function surprise1(e) {
    var ev = (e) ? e : window.event;
    var k = ev.keyCode;
    if (k > 36 && k < 41) {
        typekeys[k] = isNaN(typekeys[k]) ? 0 : typekeys[k];
        typekeys[k]++;
        if (typekeys[37] == 3) {
            if (typekeys[37] == 3 && typekeys[39] == 3) {
                if (typekeys[37] == 3 && typekeys[39] == 3 && typekeys[38] == 3) {
                    if (typekeys[37] == 3 && typekeys[39] == 3 && typekeys[38] == 3 && typekeys[40] == 3) {
                        alert("You ve found Surprise 1! Contact the site admin ASAP to get your prize!");
                        typekeys[37] = typekeys[39] = typekeys[38] = typekeys[40] = 0;
                    }
                } else {
                    typekeys[40] = 0;
                }
            } else {
                typekeys[38] = typekeys[40] = 0;
            }
        } else {
            if (typekeys[37] > 3) {
                typekeys[37] = 0;
            }
            typekeys[39] = typekeys[38] = typekeys[40] = 0;
        }
    } else {
        typekeys[37] = typekeys[39] = typekeys[38] = typekeys[40] = 0;
    }
};

function surprise2(e) {
    var ev = (e) ? e : window.event;
    var k = ev.keyCode;
    if (k > 8 && k < 18) {
        typekeys[k] = isNaN(typekeys[k]) ? 0 : typekeys[k];
        typekeys[k]++;
        if (typekeys[16] == 3) {
            if (typekeys[9] == 3) {
                if (typekeys[16] == 3 && typekeys[9] == 3 && typekeys[17] == 3) {
                    alert("You ve found Surprise 2! Contact the site admin ASAP to get your prize!");
                    typekeys[16] = typekeys[9] = typekeys[17] = 0;
                }
            }
        } else {
            if (typekeys[16] > 3) {
                typekeys[16] = 0;
            }
            typekeys[9] = typekeys[17] = 0;
        }
    } else {
        typekeys[16] = typekeys[9] = typekeys[17] = 0;
    }
};

告诉我为什么不工作? 在我看来,它应当努力。

Answer 1

Try this: I m using jquery plugin

$(function () {
   var combination =   
   key( left , function(){ 
       combination =  left ;
       checkCombination();
   });
   key( right , function(){ 
       combination+=  right ;
       checkCombination();
   });
   key( up , function(){ 
       combination+=  up ;
       checkCombination();
   });
   key( down , function(){ 
       combination+=  down ;
       checkCombination();
   });

   function checkCombination() {
      if(combination ===  leftrightupdown ) {
        alert( surprise 1 );  
      } 
   }
});

Demo: http://jsfiddle.net/codef0rmer/BSdCq/

Answer 2

这就是我的解决办法。我不得不做一些细微的工作来比较以下描述的阵列:here 。我确信,你能够适应这一文字的一般形象,以适应你们的需要。

var seqs = [ [37,37,37,38,38,38,39,39,39,40,40,40], [9,9,9,16,16,16,17,17,17] ];
var seq  = [];

var messages=["You ve found Surprise 1! Contact the site admin ASAP to get your prize!", "You ve found Surprise 2! Contact the site admin ASAP to get your prize!"];

window.addEventListener("keydown", function(e){
    seq.push(e.keyCode);
    var eq = function(a,b){ return !( a<b || b<a ); };
    for ( var i = 0; i < seqs.length; i++ ) {
        if ( eq( seq, seqs[i].slice(0,seq.length) )) {
          if ( eq(seq, seqs[i]) ) {
            alert( messages[i] );
            seq = [];
          }
        return;
        }
    }
    seq = [];
});

Answer 3

这里我是如何解决这一问题的。

var nums = [38, 38, 40, 40, 37, 39, 37, 39, 66, 65];
var n = nums.slice();
$(document).keydown(function(e){
  if(e.which == n[0]){
    n.shift();
  } else n = nums.slice();
  if(n.length == 0) {
    //success!
    n = nums.slice();
  }
});

如果你知道顺序是多少,就会ku;

Answer 4

这一点更好:

$(function () {
  var combination =   
   key( left , function(){ 
   combination =  left ;
   checkCombination();
 });
key( right , function(){ 
    combination+=  right ;
    checkCombination();
});
key( up , function(){ 
    combination+=  up ;
    checkCombination();
});
key( down , function(){ 
    combination+=  down ;
    checkCombination();
});

key(! down  && ! left  && ! right  && ! up ,function() {
    combination =   ;
});

function checkCombination() {
   if(combination ===  leftrightupdown ) {
     alert( surprise 1 );  
   } 
}
});

友情链接