回到事件监听器? (Code.org - Javascript)

时间:2017-04-21 19:10:47

标签: javascript code.org

我对编程相对较新,并且正在使用Code.org进行教学。我相信代码是javascript,虽然我不太确定。我所做的程序是一个简单的连接点程序,我现在正试图实现级别。然而,一旦我在一个级别结束并且所有变量都被重置并且屏幕被更改,我就无法点击这些点。

这是因为在此部分代码运行之后:



//Finish the level, change the level 
  if(clickedX.length >= (level*2+4) - 1 && clickedY.length >= (level*2+4) - 1){
    for(var i = 1; i < level*2+4; i++){
      deleteElement(i + "Text");
    }
    clearCanvas("mainCanvas" + level);
    level++;
    setScreen("level" + level);
    setActiveCanvas("mainCanvas" + level);
    dotsX = [];
    dotsY = [];
    levelSector = [10];
    levelDiv = 300/(level*2+4);
    clickedX = [];
    clickedY = [];
    createDotCoords();
    }
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它不会返回它最初到达那里的点击事件监听器。这是整个计划,不是很长。我已经用尽了自己的解决方案,发现我现在卡住了。任何帮助是极大的赞赏。感谢。

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var dotsX = [];
var dotsY = [];
var level = 1;
var clickedX = [];
var clickedY = [];
var levelSector = [10];
var levelDiv = (300/(level*2+4));
setActiveCanvas("mainCanvas" + level);

createDotCoords();

//Searches an array for a specific number
//Adds and subtracts 4 to allow for the entire circle to be clicked
function searchVar(variable,needle) {
    for(var i = 0; i <  variable.length; i++){
      if(variable[i] === needle){
        return true;
      }
     else{
      for(var x = 1; x < 4; x++){
        if(variable[i] + x === needle){
          return true;
        }
        if(variable[i] - x === needle){
          return true;
        }
      }
      }
  }
}

//Create Dot Coordinates and Draw
function createDotCoords(){
  levelDiv = Math.round(levelDiv);

  for(var i = 1; i < level*2+4; i++){
    appendItem(levelSector,levelDiv * i);
  
    appendItem(dotsX,randomNumber(levelSector[i - 1],levelSector[i]));
    appendItem(dotsY,randomNumber(40,410));
  
    circle(dotsX[i - 1],dotsY[i - 1],4);
    textLabel(i + "Text", i);
    setPosition(i + "Text",dotsX[i - 1],dotsY[i - 1],3,4);
  }
}


//Get mouse click location and check if the location is the same
//as the dot location. Create a line between that one and the 
//previous dot if clicked already
onEvent("mainCanvas" + level, "click", function(event){
 var clickX = event.offsetX;
 var clickY = event.offsetY;
 var booX = false;
 var booY = false;
 console.log("X " + clickX + "  Y " + clickY);
 if (searchVar(dotsX,clickX)){
      booX = true;
      console.log("x is true");

      }
 if (searchVar(dotsY,clickY)){
      booY = true;
      console.log("y is true");

      }
      
if(booY && booX){
  if(clickedY.length === 0){
     appendItem(clickedY,clickY);
     }
    if(clickedY.length >= 1){
       if(searchVar(clickedY,clickY)){
          console.log("already in Y");
       }
       else{
          appendItem(clickedY,clickY);
       }
    }
        
    if(clickedX.length === 0){
        appendItem(clickedX,clickX);
     }
    if(clickedX.length >= 1){
      if(searchVar(clickedX,clickX)){
        console.log("already in X");
         }
         else{
         appendItem(clickedX,clickX);
         }
        }
        
        
    if(clickedX.length === 1 && clickedY.length === 1){
    console.log("only one dot clicked");
    setStrokeColor("green");
    circle(dotsX[0],dotsY[0],4);
    setStrokeColor("black");
    }
    else{
    setStrokeColor("green");
    circle(dotsX[clickedX.length - 1], dotsY[clickedY.length - 1], 4);
    setStrokeColor("black");
    line(dotsX[clickedX.length - 2], dotsY[clickedY.length - 2], dotsX[clickedX.length - 1], dotsY[clickedY.length - 1]);
    }
  }
//Finish the level, change the level 
  if(clickedX.length >= (level*2+4) - 1 && clickedY.length >= (level*2+4) - 1){
    for(var i = 1; i < level*2+4; i++){
      deleteElement(i + "Text");
    }
    clearCanvas("mainCanvas" + level);
    level++;
    setScreen("level" + level);
    setActiveCanvas("mainCanvas" + level);
    dotsX = [];
    dotsY = [];
    levelSector = [10];
    levelDiv = 300/(level*2+4);
    clickedX = [];
    clickedY = [];
    createDotCoords();
    }
});
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关于格式化/一般逻辑/编程礼仪的任何提示,这样的事情,也会非常有用。

1 个答案:

答案 0 :(得分:0)

我的猜测是,在你重置变量和画布时,某些东西会妨碍点击监听器,或者响应中显示的内容。你能在这里发布项目的分享链接吗?我可以看看代码。 (我每天都在App Lab工作。所以我可以帮助代码,或者弄清楚它是否是一个真正的错误。)