我正在使用jsfiddle创建一个路径，该路径根据用户输入的数字循环一定次数。例如，用户输入的任何数字小于30 1个路径（一个瓶子）将在纸上绘制，用户输入的任何数字超过30个2个路径（2个瓶子）将在纸上绘制。最终我想让它继续最多10道路（10瓶）所以范围是（0-30,31-60,61-90 ...）目前它正在工作，但如果我写出整个事情，这将是很多代码，在我这样做之前，我想知道是否有任何新的方法可以缩短它。我确信有一种方法我无法理解。
http://jsfiddle.net/anderskitson/Gqbmk/
var paper = Raphael(document.getElementById("notepad"), 500, 500);
$("input").keyup(function() {
var value = parseInt($(this).val());
paper.clear();
if (value > 30) {
for (i = 0; i < 2; i++) {
var randomNumber1 = Math.floor(Math.random()*200 +25);
var randomNumber2 = Math.floor(Math.random()*200 +25);
var randomNumber3 = Math.floor(Math.random()*25);
var path_a = paper.path("M242.07,270.119c0,0-14.596-30.606-7.625-35.793 c3.864-2.876,2.145-18.561,1.832-18.784c-0.313-0.224-1.839-0.319-1.839-0.319c-1.555-0.192-0.201-3.456-0.201-3.456 s0,0-0.598-0.352c-0.598-0.351,1.129-1.345,1.129-1.345c3.738-2.785,10.449-2.983,11.126-2.344c0.677,0.64-0.354,1.44-0.354,1.44 s0.73,0.832,1.333,2.111c0.604,1.28-0.792,1.665-0.792,1.665c1.852,6.718,9.877,14.935,9.877,14.935 c4.795,0.589,7.7,10.683,7.7,10.683l6.271,22.746C269.929,261.307,263.641,270.119,242.07,270.119z");
path_a.attr({fill: 'none',stroke: '#231F20',"stroke-width": '3',"stroke-miterlimit": '10','stroke-opacity': '1'}).data('id', 'path_a');
path_a.translate(i*30, 10);
path_a.rotate(randomNumber3);
}//end of for statement
}else{
for (i = 0; i < 1; i++) {
var randomNumber1 = Math.floor(Math.random()*200 +25);
var randomNumber2 = Math.floor(Math.random()*200 +25);
var randomNumber3 = Math.floor(Math.random()*25);
var path_a = paper.path("M242.07,270.119c0,0-14.596-30.606-7.625-35.793 c3.864-2.876,2.145-18.561,1.832-18.784c-0.313-0.224-1.839-0.319-1.839-0.319c-1.555-0.192-0.201-3.456-0.201-3.456 s0,0-0.598-0.352c-0.598-0.351,1.129-1.345,1.129-1.345c3.738-2.785,10.449-2.983,11.126-2.344c0.677,0.64-0.354,1.44-0.354,1.44 s0.73,0.832,1.333,2.111c0.604,1.28-0.792,1.665-0.792,1.665c1.852,6.718,9.877,14.935,9.877,14.935 c4.795,0.589,7.7,10.683,7.7,10.683l6.271,22.746C269.929,261.307,263.641,270.119,242.07,270.119z");
path_a.attr({fill: 'none',stroke: '#231F20',"stroke-width": '3',"stroke-miterlimit": '10','stroke-opacity': '1'}).data('id', 'path_a');
path_a.translate(i*30, 10);
path_a.rotate(randomNumber3);
}//end of for statement
}//end of else statement
}）;
答案 0 :(得分：2)
将冗余代码移动到单独的函数中。
此外，randomNumber1和randomNumber2尚未使用。
var paper = Raphael(document.getElementById("notepad"), 500, 500);
function drawBottles(count) {
for (i = 0; i < count; i++) {
var randomNumber3 = Math.floor(Math.random()*25);
var path_a = paper.path("M242.07,270.119c0,0-14.596-30.606-7.625-35.793 c3.864-2.876,2.145-18.561,1.832-18.784c-0.313-0.224-1.839-0.319-1.839-0.319c-1.555-0.192-0.201-3.456-0.201-3.456 s0,0-0.598-0.352c-0.598-0.351,1.129-1.345,1.129-1.345c3.738-2.785,10.449-2.983,11.126-2.344c0.677,0.64-0.354,1.44-0.354,1.44 s0.73,0.832,1.333,2.111c0.604,1.28-0.792,1.665-0.792,1.665c1.852,6.718,9.877,14.935,9.877,14.935 c4.795,0.589,7.7,10.683,7.7,10.683l6.271,22.746C269.929,261.307,263.641,270.119,242.07,270.119z");
path_a.attr({fill: 'none',stroke: '#231F20',"stroke-width": '3',"stroke-miterlimit": '10','stroke-opacity': '1'}).data('id', 'path_a');
path_a.translate(i*30, 10);
path_a.rotate(randomNumber3);
}//end of for statement
}
$("input").keyup(function() {
var value = parseInt($(this).val());
paper.clear();
if (value > 30) {
drawBottles(2);
} else {
drawBottles(1);
}//end of else statement
});
答案 1 :(得分：1)
两个for循环看起来都一样。如果循环中的终点是唯一的diff，那么将其设置为变量并且只写入for循环1次。
$("input").keyup(function() {
var value = parseInt($(this).val());
paper.clear();
var endVal = value > 30 ? 2 : 1; //all values that change between the two define as such
for (i = 0; i < endVal; i++) {
var randomNumber1 = Math.floor(Math.random()*200 +25);
var randomNumber2 = Math.floor(Math.random()*200 +25);
var randomNumber3 = Math.floor(Math.random()*25);
var path_a = paper.path("M242.07,270.119c0,0-14.596-30.606-7.625-35.793 c3.864-2.876,2.145-18.561,1.832-18.784c-0.313-0.224-1.839-0.319-1.839-0.319c-1.555-0.192-0.201-3.456-0.201-3.456 s0,0-0.598-0.352c-0.598-0.351,1.129-1.345,1.129-1.345c3.738-2.785,10.449-2.983,11.126-2.344c0.677,0.64-0.354,1.44-0.354,1.44 s0.73,0.832,1.333,2.111c0.604,1.28-0.792,1.665-0.792,1.665c1.852,6.718,9.877,14.935,9.877,14.935 c4.795,0.589,7.7,10.683,7.7,10.683l6.271,22.746C269.929,261.307,263.641,270.119,242.07,270.119z");
path_a.attr({fill: 'none',stroke: '#231F20',"stroke-width": '3',"stroke-miterlimit": '10','stroke-opacity': '1'}).data('id', 'path_a');
path_a.translate(i*30, 10);
path_a.rotate(randomNumber3);
}//end of for statement
})
答案 2 :(得分：1)
GGG代码的Mod。使您能够以30步为单位继续运行
var stepping = 30;
var paper = Raphael(document.getElementById("notepad"), 500, 500);
function drawBottles (count) {
for (i = 0; i < count; i++) {
var randomNumber1 = Math.floor(Math.random()*200 +25);
var randomNumber2 = Math.floor(Math.random()*200 +25);
var randomNumber3 = Math.floor(Math.random()*25);
var path_a = paper.path("M242.07,270.119c0,0-14.596-30.606-7.625-35.793 c3.864-2.876,2.145-18.561,1.832-18.784c-0.313-0.224-1.839-0.319-1.839-0.319c-1.555-0.192-0.201-3.456-0.201-3.456 s0,0-0.598-0.352c-0.598-0.351,1.129-1.345,1.129-1.345c3.738-2.785,10.449-2.983,11.126-2.344c0.677,0.64-0.354,1.44-0.354,1.44 s0.73,0.832,1.333,2.111c0.604,1.28-0.792,1.665-0.792,1.665c1.852,6.718,9.877,14.935,9.877,14.935 c4.795,0.589,7.7,10.683,7.7,10.683l6.271,22.746C269.929,261.307,263.641,270.119,242.07,270.119z");
path_a.attr({fill: 'none',stroke: '#231F20',"stroke-width": '3',"stroke-miterlimit": '10','stroke-opacity': '1'}).data('id', 'path_a');
path_a.translate(i*30, 10);
path_a.rotate(randomNumber3);
}//end of for statement
}
$("input").keyup(function() {
var value = parseInt($(this).val());
paper.clear();
drawBottles(Math.floor(value/stepping));
});