有没有更好的方式来写这个,或更短的更紧凑的方式
我正在创建一个jsfiddle根据用户输入的数字循环一定次数的路径。例如,用户输入任何小于30 1的路径(一个瓶子)将在用户输入的纸张上绘制30个以上的路径(2个瓶子)将在纸张上绘制。最终我想继续使用10条路径(10瓶),所以范围应该是(0-30,31-60,61-90 ...)。目前它正在工作,但是如果我写出全部内容,它会有很多代码,而在我这样做之前,我想知道是否有人用新的方法缩短了这一点。我确信有一种方法我无法弄清楚。有没有更好的方式来写这个,或更短的更紧凑的方式
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
});
将冗余代码移入单独的函数中。
而且,randomNumber1
和randomNumber2
未被使用。
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
});
这适用于我,除了当输入字段为空或删除空白paper.clear不像以前那样运行。 – 2012-02-28 02:06:38
@AndersKitson它似乎没有清除您原来的小提琴中的纸张。我没有改变那部分。 – 2012-02-28 02:16:15
哦,你的权利,这是我的任何建议,让我工作的最后一块。 – 2012-02-28 02:31:17
两个for循环看起来是一样的。如果循环中的终点是唯一的diff,那么将其设置为一个变量,并且只写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
})
GGG代码的Mod。使您能够保持30
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));
});
啊,是的,我错过了那部分。当你在它时,不妨将它盖在十个瓶子上;) – 2012-02-27 23:47:41
有一件事我仍然试图得到。当输入30以下的数字时,我想要有一瓶,但现在显示0。 – 2012-02-28 01:43:10
对不起,只是一个小错误。你需要这个:'drawBottles(Math.floor(value/stepping)+ 1);' – 2012-02-28 18:55:42
这听起来像http://codereview.stackexchange.com/ – 2012-02-27 23:46:08
@MikeSamuel一个问题,它像是*听起来像*一个步骤运行,但IMO它太琐碎CR。我宁愿亲自看到它在这里问及回答。 – 2012-02-28 03:41:36
这个问题似乎是无关紧要的,因为它没有显示对问题的最小理解。 – 2014-03-03 05:42:42