Perl的打印功能：打印到最后的文件句柄？(Perl's print function: print to last filehandle?)

 这里很棒的教程 
 
 想法是在第一次转换的持续时间内延迟第二次转换。 
 
 因此，如果你有1个转换，每个持续时间为1秒，则将第二个延迟1秒，第三个延迟2秒，因为我们必须等待第一个和第二个转换完成。 希望你能得到这个想法。 
 
 
 
 
 
  
var canvas = d3.select('body')
						.append("svg")
						.attr("width",500)
						.attr("height",500);
		var addcircle = canvas.append("circle")
							.attr("cx",50)
							.attr("cy",50)
							.attr("r",25);		
		
		var circles = d3.select('circle');
		
		// first transition
		circles.transition().duration(1000)
				.attr("cx",250);
		
		// 2nd
		circles.transition().delay(1000)
				.duration(1000)
				.attr("cy",250)
		// 3rd
		circles.transition().delay(2000)
				.duration(1000)
				.attr("cx",50)
		// 4th
		circles.transition().delay(3000)
				.duration(1000)
				.attr("cy",50);
 
  
<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>
 
 
 
 

Excellent tutorial here
 
Idea is delay the second transition by the duration of first transition.
 
So if you have 3 transitions each of duration 1 sec then, delay the second by 1 sec and third by 2 sec, because we have to wait for both first and second transitions to complete. Hope you get the idea.
 
 
 
 
 
  
var canvas = d3.select('body')
						.append("svg")
						.attr("width",500)
						.attr("height",500);
		var addcircle = canvas.append("circle")
							.attr("cx",50)
							.attr("cy",50)
							.attr("r",25);		
		
		var circles = d3.select('circle');
		
		// first transition
		circles.transition().duration(1000)
				.attr("cx",250);
		
		// 2nd
		circles.transition().delay(1000)
				.duration(1000)
				.attr("cy",250)
		// 3rd
		circles.transition().delay(2000)
				.duration(1000)
				.attr("cx",50)
		// 4th
		circles.transition().delay(3000)
				.duration(1000)
				.attr("cy",50);
 
  
<script src="https://cdnjs.cloudflare.com/ajax/libs/d3/3.4.11/d3.min.js"></script>