jQuery递减循环

Question

我需要创建一个jQuery递增/递减循环。

因此，我在这里张贴的香草JavaScript，谁可以为我更改为jQuery代码。

​

var incomeTicker = 60;

window.setInterval(function(){
	
 	if (incomeTicker > 0){
	 	incomeTicker--;
		document.getElementById("incomeTicker").innerHTML = "Next Profit In : " + incomeTicker + " seconds";
// other code implemented as long as incomTicker > 0

	}
	
	if (incomeTicker <= 1){
  
  //code that is implemented when incomeTicker <=1
	
  incomeTicker = 60;
		
  //code that is implemented when incomeTicker <=1
	}
}, 1000);

<span class = "incomeTicker" id = "incomeTicker" > Next Profit In : 100 seconds</span>

​

有人能帮我把它翻译成jQuery吗？它必须递减，然后在循环完成后重置，如代码片段所示

Answer 1

在这里，您可以使用解决方案https://jsfiddle.net/vh0obhz6/

​

var incomeTicker = 60;

window.setInterval(function(){
   if (incomeTicker > 1){
      incomeTicker--;
      $("#incomeTicker").html(`Next Profit In : ${incomeTicker} seconds`);
   }else{
      incomeTicker = 60;
   }
}, 1000);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<span class = "incomeTicker" id = "incomeTicker" > Next Profit In : 100 seconds</span>

​

Answer 2

你应该使用更多的jQuery，它可以做所有的事情。

下面是添加更多jQuery的方法

​

$.incomeTicker = 60;

(function rec() {
  $.each((new Array($.incomeTicker)).fill(0), function(i) {
    var sec = Math.abs(i - $.incomeTicker);
    $('#incomeTicker').delay(1000).queue(function(n) {
      $(this).html("Next Profit In : " + sec + " seconds"); 
      n(); if (sec === 1) rec();
    });
  });
})();

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="incomeTicker"></div>

​

Answer 3

var incomeTicker = 60;

window.setInterval(function(){

    if (incomeTicker > 0){
        incomeTicker--;
        $("#incomeTicker").html("Next Profit In : " + incomeTicker + " seconds");
// other code implemented as long as incomTicker > 0

    }

    if (incomeTicker <= 1){

  //code that is implemented when incomeTicker <=1

  incomeTicker = 60;

  //code that is implemented when incomeTicker <=1
    }
}, 1000);