Javascript 来自指定变量的jQuery计数器

如果我有下面的变量

$(document).ready(function() {
    var minsecond = 1;
    var maxseconds = 4;
    var updateinterval = 1;
    var interval = updateinterval*1000;
    window.setInterval(function(){
    var currentseconds = ; //value increasing by 1 every var updateinterval (value) in seconds  from var minsecond value until it reaches var maxseconds and resetting to var minsecond value and cycling again
    $("span").html(currentseconds);
    }, interval);
});
​

如何创建一个计数器，该计数器从值

var minsecond

开始，每隔

var updateinterval

秒递增1，直到值

var maxsecond

然后在

var minsecond

重新启动

---

您喜欢这项工作吗

var currentseconds = 0;
setInterval(yourFunction(), interval);

function yourFunction() {
  if (minsecond >= maxsecond) {
    currentseconds = minsecond;
  }
  else {
    currentseconds++;
  }
  $("span").html(currentseconds);
}

---

设置一个测试，检查它是否达到您的

maxsecond

var，重置它。

我想下面的内容就是您想要的

这个怎么样：

var maxseconds = 4;
var updateinterval = 1;
var interval = updateinterval * 1000;

var counter = 1;

window.setInterval(function() {
    $("span").html(counter);
    counter = (counter % maxseconds == 0) ? 1 : counter + 1;
}, interval);​

---

示例

这应该可以。棘手的部分是时间间隔，因为它是可变的。在我的回答中，我假设你不是按整数递增，而是按区间数递增。希望你也是这么想的

---

注意调用-

setInterval

gotcha.Not currentseconds+=updateinterval From OP post“计数器，从值var minsecond开始，每一个值var updateinterval（以秒为单位）上升1到值var maxsecond，然后在值var minsecond重新启动”

var maxseconds = 4;
var updateinterval = 1;
var interval = updateinterval * 1000;

var counter = 1;

window.setInterval(function() {
    $("span").html(counter);
    counter = (counter % maxseconds == 0) ? 1 : counter + 1;
}, interval);​

$(document).ready(function() {
    var minsecond = 1;
    var maxseconds = 4;
    var updateinterval = 1;
    var interval = updateinterval * 1000;
    var currentseconds = minsecond;

    window.setInterval(function() {
        currentseconds += updateinterval;
        if (currentseconds > maxseconds) {
            currentseconds = minsecond;
        }
        $("span").html(currentseconds);
    }, interval);
});​