在 Leaflet 中以特定值启动 JQuery UI Slider

Posted 2023-05-08

【中文标题】在 Leaflet 中以特定值启动 JQuery UI Slider

【英文标题】：start JQuery UI Slider at specific value in Leaflet

【发布时间】：2017-10-05 22:47:04

【问题描述】：

我正在使用这个 LeafletSlider 库在 Leaflet 中滑过多个层，每个层都有一个最接近半小时的关联时间。我想在最接近最近时间的数据点启动滑块。

在SliderControl.js，我替换了

var index_start = _options.minValue;

与

    var now = new Date();
    var hour = now.getHours();
    var minutes = now.getMinutes();
    if (minutes < 15) 
        minutes = "00";
     else if (minutes < 45)
        minutes = "30";
     else 
        minutes = "00";
        ++hour;
    
    var time = hour + minutes;
    var times = ["000", "030", "100", "130", "200", "230", "300", "330", "400", "430", "500", "530", "600", "630", "700", "730", "800", "830", "900", "930", "1000", "1030", "1100", "1130", "1200", "1230", "1300", "1330", "1400", "1430", "1500", "1530", "1600", "1630", "1700", "1730", "1800", "1830", "1900", "1930", "2000", "2030", "2100", "2130", "2200", "2230", "2300", "2330"];
    var index_start = times.indexOf(time);

但是，这不起作用。如何以特定值启动滑块？

【参考方案1】：

不知道你为什么用字符串而不是整数来做。无论哪种方式，您都可以这样做：

JavaScript

  var now = new Date();
  var hour = now.getHours() * 100;
  var minutes = now.getMinutes();
  var index_start = 0;
  if (minutes < 15) 
    minutes = 0;
   else if (minutes < 45) 
    minutes = 30;
   else 
    minutes = 100;
  
  var t = hour + minutes;
  var times = [0, 30, 100, 130, 200, 230, 300, 330, 400, 430, 500, 530, 600, 630, 700, 730, 800, 830, 900, 930, 1000, 1030, 1100, 1130, 1200, 1230, 1300, 1330, 1400, 1430, 1500, 1530, 1600, 1630, 1700, 1730, 1800, 1830, 1900, 1930, 2000, 2030, 2100, 2130, 2200, 2230, 2300, 2330];
  index_start = times.indexOf(t);

测试示例：https://jsfiddle.net/Twisty/qmsux550/