Angularjs 滑块、customValueToPosition 无法使用范围选择

Question

我正在尝试将 angularjs 滑块用于范围 selection 和自定义缩放。

这是滑块选项对象...

vm.priceSlider = {  
    min: 3,  
    high: 10,  
    options: {  
        floor: 0,  
        ceil: 3600,  
        showTicksValues: true,  
        step: 1,  
        ticksArray: ticksArray,  

        customValueToPosition: function(val, minVal, maxVal) {  
            var pos =  1/16 * (ticksArray.indexOf(val));  
            if (positions.indexOf(pos) === -1){  
                positions.push(pos);  
            }  
            return pos;  
          },  

          customPositionToValue: function(percent, minVal, maxVal) {  
            var index = Math.round(percent * 16);  

            var min = (ticksArray[index - 1]);  
            var max = (ticksArray[index]);  

            var minPos = positions[index -1];  
            var maxPos = positions[index];  

            var value = max - ((maxPos-percent)/(maxPos-minPos)) * (max - min);   
            return Math.round(value);  
          }  
        }  
    }

jsfiddle http://jsfiddle.net/cwhgLcjv/2527/

我面临的问题是
1) 滑块指针在不包括刻度的值之间不可见
2) 无法 select 范围正确

如有格式错误，请见谅..

Answer 1

为此，我们需要正确实施 customValueToPosition 和 customPositionToValue 以进行自定义缩放。

      customValueToPosition: function(val, minVal, maxVal) {
        if(!ticksPositions[val]){
            console.log("value => " + val);
            ticksPositions[val] = calculateValuePosition(val);
        }
        return ticksPositions[val];
      },  

      customPositionToValue: function(percent, minVal, maxVal) {  
          var minPos;
          var maxPos;      
        var index = Math.round(percent * 16);
        var min = (ticksArray[index - 1]);
        var max = (ticksArray[index]);
        minPos = ticksPositions[min];
            maxPos = ticksPositions[max];            
        var value = max - ((maxPos-percent)/(maxPos-minPos)) * (max - min); 
        return Math.round(value);
      }

这是更新后的 jsfiddle http://jsfiddle.net/cwhgLcjv/2682/

Angularjs 滑块、customValueToPosition 无法使用范围选择

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