Question

我使用flexselect jQuery plugin来简化大型<select>列表的选择，这些列表采用Quicksilver algorithm灵活匹配（模糊匹配？）选项。

例如： 3gscam的输入与iPhone 3GS Camera快速匹配，但cam3gs的输入不匹配。

Quicksilver算法的修改版本是否适用于两个方向，并且优先选择前向匹配？

现有的jQuery插件会有所帮助，所以我不必自己动手。

Answer 1

我们可以看到方向是由for循环从长度设置为0 如果我们没有找到任何选项，我们可以从0到长度的另一种方式运行相同的代码

我希望我做对了

     String.prototype.score = function (abbreviation, offset) {
        offset = offset || 0 // TODO: I think this is unused... remove

        if (abbreviation.length == 0) return 0.9
        if (abbreviation.length > this.length) return 0.0

        for (var i = abbreviation.length; i > 0; i--) {
            var sub_abbreviation = abbreviation.substring(0, i)
            var index = this.indexOf(sub_abbreviation)


            if (index < 0) continue;
            if (index + abbreviation.length > this.length + offset) continue;

            var next_string = this.substring(index + sub_abbreviation.length)
            var next_abbreviation = null

            if (i >= abbreviation.length)
                next_abbreviation = ''
            else
                next_abbreviation = abbreviation.substring(i)

            var remaining_score = next_string.score(next_abbreviation, offset + index)

            if (remaining_score > 0) {
                var score = this.length - next_string.length;

                if (index != 0) {
                    var j = 0;

                    var c = this.charCodeAt(index - 1)
                    if (c == 32 || c == 9) {
                        for (var j = (index - 2); j >= 0; j--) {
                            c = this.charCodeAt(j)
                            score -= ((c == 32 || c == 9) ? 1 : 0.15)
                        }
                    } else {
                        score -= index
                    }
                }

                score += remaining_score * next_string.length
                score /= this.length;
                return score
            }
        }

        for (var i = 0; i < abbreviation.length;  i++) { //Cangee 
            var sub_abbreviation = abbreviation.substring(i, abbreviation.length-1) //Change
            var index = this.indexOf(sub_abbreviation)


            if (index < 0) continue;
            if (index + abbreviation.length > this.length + offset) continue;

            var next_string = this.substring(index + sub_abbreviation.length)
            var next_abbreviation = null

            if (i >= abbreviation.length)
                next_abbreviation = ''
            else
                next_abbreviation = abbreviation.substring(i)

            var remaining_score = next_string.score(next_abbreviation, offset + index)

            if (remaining_score > 0) {
                var score = this.length - next_string.length;

                if (index != 0) {
                    var j = 0;

                    var c = this.charCodeAt(index - 1)
                    if (c == 32 || c == 9) {
                        for (var j = (index - 2); j >= 0; j--) {
                            c = this.charCodeAt(j)
                            score -= ((c == 32 || c == 9) ? 1 : 0.15)
                        }
                    } else {
                        score -= index
                    }
                }

                score += remaining_score * next_string.length
                score /= this.length;
                return score
            }
        }

        return 0.0
    }

用于柔性匹配的双向Quicksilver算法

1 个答案: