JSON表格近似字符串搜索功能实现 匹配相近键返回对应条目数据

需求说明 #

• 入参：用户输入的搜索字符串
• 数据源：固定格式的JSON条目表，键为条目名称，值为条目属性
• 出参：返回和输入字符串相似度最高的键对应的完整条目，相似度过低时返回空

数据源 #

{
    "Death": { "Value": 190000, "Demand": 4, "Trend": 1, "Tier": "S" },
    "Legendary Borul (Alternative)": { "Value": 125000, "Demand": 2, "Trend": 1, "Tier": "S" },
    "Club Beast": { "Value": 50000, "Demand": 6, "Trend": 1, "Tier": "S" },
    "Dark Wing": { "Value": 19800, "Demand": 6, "Trend": 1, "Tier": "S" },
    "Zaruto (GRR III)": { "Value": 18800, "Demand": 6, "Trend": 1, "Tier": "S" },
    "First Wood Bender": { "Value": 17000, "Demand": 8, "Trend": 1, "Tier": "S" },
    "Old Will": { "Value": 14500, "Demand": 5, "Trend": 1, "Tier": "S" },
    "Martial Artist": { "Value": 12000, "Demand": 3, "Trend": 6, "Tier": "S" },
    "Expert Sorcerer": { "Value": 11250, "Demand": 5, "Trend": 1, "Tier": "S" },
    "Zaruto (GRR II)": { "Value": 9600, "Demand": 8, "Trend": 7, "Tier": "S" },
    "Sandwhich Leader / Mecha Frieza": { "Value": 1050000, "Demand": 0, "Trend": 4, "Tier": "S" },
    "First Wood Bender (Sage)": { "Value": "N/A", "Demand": 0, "Trend": 4, "Tier": "S" }
}

实现方案 #

采用编辑距离+规则加权的轻量实现，不需要引入第三方依赖，对短文本搜索场景准确率足够，匹配时统一转小写实现大小写不敏感。

完整代码 #

// 先把数据源存为常量
const itemData = {
    "Death": { "Value": 190000, "Demand": 4, "Trend": 1, "Tier": "S" },
    "Legendary Borul (Alternative)": { "Value": 125000, "Demand": 2, "Trend": 1, "Tier": "S" },
    "Club Beast": { "Value": 50000, "Demand": 6, "Trend": 1, "Tier": "S" },
    "Dark Wing": { "Value": 19800, "Demand": 6, "Trend": 1, "Tier": "S" },
    "Zaruto (GRR III)": { "Value": 18800, "Demand": 6, "Trend": 1, "Tier": "S" },
    "First Wood Bender": { "Value": 17000, "Demand": 8, "Trend": 1, "Tier": "S" },
    "Old Will": { "Value": 14500, "Demand": 5, "Trend": 1, "Tier": "S" },
    "Martial Artist": { "Value": 12000, "Demand": 3, "Trend": 6, "Tier": "S" },
    "Expert Sorcerer": { "Value": 11250, "Demand": 5, "Trend": 1, "Tier": "S" },
    "Zaruto (GRR II)": { "Value": 9600, "Demand": 8, "Trend": 7, "Tier": "S" },
    "Sandwhich Leader / Mecha Frieza": { "Value": 1050000, "Demand": 0, "Trend": 4, "Tier": "S" },
    "First Wood Bender (Sage)": { "Value": "N/A", "Demand": 0, "Trend": 4, "Tier": "S" }
}

// 计算两个字符串的Levenshtein编辑距离
function getEditDistance(str1, str2) {
    const len1 = str1.length
    const len2 = str2.length
    const dp = Array.from({ length: len1 + 1 }, () => new Array(len2 + 1).fill(0))
    for (let i = 0; i <= len1; i++) dp[i][0] = i
    for (let j = 0; j <= len2; j++) dp[0][j] = j
    for (let i = 1; i <= len1; i++) {
        for (let j = 1; j <= len2; j++) {
            const cost = str1[i-1] === str2[j-1] ? 0 : 1
            dp[i][j] = Math.min(
                dp[i-1][j] + 1,
                dp[i][j-1] + 1,
                dp[i-1][j-1] + cost
            )
        }
    }
    return dp[len1][len2]
}

function search(input) {
    const trimmedInput = input.trim()
    if (!trimmedInput) return null
    const searchLower = trimmedInput.toLowerCase()
    let bestMatchKey = null
    let maxScore = -Infinity

    for (const itemName of Object.keys(itemData)) {
        const nameLower = itemName.toLowerCase()
        // 取键名和输入等长的前缀计算编辑距离，归一化得到基础相似度分
        const namePrefix = nameLower.slice(0, searchLower.length)
        const distance = getEditDistance(searchLower, namePrefix)
        let score = 1 - distance / searchLower.length
        // 前缀完全匹配加权重，符合用户输入习惯
        if (nameLower.startsWith(searchLower)) score += 0.5
        // 输入是键名子串加权重
        if (nameLower.includes(searchLower)) score += 0.3

        if (score > maxScore) {
            maxScore = score
            bestMatchKey = itemName
        }
    }
    // 低于相似度阈值直接返回空，避免无关结果
    return maxScore > 0.3 ? { [bestMatchKey]: itemData[bestMatchKey] } : null
}

调用示例 #

search("legen")
// 返回结果
// { "Legendary Borul (Alternative)": { "Value": 125000, "Demand": 2, "Trend": 1, "Tier": "S" } }

可调参数说明 #

• 权重值：前缀匹配、子串匹配的加分数值可根据实际测试效果调整，越看重对应规则数值设越高
• 相似度阈值：当前设为0.3，阈值越高匹配越严格，越低模糊匹配范围越大
• 若需要支持中文搜索，可在相似度计算前增加拼音转换、中文分词逻辑即可

内容的提问来源于stack exchange，提问作者pygon