URL过滤器删除功能代码失效，寻求技术解决方案

我需要实现一个函数，用来删除URL中传递的指定过滤器。举个例子：
原URL：&data=true&filterList=include|value|Location|Region|Europe|Asia;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;
当删除Employee过滤器时，输出应该是：&data=true&filterList=include|value|Location|Region|Europe|Asia;include|value|Time|Current+Month+Flag|1;&decimals=5;
原因是Employee是include|value|Employee这个主题区域下唯一的过滤器，所以要删除整个对应的filterList条目。

下面是我自己写的函数代码，但它无法实现预期需求：

function deleteUrlSection(url, key, deleteValue) {
  const sections = url.split(';');
  let updatedSections = [];
  for (let i = 0; i < sections.length; i++) {
    let section = sections[i];
    let keyIndex = section.indexOf(key);
    if (keyIndex !== -1) {
      let values = section.substring(keyIndex).split('|');
      let updatedValues = [];
      for (let j = 0; j < values.length; j++) {
        if (values[j] !== deleteValue) {
          updatedValues.push(values[j]);
        }
      }
      if (updatedValues.length > 0) {
        updatedSections.push(updatedValues.join('|'));
      }
    } else {
      updatedSections.push(section);
    }
  }
  return updatedSections.join(';');
}

//test 1
let url = "&data=true&filterList=include|value|Location|Region|Europe|Asia;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;";
let key = "Region";
let deleteV = "Asia";
console.log("New Url ", "", deleteUrlSection(url, key, deleteV));

//expected output 
//&data=true&filterList=include|value|Location|Region|Europe;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;


//test 2
let url2 = "&data=true&filterList=include|value|Location|Region|Europe;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;";
let key2 = "Region";
let deleteV2 = "Europe";
console.log("New Url2 ", "", deleteUrlSection(url2, key2, deleteV2));

//expected output 
//&data=true&filterList=include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;


//test 3
let url3 = "&data=true&filterList=include|value|Location|Region|Europe|Asia;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;";
let key3 = "Current+Month+Flag";
let deleteV3 = "1";
console.log("New Url3 ", "", deleteUrlSection(url, key3, deleteV3));

//expected output 
//&data=true&filterList=include|value|Location|Region|Europe|Asia;include|value|Employee|Employee+Type|Employee;&decimals=5;

过滤器规则说明 #

• 过滤器格式：include|value|subjectArea|key|filter1|filter2...，每个filterList条目可包含多个过滤器
• 存在过滤器与subjectArea同名的情况，但仅需删除过滤器部分
• 如果删除目标过滤器后，该条目下没有剩余过滤器，需要删除整个filterList条目

我的思路是通过key（位于第3个竖线|之后）定位filterList条目，删除目标过滤器；如果删除后只剩一个过滤器就删除整个条目，但当前代码无法实现这个逻辑，需要修正。

修正后的函数实现 #

function deleteUrlSection(url, key, deleteValue) {
  const sections = url.split(';');
  const updatedSections = [];

  for (const section of sections) {
    // 跳过空段
    if (!section) continue;
    
    // 只处理包含目标key的段
    if (!section.includes(key)) {
      updatedSections.push(section);
      continue;
    }

    // 拆分出前缀（比如"filterList="）和过滤器内容
    const [prefix, ...filterParts] = section.split('=');
    const fullFilter = filterParts.join('=');
    const filterSegments = fullFilter.split('|');

    // 确认key在规则指定的位置（第3个竖线后，索引为3）
    if (filterSegments.indexOf(key) !== 3) {
      updatedSections.push(section);
      continue;
    }

    // 提取key之后的所有过滤器值
    const filters = filterSegments.slice(4);
    // 过滤掉要删除的值
    const remainingFilters = filters.filter(filter => filter !== deleteValue);

    if (remainingFilters.length === 0) {
      // 没有剩余过滤器，直接跳过整个条目
      continue;
    } else {
      // 重新构建完整的filter条目
      const newFilter = [
        ...filterSegments.slice(0, 4),
        ...remainingFilters
      ].join('|');
      updatedSections.push(`${prefix}=${newFilter}`);
    }
  }

  return updatedSections.join(';');
}

// 测试用例执行
// test 1
let url = "&data=true&filterList=include|value|Location|Region|Europe|Asia;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;";
let key = "Region";
let deleteV = "Asia";
console.log("New Url ", deleteUrlSection(url, key, deleteV));
// 预期输出: &data=true&filterList=include|value|Location|Region|Europe;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;

// test 2
let url2 = "&data=true&filterList=include|value|Location|Region|Europe;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;";
let key2 = "Region";
let deleteV2 = "Europe";
console.log("New Url2 ", deleteUrlSection(url2, key2, deleteV2));
// 预期输出: &data=true&filterList=include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;

// test 3
let url3 = "&data=true&filterList=include|value|Location|Region|Europe|Asia;include|value|Time|Current+Month+Flag|1;include|value|Employee|Employee+Type|Employee;&decimals=5;";
let key3 = "Current+Month+Flag";
let deleteV3 = "1";
console.log("New Url3 ", deleteUrlSection(url3, key3, deleteV3));
// 预期输出: &data=true&filterList=include|value|Location|Region|Europe|Asia;include|value|Employee|Employee+Type|Employee;&decimals=5;

关键逻辑说明 #

1. 精准定位目标条目：通过检查key是否在过滤器的第4个位置（索引3），避免误匹配与subjectArea同名的过滤器
2. 独立处理过滤器部分：提取key之后的所有过滤器值，单独过滤要删除的内容
3. 判断条目保留与否：如果过滤后没有剩余过滤器，直接丢弃整个条目；否则重新构建完整的filter段
4. 保留其他URL参数：不包含目标key的段直接保留，保证URL其他部分不受影响

内容的提问来源于stack exchange，提问作者Jakub97