Firestore动态子字段查询的复合索引策略：基于关键词数组的优化方案

{
  "title": "Example Article",
  "year": 2023,
  "season": "Fall",
  "genres": {
    "Action": true,
    "Comedy": true,
    "Sci-Fi": true
  },
  "studios": {
    "Studio A": true,
    "Studio B": true
  }
}

q = query(q, where(`genres.${filters.genre}`, "==", true));
q = query(q, where(`studios.${filters.studio}`, "==", true));

{
  "title": "Example Article",
  "year": 2023,
  "season": "Fall",
  "genres": {
    "Action": true,
    "Comedy": true
  },
  "studios": {
    "Studio A": true
  }
}

{
  "title": "Example Article",
  "year": 2023,
  "season": "Fall",
  "genres": {
    "Action": true,
    "Comedy": true
  },
  "studios": {
    "Studio A": true
  },
  "keywords": [
    "Action",
    "Comedy",
    "Studio A",
    "Action, Studio A", // 组合关键词
    "Comedy, Studio A"  // 组合关键词
    // ... 更多可能的组合
  ]
}

/**
 * 从多个字符串数组中生成所有可能的单项和组合关键词。
 * 例如：combineKeywords(["apple", "banana"], ["happy", "sad"])
 * 返回：["apple", "banana", "happy", "sad", "apple, happy", "apple, sad", "banana, happy", "banana, sad"]
 */
function combineKeywords(...arrays: string[][]): string[] {
  const allKeywords: Set<string> = new Set();
  const baseItems: string[] = [];

  // 将所有独立项添加到基础列表和关键词集合中
  for (const arr of arrays) {
    for (const item of arr) {
      if (item) {
        baseItems.push(item);
        allKeywords.add(item);
      }
    }
  }

  // 生成所有两两组合
  for (let i = 0; i < baseItems.length; i++) {
    for (let j = i + 1; j < baseItems.length; j++) {
      // 确保组合的顺序一致，例如 "A, B" 而不是 "B, A"
      const combined = [baseItems[i], baseItems[j]].sort().join(", ");
      allKeywords.add(combined);
    }
  }

  // 如果需要，可以扩展到三项或更多项组合，但通常两项组合已能满足大部分需求
  // 复杂的组合可能会导致关键词数组过大

  return Array.from(allKeywords);
}

// 示例用法：
const docGenres = ["Action", "Comedy", "Drama"];
const docStudios = ["Studio A", "Studio B"];

// 在保存文档时调用此函数
const keywordsForDocument = combineKeywords(docGenres, docStudios);
// keywordsForDocument 结果可能包含:
// ["Action", "Comedy", "Drama", "Studio A", "Studio B",
//  "Action, Comedy", "Action, Drama", "Action, Studio A", "Action, Studio B",
//  "Comedy, Drama", "Comedy, Studio A", "Comedy, Studio B",
//  "Drama, Studio A", "Drama, Studio B"]

/**
 * 根据用户选择的筛选值生成查询目标关键词。
 * 如果只选择了一个值，则返回该值；如果选择了多个，则返回它们的组合。
 * 注意：组合顺序应与生成文档关键词时的顺序保持一致。
 */
function generateQueryTarget(...values: string[]): string | null {
  const filteredValues = values.filter(v => v); // 过滤空值
  if (filteredValues.length === 0) {
    return null; // 没有有效的筛选条件
  }
  // 确保组合的顺序与文档中存储的关键词一致
  return filteredValues.sort().join(", ");
}

// 示例用法：
const userSelectedGenre = filters.genre; // 例如 "Action"
const userSelectedStudio = filters.studio; // 例如 "Studio A"

const targetKeyword = generateQueryTarget(userSelectedGenre, userSelectedStudio);
// 如果 userSelectedGenre="Action", userSelectedStudio="Studio A", 则 targetKeyword="Action, Studio A"
// 如果 userSelectedGenre="Action", userSelectedStudio="", 则 targetKeyword="Action"

import { collection, query, where, orderBy, limit, Query, DocumentData } from "firebase/firestore";
import { firestore } from "./firebaseConfig"; // 假设你的Firestore实例

export function generateSearchQuery(
  searchTerms: string,
  filters: {
    year: number | "";
    season: string;
    genre: string;
    studio: string;
  }
): Query<DocumentData> {
  const docRef = collection(firestore, "example");
  let q = query(docRef);

  if (searchTerms) q = query(q, where("title", "==", searchTerms));
  if (filters.year) q = query(q, where("year", "==", filters.year));
  if (filters.season) q = query(q, where("season", "==", filters.season));

  // --- 关键修改部分 ---
  const genreAndStudioKeywords: string[] = [];
  if (filters.genre) genreAndStudioKeywords.push(filters.genre);
  if (filters.studio) genreAndStudioKeywords.push(filters.studio);

  const targetKeyword = generateQueryTarget(...genreAndStudioKeywords);

  if (targetKeyword) {
    q = query(q, where("keywords", "array-contains", targetKeyword));
  }
  // --- 关键修改部分结束 ---

  q = query(q, orderBy("id", "desc"), limit(20)); // id用于分页

  return q;
}