查询两个参数的mongodb

Question

有人有想法吗，我该如何修改这段代码以便可以根据标题或评分或两者都进行搜索？我已经尝试了几种方法，包括使用$and运算符，但都没有成功

static async getMovies({
    filters = null,
    page = 0,
    moviesPerPage = 20,
  } = {}) {
    let query = {}
    if (filters) {
      if ("title" in filters) {
        console.log(filters["title"]);
        query.title = { $regex: filters["title"], $options: "i" };
      }  if ("rated" in filters) {
        query.rated = { $eq: filters["rated"] };
      }
    }

    let cursor
    try {
      cursor = await movies.find(query)
        .limit(moviesPerPage)
        .skip(moviesPerPage * page);
      const moviesList = await cursor.toArray();

      for (let i = 0; i < moviesList.length; i++) {
        const movie = moviesList[i];
        const movieReviews = await reviews
          .find({ movie_id: movie._id })
          .toArray();
        movie.review = movieReviews;
      }

      const totalNumMovies = await movies.countDocuments(query);
      return { moviesList, totalNumMovies };
    } catch (e) {
      console.error(`无法发出查找命令，${e}`);
      return { moviesList: ["error"], totalNumMovies: 0 };
    }
  }

Answer 1

如果您想要根据名称或评分进行搜索，或者同时使用两者，您可以简单地使用下面的查询。您不需要使用$eq运算符与rated的过滤器一起使用。mongoose的find函数会直接匹配值，而不使用$eq运算符。

if (filters?.title) {
  query.title = { $regex: `^${filters.title.replace(/[-[\]{}()*+?.,\/^$|#\s]/g, "\$&")}`, $options: "i" }
}  
if (filters?.rated) {
  query.rated = filters.rated
}
console.log(query)