如何使用Mongoose获取数组中各ID对应的最后一条文档？

Absolutely, you can pull this off entirely with Mongoose (thanks to MongoDB's aggregation framework)—no manual post-processing of results needed. Here's how to fetch the latest QuizResult document for each levelAnswered ID in your levelIds array:

Step-by-Step Aggregation Pipeline #

We'll use a series of aggregation stages to filter, sort, group, and reshape your data to get exactly what you need:

QuizResult.aggregate([
  // 1. Narrow down to only the relevant documents for the user and target levels
  {
    $match: {
      levelAnswered: { $in: levelIds },
      answeredByUser: result.applicant._id
    }
  },
  // 2. Sort documents so the newest entry for each level comes first
  {
    $sort: { created: -1 }
  },
  // 3. Group by level ID, keeping only the latest document in each group
  {
    $group: {
      _id: "$levelAnswered",
      latestResult: { $first: "$$ROOT" } // $$ROOT refers to the entire document object
    }
  },
  // 4. Optional: Reshape output to match the structure of a regular find() result
  {
    $replaceRoot: { newRoot: "$latestResult" }
  }
])
.then(latestResults => {
  // latestResults is an array of the most recent QuizResult per levelId
  console.log(latestResults);
})
.catch(err => {
  console.error("Error fetching latest results:", err);
});

Breakdown of Each Stage: #

• $match: Filters out irrelevant documents upfront, reducing the data we process in later stages.
• $sort: Orders matched documents by created date in descending order, ensuring the newest entry for each level is at the top of its group.
• $group: Clusters documents by levelAnswered ID. Using $first: "$$ROOT" grabs the very first (and thus latest) document from each sorted group.
• $replaceRoot: Optional but convenient—this moves the nested latestResult object up to the root of each result entry, making the output look like what you'd get from a standard find() query.

Handling Populated Fields #

If you need to resolve the referenced fields (like answeredByUser or levelAnswered), you have two straightforward options:

Option 1: Use Mongoose's populate After Aggregation #

Aggregation results don't auto-populate, but you can run populate on the returned array:

QuizResult.aggregate([/* ... pipeline as above ... */])
.then(results => {
  return QuizResult.populate(results, { path: 'levelAnswered answeredByUser' });
})
.then(populatedResults => {
  // Now populatedResults includes the full referenced documents
});

Option 2: Add $lookup to the Aggregation Pipeline #

You can directly join with referenced collections in the pipeline using MongoDB's $lookup stage. For example, to populate levelAnswered:

QuizResult.aggregate([
  // ... match, sort, group, replaceRoot stages ...
  {
    $lookup: {
      from: "quizlevels", // Name of your QuizLevel collection (Mongoose pluralizes model names by default)
      localField: "levelAnswered",
      foreignField: "_id",
      as: "levelAnswered"
    }
  },
  { $unwind: "$levelAnswered" } // Convert the array from $lookup to a single object
])
// ... rest of your code

Why This Is Better Than Multiple Queries #

Using aggregation is way more efficient than running a separate findOne for each levelId (which would trigger N database calls for N IDs). The pipeline handles everything in one round trip, cutting down on network overhead and improving performance.

内容的提问来源于stack exchange，提问作者claudiomatiasrg