Mongoose查询：如何匹配数组含另一数组任意元素的文档

Got it, let's work through this for you. First, let's fix a couple of small issues from your original code, then land on the correct query for matching multiple order IDs.

1. Why Your Original Approach Needs Tweaks #

Your initial query used $all—which checks if the array contains every single ID you pass in. That works great for a single ID, but not when you want to match any ID from a list. Also, putting $populate inside the query object isn't how Mongoose operates—populate is a method you chain after find, not part of the query conditions.

2. The Correct Query for Multiple Order IDs #

To find all Combined documents where either done or notDone includes any ID from your list, swap $all for $in and keep using $or to cover both arrays. Here's the full working code:

First, define your list of order IDs (if you have string IDs, Mongoose will auto-convert them to ObjectIds, but you can explicitly cast them if you want):

const orderIds = ['60d21b4667d0d8992e610c85', '60d21b8967d0d8992e610c86']; // Example ID list
// Optional: Explicitly cast strings to ObjectIds
// const orderIds = ['your-id-1', 'your-id-2'].map(id => mongoose.Types.ObjectId(id));

Then run the query with proper population:

const matchingCombinedDocs = await Combined.find({
  $or: [
    { done: { $in: orderIds } }, // Match if `done` has ANY of the order IDs
    { notDone: { $in: orderIds } } // Match if `notDone` has ANY of the order IDs
  ]
})
.populate('done notDone'); // Chain populate to load the referenced order documents

Breakdown of Key Changes #

• $in: This operator checks if the array contains at least one element from your list—exactly what you need for "any ID in the list" matching.
• $or: Ensures we check both the done and notDone arrays for matches.
• Proper populate: Mongoose's populate is a query method, not part of the filter object. You can pass multiple fields as a space-separated string or an array.

Quick Side Note #

If you ever need to match documents where done or notDone contains all IDs from your list, you'd switch back to $all—but for your current use case, $in is the right tool.

内容的提问来源于stack exchange，提问作者Rares P