Mongoose查询：找出数组字段完全包含于指定数组的所有文档

Hey there! I get exactly what you're trying to do here — you want to pull all recipes from your Recipe collection where every ingredient in the ingredients array is part of your target list (['banana', 'apple', 'orange', 'mango']), right? That means excluding any recipe that has an ingredient outside that list (like the kiwi in your third test document).

The Cleanest Solution (MongoDB 3.2+) #

Starting with MongoDB 3.2, you can use the $setIsSubset aggregation operator wrapped in $expr to make this query super straightforward. Here's the code:

db.Recipe.find({
  $expr: {
    $setIsSubset: ["$ingredients", ["banana", "apple", "orange", "mango"]]
  }
})

How this works:

• $setIsSubset checks if the first array (the document's ingredients field) is a full subset of the second array (your target ingredient list). In plain terms: every item in the recipe's ingredients has to exist in your allowed list.
• $expr lets you use aggregation expressions directly in a query filter, which is necessary here since $setIsSubset is an aggregation operator.

Testing this against your sample data:

• Document 1 (['banana', 'apple']): All ingredients are in the target list → returned
• Document 2 (['banana', 'apple', 'orange']): Same deal, all ingredients are allowed → returned
• Document 3 (['apple', 'orange', 'kiwi']): Kiwi isn't in your target list → excluded

Perfect match for what you need!

For Older MongoDB Versions (Pre-3.2) #

If you're stuck on an older version where $expr isn't available, you can use a combination of $not and $elemMatch to get the same result:

db.Recipe.find({
  ingredients: {
    $not: {
      $elemMatch: {
        $nin: ["banana", "apple", "orange", "mango"]
      }
    }
  }
})

This logic flips the problem on its head: it finds recipes where no ingredient in the ingredients array is not in your target list. Which is exactly the same as saying all ingredients are in the list.

内容的提问来源于stack exchange，提问作者Gusti