匹配包含MongoDB中提供的数组的任意组合的数组字段 [英] Matching an array field which contains any combination of the provided array in MongoDB

问题描述

我想查询指定的数组元素列表，以便返回的文档只能包含我传递的元素，而不必包含所有元素.

I would like to query with a specified list of array elements such that documents returned can only contain the elements I pass, but need not contain all of them.

给出如下文件:

{
  name: "Article 1",
  tags: ["Funny", "Rad"]
}

{
  name: "Article 2",
  tags: ["Cool", "Rad"]
}

{
  name: "Article 3",
  tags: ["Rad"]
}

这是一些示例数组及其各自的结果.

Here are some example arrays and their respective results.

• ["Rad"]应返回第3条
• ["Rad", "Cool"]应返回第2条和第3条
• ["Funny", "Cool"]应该不返回任何内容，因为没有文章中只有一个或两个标签

• ["Rad"] should return Article 3
• ["Rad", "Cool"] should return Article 2 and Article 3
• ["Funny", "Cool"] should return nothing, since there are no articles with only one of those tags or both

我敢肯定我可以用$where做到这一点，但是出于明显的原因，我想避免这种情况.

I'm sure I can pull this off with $where but I'd like to avoid that for obvious reasons.

推荐答案

您可以通过组合多个运算符来做到这一点:

You can do this by combining multiple operators:

db.test.find({tags: {$not: {$elemMatch: {$nin: ['Rad', 'Cool']}}}})

带有$nin的$elemMatch正在查找单个tags元素既不是'Rad'也不是'Cool'的文档，然后父级$not将该匹配取反以返回所有文档，其中没有任何元素.

The $elemMatch with the $nin is finding the docs where a single tags element is neither 'Rad' nor 'Cool', and then the parent $not inverts the match to return all the docs where that didn't match any elements.

但是，这也会返回缺少tags或没有元素的文档.要排除那些元素，您需要添加一个限定符，以确保tags具有至少一个元素:

However, this will also return docs where tags is either missing or has no elements. To exclude those you need to add a qualifier that ensures tags has at least one element:

db.test.find({
    tags: {$not: {$elemMatch: {$nin: ['Rad', 'Cool']}}},
    'tags.0': {$exists: true}
})

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