检查数组中的每个元素是否都符合条件 [英] Check if every element in array matches condition

问题描述

我有一组文件:

date: Date
users: [
  { user: 1, group: 1 }
  { user: 5, group: 2 }
]

date: Date
users: [
  { user: 1, group: 1 }
  { user: 3, group: 2 }
]

我想查询该集合以查找所有文档，其中我的用户数组中的每个用户ID均位于另一个数组[1、5、7]中.在此示例中，仅第一个文档匹配.

I would like to query against this collection to find all documents where every user id in my array of users is in another array, [1, 5, 7]. In this example, only the first document matches.

我能够找到的最佳解决方案是:

The best solution I've been able to find is to do:

$where: function() { 
  var ids = [1, 5, 7];
  return this.users.every(function(u) { 
    return ids.indexOf(u.user) !== -1;
  });
}

不幸的是， $ where docs中指出: /p>

Unfortunately, this seems to hurt performance is stated in the $where docs:

$ where评估JavaScript，不能利用索引.

$where evaluates JavaScript and cannot take advantage of indexes.

如何改善此查询?

推荐答案

您想要的查询是这样:

db.collection.find({"users":{"$not":{"$elemMatch":{"user":{$nin:[1,5,7]}}}}})

这说我找到了所有文档，这些文档没有元素不在列表1,5,7之外.

This says find me all documents that don't have elements that are outside of the list 1,5,7.

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