$setIntersection (aggregation)

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Definition定义

$setIntersection

Takes two or more arrays and returns an array that contains the elements that appear in every input array.获取两个或多个数组并返回一个数组，该数组包含每个输入数组中出现的元素。

$setIntersection has the following syntax:语法如下所示：

{ $setIntersection: [ <array1>, <array2>, ... ] }

The arguments can be any valid expression as long as they each resolve to an array. 参数可以是任何有效的表达式，只要它们各自解析为一个数组。For more information on expressions, see Expressions.有关表达式的详细信息，请参阅表达式。

Behavior行为

$setIntersection performs set operation on arrays, treating arrays as sets. 对数组执行集合操作，将数组视为集合。If an array contains duplicate entries, $setIntersection ignores the duplicate entries. 如果数组包含重复项，$SetCrossion将忽略重复项。$setIntersection ignores the order of the elements.$setIntersection忽略元素的顺序。

$setIntersection filters out duplicates in its result to output an array that contain only unique entries. 筛选掉结果中的重复项，以输出只包含唯一项的数组。The order of the elements in the output array is unspecified.未指定输出数组中元素的顺序。

If a set contains a nested array element, $setIntersection does not descend into the nested array but evaluates the array at top-level.如果集合包含嵌套数组元素，$setIntersection不会下降到嵌套数组中，而是在顶层计算数组。

Example示例Result结果
 
{ $setIntersection: [ [ "a", "b", "a" ], [ "b", "a" ] ] }
 
[ "b", "a" ]
{ $setIntersection: [ [ "a", "b" ], [ [ "a", "b" ] ] ] }
 
[ ]

Example示例

Consider an experiments collection with the following documents:考虑一个包含以下文档的experiments集合：

{ "_id" : 1, "A" : [ "red", "blue" ], "B" : [ "red", "blue" ] }
{ "_id" : 2, "A" : [ "red", "blue" ], "B" : [ "blue", "red", "blue" ] }
{ "_id" : 3, "A" : [ "red", "blue" ], "B" : [ "red", "blue", "green" ] }
{ "_id" : 4, "A" : [ "red", "blue" ], "B" : [ "green", "red" ] }
{ "_id" : 5, "A" : [ "red", "blue" ], "B" : [ ] }
{ "_id" : 6, "A" : [ "red", "blue" ], "B" : [ [ "red" ], [ "blue" ] ] }
{ "_id" : 7, "A" : [ "red", "blue" ], "B" : [ [ "red", "blue" ] ] }
{ "_id" : 8, "A" : [ ], "B" : [ ] }
{ "_id" : 9, "A" : [ ], "B" : [ "red" ] }

The following operation uses the $setIntersection operator to return an array of elements common to both the A array and the B array:以下操作使用$setIntersection运算符返回A数组和B数组共有的元素数组：

db.experiments.aggregate(
   [
     { $project: { A: 1, B: 1, commonToBoth: { $setIntersection: [ "$A", "$B" ] }, _id: 0 } }
   ]
)

The operation returns the following results:操作返回以下结果：

{ "A" : [ "red", "blue" ], "B" : [ "red", "blue" ], "commonToBoth" : [ "blue", "red" ] }
{ "A" : [ "red", "blue" ], "B" : [ "blue", "red", "blue" ], "commonToBoth" : [ "blue", "red" ] }
{ "A" : [ "red", "blue" ], "B" : [ "red", "blue", "green" ], "commonToBoth" : [ "blue", "red" ] }
{ "A" : [ "red", "blue" ], "B" : [ "green", "red" ], "commonToBoth" : [ "red" ] }
{ "A" : [ "red", "blue" ], "B" : [ ], "commonToBoth" : [ ] }
{ "A" : [ "red", "blue" ], "B" : [ [ "red" ], [ "blue" ] ], "commonToBoth" : [ ] }
{ "A" : [ "red", "blue" ], "B" : [ [ "red", "blue" ] ], "commonToBoth" : [ ] }
{ "A" : [ ], "B" : [ ], "commonToBoth" : [ ] }
{ "A" : [ ], "B" : [ "red" ], "commonToBoth" : [ ] }