选择与mongodb中两个字段的concat值匹配的记录

问题描述

是否可以在MongoDB上执行类似的操作？

Is there a way to do something like this on MongoDB?

select * from table where concat(field1, field2) = 'value'

为了澄清，我有一个全名数组，但是文档的名字和姓氏分开，所以我想做类似的事情：

To clarify, I have an array of full names, but the documents have firstname and lastname separate, so I want to do something like:

select * from table where concat(firstname, lastname) in ([ARRAY OF NAMES])

推荐答案

您只能使用聚合框架，而不能使用常规查找。

You can only do it with aggregation framework, not with regular find.

db.coll.aggregate({$project:{newField:{$concat:["$field1","$field2"]}}},
                  {$match:{newField:"value"}}
);

请注意，这将无法使用任何索引，因为不支持计算索引

Note that this will not be able to use any indexes, since there is no support for indexes on computed values in MongoDB (yet).

如果您在 field1 上有一个索引，并且您知道期望多少个字符field1有助于产生值，您可以像这样提高聚合的性能：

If you have an index on field1 and you know how many characters you expect field1 to contribute to value you can improve performance of this aggregation like this:

db.coll.aggregate({$match:{field1:/^val/}},
                  {$project:{newField:{$concat:["$field1","$field2"]}}},
                  {$match:{newField:"value"}}
);

其中 val 是值的第一部分字符串（尽管您不能比较的字符数不能超过 field1 的最短值。

where val is first part of "value" string (you must not compare more characters than the shortest possible value of field1 though.

编辑从3.6版开始，您可以使用 $ expr 表达式进行查找：

EDIT as of version 3.6 you can do this in find using the $expr expression:

db.coll.find({$expr:{$eq:["value", {$concat:["$field1", "$field2"]}]}})

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