MongoDB：使用两个相等的字段查询文档，$ match和$ eq

What is the best way to return all documents in a collection if I want document.a == document.b?

如果我想要document.a == document.b，那么返回集合中所有文档的最佳方法是什么？

I've tried

我试过了

db.collection.aggregate([ { $match: { $eq: [ '$a', '$b' ] } }])

But it returns with no errors or results, because I assume it is literally matching strings "$a" and "$b". Is there a different way to specify that these are fields?

但它返回没有错误或结果，因为我认为它字面上匹配字符串“$ a”和“$ b”。是否有不同的方法来指定这些是字段？

db.collection.aggregate([    { $project: { 
    eq: { $cond: [ { $eq: [ '$a', '$b' ] }, 1, 0 ] } 
} },
{ $match: { eq: 1 } }])

The above works, but requires the additional step of querying again with whatever documents it found or projecting all possible fields.

上述工作，但需要额外的步骤，使用它找到的任何文件或预测所有可能的字段再次查询。

Is there a better way for achieving this query?

有没有更好的方法来实现此查询？

2 个解决方案

#1

Basically, you are trying to perform a self join. An operation not supported by MongoDB.

基本上，您正在尝试执行自联接。 MongoDB不支持的操作。

Concerning the $eq operator, as you guessed:

关于$ eq运算符，正如您猜测的那样：

By design, the $eq comparison query operator match a field against a value.
根据设计，$ eq比较查询运算符将字段与值进行匹配。
But the $eq comparison aggregation operator compare the value of two expressions.
但$ eq比较聚合运算符比较两个表达式的值。

I don't know any other way to perform what you need than using an extra $project step as you suggested.

我不知道如你所建议的那样使用额外的$ project步骤来执行你需要的任何其他方法。

Please note this is not significantly more expensive as, anyway, your query cannot use any index and MongoDB will do a full scan.

请注意，这并不是特别昂贵，因为无论如何，您的查询不能使用任何索引，MongoDB将进行全面扫描。

#2

If I understood your question right you want those documents that have same values in field1 and field2.

如果我理解你的问题，你想要那些在field1和field2中具有相同值的文档。

For this try

为此尝试

db.coll.find({$where: function() { return this.field1 == this.field2 } } );

or more compact

或者更紧凑

db.coll.find({ $where : "this.field1 == this.field2" } );

#1