如何获得蒙古式查询的否定? [英] How do I get the negation of a mongoid query?

问题描述

在我的应用中，我查询:

In my app I have the query:

Model.where(flagged: true, saved: false)

或者我可以使用Model.all(:conditions => {flagged: true, saved: false})

没关系.

现在如何获得该查询的否定?即所有不满足flagged: true和saved: false(相当于flagged != true或saved != false)的Model实例

Now how do I get the negation of that query? i.e. all the instances of Model which do not satisfy flagged: true AND saved: false (which is the equivalent of flagged != true OR saved != false)

Model.excludes(flagged: true, saved: false)向我返回满足flagged != true和saved != false)的实例

Model.excludes(flagged: true, saved: false) returns me the instances which satisfy flagged != true AND saved != false)

推荐答案

我发现我可以使用所以Model.where(flagged: true, saved: false)的取反-可以表示为Model.where("this.flagged == true && this.saved == false")

So the negation of Model.where(flagged: true, saved: false) - which can be expressed as Model.where("this.flagged == true && this.saved == false")

是

Model.where("!(this.flagged == true && this.saved == false)")

正如罗素(Russell)的评论所暗示的那样，这比使用本机查询语言效率低，但是出于我的目的，这不是问题，因此我现在将其标记为正确-但对于未来的读者，请记住性能的影响.

As Russell's comment suggests this is less efficient than using the native query language, but for my purposes that is not a problem so I am marking this correct for now - but for future readers bear in mind the performance implications.

如果某人可以给出仅使用本机查询语言的答案，而无需进行显式的布尔逻辑扩展，则我将其标记为正确(也许在将来的Mongodb/Mongoid版本中可以这样做-如果可以的话)现在就完成.)

If someone can give an answer that uses only the native query language, without doing an explicit boolean logic expansion, I'll mark that correct (maybe that will be possible in a future version of Mongodb/Mongoid - if it can't be done now.)

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