使用Mongoid进行不区分大小写的排序

Question

Right now I got:

@directories = collection.directories.all.asc(:name)

But it's case-sensitive, how do I do case-insensitive sorting?

Answer 1

Currently you cannot create case insensitive indexes in MongoDB see ...

http://jira.mongodb.org/browse/SERVER-90

So, it seems that means you cannot do case insensitive "sorting" either.

You can upvote the feature for future inclusion in MongoDB via the link above if you find it useful.

Eliot Horowitz from 10Gen (the supporters of MongoDB) suggest this in the meantime:

For short term - I would just add a 2nd field that you call .toLower() on before inserting. Then you can sort on that.

Answer 2

You will probably have to store the field twice, once with its real value, and again in all lowercase. You can then query the lowercased version for case-insensitive search (don't forget to also lowercase the query string).

This approach works (or is necessary) for many database systems, and it should perform better than regular expression based techniques (at least for prefix or exact matching).

Check this answer

Answer 3

如果您的集合不会崩溃内存（99％的时间），只需在那里排序：

Blah.all.sort_by{|i| i.blah_field.downcase}

Answer 4

I think you will have to use regex

Edit: maybe not, the feature has been requested though https://jira.mongodb.org/browse/SERVER-90

Answer 5

regexsearch = Regexp.new(params[:search], true)

@users = User.where(:email => regexsearch).all

That should do it :)