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Home/ Questions/Q 8362881
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T12:02:44+00:00 2026-06-09T12:02:44+00:00

I am using Play 2.0.2 with ebean. In Info class, I defined @ManyToMany(fetch=FetchType.EAGER) private

  • 0

I am using Play 2.0.2 with ebean.

In Info class, I defined

@ManyToMany(fetch=FetchType.EAGER)
private Set<MemberInfo> members;

private Date createdDate = new Date();

And MemberInfo has memberId field.

When I do

public static Finder<Long,Info> find 
        = new Finder<Long,Info>(Long.class, Info.class);

find.fetch("members")
    .where().filterMany("members").eq("memberId", memberId)
    .order().desc("createdDate")
    .findList();

It returns all Info, without checking memberId of members.

What did I do wrong? Thanks.

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  1. Editorial Team
    Editorial Team
    2026-06-09T12:02:45+00:00Added an answer on June 9, 2026 at 12:02 pm

    filterMany() doesn’t filter parent results by children’s expressions (both has separate ‘ranges’).

    As descriped in its API it will find all Info objects and filtered members for each.

    There is also very similar topic on Google Groups where author of the question gives his own workaround for this.

    Examine the difference between:

    find.fetch("members")
        .where().filterMany("members").eq("memberId", 1L)
        .findList();
    

    and

    find.fetch("members")
        .where().eq("members.memberId", 1L)
        .findList();
    
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