2 
class File {
   String name
   File parent;    
   static belongsTo =[parent:File ]   
   static hasMany = [childrens:File];   
   static mapping = {   
      table 'Info_File'
      id(generator:'sequence', params: [sequence: 'seq_file'])
      parent:[lazy:"true",cascade:"none"]   
      children joinTable:[name:'children', key:'parent_Id', column:'Id',lazy:"true",inverse:"false",cascade:"none"]   
    }   
    static constraints = {   
        parent(nullable:true)   
    }   
}

现在我想获取父 id = 1 的所有文件我该怎么做？

我尝试使用

def fileList = File.findAllByParent(File.get(1L))

但它会发送 2 个 sql，第一个是获取我不想要的父文件信息。

是否有诸如 File.findAllByParentId(1L) 之类的方法

3x

谢谢。

parent{eq('key', 1)} 
//sql:from Info_File this_ left outer join Info_File parent_ali1_ 
//on this_.parent_id=parent_ali1_.id where (parent_ali1_.id=?)

但我不需要加入表。所以我试试

eq('parent.id',1L)
//it's what i need:
//from Info_File this_ where this_.parent_id=?
4

1 回答 1

4

I do not think that you can do it through the dynamic finders, but you should be able to use Hibernate to do a createCriteria 

File.createCriteria().list{
    parent{
        eq('key', 1)
    }
}

Perhaps this may help: https://docs.grails.org/4.0.1/ref/Domain%20Classes/createCriteria.html

于 2010-12-10T05:04:43.227 回答