HQL: заменить на элементы()

Question

У меня есть следующий HQL-запрос:

from Paperboy as paperboy where replace(replace(paperboy._mobile, ' ', ''), '/', '') = :nr or replace(replace(paperboy._phone, ' ', ''), '/', '') = :nr or :nr in elements(paperboy._additionalPhoneNumbers)

Теперь проблема в том, что я также должен заменить пробелы и slashs в _additionalPhoneNumbers. Возможно ли это через HQL?

Я пытался что-то вроде

from Paperboy as paperboy where replace(replace(paperboy._mobile, ' ', ''), '/', '') = :nr or replace(replace(paperboy._phone, ' ', ''), '/', '') = :nr or :nr in elements(replace(replace(paperboy._additionalPhoneNumbers, ' ', ''), '/', ''))

, но это бросает исключение. Отображение Paperboy как это:

<class name="Paperboy" 
    table="tblPaperboy" 
    lazy="false"> 
    <id name="_key" 
    column="Id" 
    access="field" > 
     <generator class="assigned"/> 
    </id> 
    <property name ="_businessId" column="BusinessId" access="field" /> 
    <many-to-one name="_agency" class="Agency" access="field" column="AgencyKey" /> 
    <property name ="_lastname" column="Lastname" access ="field" /> 
    <property name ="_firstname" column="Firstname" access ="field" /> 
    <property name ="_phone" column="Phone" access ="field" /> 
    <property name ="_mobile" column="Mobile" access ="field" /> 

    <bag name="_additionalPhoneNumbers" 
    access="field" 
    fetch="subselect" 
    lazy="false" 
    table="tblPaperboyAdditionalPhoneNumbers"> 
    <key column="PaperboyId"/> 
    <element column="PhoneNumber" 
      type="string"/> 
    </bag> 
</class> 

Answer 1

это может работать

select 
    paperboy 
from 
    Paperboy as paperboy left join paperboy._additionalPhoneNumbers number 
where 
    replace(replace(paperboy._mobile, ' ', ''), '/', '') = :nr or 
    replace(replace(paperboy._phone, ' ', ''), '/', '') = :nr or 
    replace(replace(number, ' ', ''), '/', '') = :nr