HQL：replace on elements（）

本文介绍了HQL：replace on elements（）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有以下HQL查询：

来自Paperboy作为paperboy 其中 replace ._mobile，''，''），'/'，''）=：nr或replace（replace（paperboy._phone，''，''），'/'，''）=：nr或：元素中的nr（paperboy._additionalPhoneNumbers）

现在的问题是，我还必须替换空格和斜杠在_additionalPhoneNumbers。这是否可以通过HQL？

我试过像Paperboy的

其中替换（替换（paperboy._mobile，''，''），'/'，''）=：nr或替换（paperboy._phone，''，' '/'，''））： c $ c>

 
 
 但这会抛出异常。纸工的映射如下：
 < class name =Paperboy
 table =tblPaperboy
 lazy =false> 
< id name =_ key
 column =Id
 access =field> 
< generator class =assigned/> 
< / id> 
< property name =_ businessIdcolumn =BusinessIdaccess =field/> 
< many-to-one name =_ agencyclass =Agencyaccess =fieldcolumn =AgencyKey/> 
< property name =_ lastnamecolumn =Lastnameaccess =field/> 
< property name =_ firstnamecolumn =Firstnameaccess =field/> 
< property name =_ phonecolumn =Phoneaccess =field/> 
< property name =_ mobilecolumn =Mobileaccess =field/> 
 
< bag name =_ additionalPhoneNumbers
 access =field
 fetch =subselect
 lazy =false
 table = tblPaperboyAdditionalPhoneNumbers> 
< key column =PaperboyId/> 
< element column =PhoneNumber
 type =string/> 
< / bag> 
< / class> 
  
 
 
解决方案
这可能有用
  select 
 paperboy 
从
纸工作为paperboy left join paperboy._additionalPhoneNumbers数字
其中
 （replace（paperboy._phone，''，''），'/'，''）=：nr或
 replace ）=：nr或
 replace（replace（number，''，''），'/'，''）=：nr 
  I have the following HQL-Query:

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


Now the problem is, that I also have to replace spaces and slashs in the _additionalPhoneNumbers. Is this possible via HQL?

I tried things like 

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


but this throws an exception. Mapping of Paperboy is like this:
<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>
 解决方案 
this might work
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
                        
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