auto generate a property field

cyber · **Joined:** Mon Jan 23, 2006 8:26 am **Posts:** 6

how can i make a mapping property (lpos) to become incremental by not using id because i already have an id that is assigned but i need another property to be incremental. making lpos an id will throw an duplicate id constraint error. is there a way to put a generator for a property mapping?

this existing can save my data up to the 2nd level. but i cannot load it properly because lpos is always zero (0), but when i edit the records directly can modify the lpos 0..N respectively, i can load the record properly. please help. thx...

Hibernate version:
3.1

Mapping documents:

Code:

<class name="billing.collections.GeneralCollections" table="gc">
   <id name="id"><generator class="assigned"/></id>   
   <property name="payor" type="string"/>
   <property name="tenantId" type="string"/>
   <property name="leaseId" type="string"/>
   <property name="trade" type="string"/>
   <property name="business" type="string"/>
   <property name="lname" type="string"/>
   <property name="fname" type="string"/>
   <property name="mi" type="string"/>
   <property name="date" type="string"/>
   <property name="invoice" type="string"/>
   <property name="particulars" type="string"/>
   <property name="periodStart" type="string"/>
   <property name="periodEnd" type="string"/>   
   <list name="collections" table="gc_col" lazy="true"
      inverse="true" cascade="all">      
      <key column="gcId"/>
      <index column="lpos" />
      <one-to-many class="billing.collections.GCCollection" />
   </list>
   <list name="orVat" table="ors" lazy="true"
      inverse="true" cascade="all" where="type=1">
      <key column="ref"/>
      <index column="lpos" />
      <one-to-many class="com.payment.OR" />
   </list>
   <list name="orNonVat" table="ors" lazy="true"
      inverse="true" cascade="all" where="type=0">
      <key column="ref"/>
      <index column="lpos" />
      <one-to-many class="com.payment.OR" />
   </list>
</class>

<class name="billing.collections.GCCollection" table="gc_col">
   <id name="id"><generator class="assigned"/></id>
   <property name="lpos"/>
   <property name="gcId" type="string"/>
   <property name="defId" type="string"/>     
   <property name="desc" column="description" type="string"/>   
   <property name="payment" type="string"/>
   <property name="type" type="string"/>
   <property name="vat"/>
   <property name="entryAmount" type="string"/>
   <property name="cashAmount" type="string"/>
   <property name="checkAmount" type="string"/>
</class>

<class name="com.payment.OR" table="ors">
   <id name="id" type="string"><generator class="assigned"/></id>
   <property name="ref" type="string"/>
   <property name="type" type="string"/>
   <property name="orNumber" type="string"/>
   <property name="status" type="string"/>      
   <list name="entries" table="or_entry" lazy="true"
      inverse="true" cascade="all">
      <key column="orId"/>
      <index column="lpos" />
      <one-to-many class="com.payment.OREntry" />
   </list>
</class>

<class name="com.payment.OREntry" table="or_entry">
   <id name="id" type="string"><generator class="assigned"/></id>      
   <property name="orId" type="string"/>
   <property name="ref" type="string"/>
   <property name="amount" type="string"/>
   <!--<property name="isDeduction"/>-->
   <property name="endPeriod" type="string"/>
   <property name="itemId" type="string"/>
</class>

Code between sessionFactory.openSession() and session.close():

Code:

n/a

Full stack trace of any exception that occurs:

Code:

n/a

Name and version of the database you are using:

Code:

Mysql 5.0.1

The generated SQL (show_sql=true):

Debug level Hibernate log excerpt:

sarahsmith_books · **Joined:** Thu Oct 20, 2005 1:03 pm **Posts:** 38

i believe that you are misusing the id tag. the id tag can be any field in the database - if it's lpos, then the column value should be lpos for the id tag and the attribute generated - like this and the id property is simply a property:

<class name="billing.collections.GeneralCollections" table="gc">
<id name="lpos"><generator class="generated"/></id>
<property name="id" column="ID"/>

. . .

steve · **Posted:** Wed Jan 25, 2006 3:17 pm

the same column cannot be used for multiple list-index values...

cyber · **Joined:** Mon Jan 23, 2006 8:26 am **Posts:** 6

i need the column id to be the id because the one-to-many maps to the defined id, right? the id is what i link to the other table, lpos however, must be a integer from 0...N,

example:
GeneralCollection has 5 GCCollection object, lpos of the records should 0,1,2,3,4. i did a test and directly edit the lpos and put 10, there will be 6 extra objects that is created.

just a question, what is the right way to do list mapping?

any help or example will be greatly appreciated. thx...

tenwit · **Joined:** Thu Dec 23, 2004 9:08 pm **Posts:** 2008

For indexed lists like you have, the index is accessible from the containing class. You've already got this: the value of the lpos column is accessible as

Code:

genCol.getCollections().indexOf(collection);

To make this lpos column available (read-only) from inside the Collection list, the easiest thing to do is to map lpos as a formula:

Code:

This will allow hibernate to maintain lpos (as the index of the list) and also allows the user to find out a list element's index without having the list.

cyber · **Joined:** Mon Jan 23, 2006 8:26 am **Posts:** 6

thanks for the help... its working now

tenwit · **Joined:** Thu Dec 23, 2004 9:08 pm **Posts:** 2008

Excellent. Who gets the credit?

GDW · **Joined:** Tue Jun 20, 2006 11:51 am **Posts:** 14

tenwit wrote:

For indexed lists like you have, the index is accessible from the containing class. You've already got this: the value of the lpos column is accessible as

Code:

genCol.getCollections().indexOf(collection);

To make this lpos column available (read-only) from inside the Collection list, the easiest thing to do is to map lpos as a formula:

Code:

This will allow hibernate to maintain lpos (as the index of the list) and also allows the user to find out a list element's index without having the list.

Tenwit, could you elaborate on this strategy? (possibly with a simple example including mappings & code)

I'm not sure where the following goes or what the objects are in relation to cyber's example:

Code:

genCol.getCollections().indexOf(collection);

I'm also not clear as to whether the parent should be marked inverse or whether there should be some other type of mapping.

Thanks,
Gary

tenwit · **Joined:** Thu Dec 23, 2004 9:08 pm **Posts:** 2008

Code:

<class name="ContainingClass" ...>
  ...
  <list name="Things" ...>
    <key column="keyCol"/>
    <list-index column="indexCol"/>
    <one-to-many class="Thing"/>
  ...
</class>
<class name="Thing" ...>
  ...
  <property name="MyIndexInContainingClassList" formula="indexCol"/>
  ...
</class>

Your java code could now safely include code like

Code:

for (int i = 0; i < containingObject.getThings().size(); ++i)
{
  Thing t = containingObject.getThings().get(i);
  if (t.getMyIndexInContainingClassList() != i)
  {
    App.getDeveloper().setSalary(0);
  }
}

And you'd never have to worry about working for free :)

GDW · **Joined:** Tue Jun 20, 2006 11:51 am **Posts:** 14

Thanks for the quick reply. I'm still not getting it to work (list-index column is null). Can you take a look at the following and see what I'm doing wrong?

Thx,
Gary

parent mapping

Code:

<hibernate-mapping>
   <class name="mil.jfcom.model.Survey" table="SURVEY">
      <id name="id" column="id" unsaved-value="null">
         <generator class="increment" />
      </id>
      <property name="title" length="50" />
      <property name="description" length="150" />
      <list name="pages" table="PAGE" lazy="true" inverse="true" cascade="all-delete-orphan">
         <key column="SURVEY_ID" />
         <list-index column="PAGE_NUM" base="1" />
         <one-to-many class="mil.jfcom.model.Page" />
      </list>
   </class>
</hibernate-mapping>

child mapping

Code:

<hibernate-mapping>
   <class name="mil.jfcom.model.Page" table="PAGE">
      <id name="id" column="id" unsaved-value="null">
         <generator class="increment" />
      </id>
      <property name="pageNumber" formula="PAGE_NUM" />
      <property name="title" length="50" />
      <many-to-one name="survey" column="SURVEY_ID" not-null="true" />
   </class>
</hibernate-mapping>

parent code

Code:

public class Survey {
   
   private Long id;
   private String title;
   private String description;
   private List pages = new ArrayList();
   
   public Survey() {
   }
   
   public Long getId() {
      return id;
   }
   public void setId(Long id) {
      this.id = id;
   }
   
   public String getTitle() {
      return title;
   }
   public void setTitle(String title) {
      this.title = title;
   }
   
   public String getDescription() {
      return description;
   }
   public void setDescription(String description) {
      this.description = description;
   }

   public List getPages() {
      return pages;
   }
   
   public void setPages(List pages) {
      this.pages = pages;
   }
   
   public void addPage(Page page) {
      page.setSurvey(this);
      pages.add(page);
   }
}

child code

Code:

public class Page {
   
   private Long id;
   private Integer pageNumber;
   private String title;
   private Survey survey;
   
   public Page() {
   }
   
   public Page(String title) {
      this.title = title;
   }
   
   public Long getId() {
      return id;
   }

   public void setId(Long id) {
      this.id = id;
   }

   public Integer getPageNumber() {
      return pageNumber;
   }
   public void setPageNumber(Integer pageNumber) {
      this.pageNumber = pageNumber;
   }
   
   public String getTitle() {
      return title;
   }
   public void setTitle(String title) {
      this.title = title;
   }

   public Survey getSurvey() {
      return survey;
   }
   public void setSurvey(Survey survey) {
      this.survey = survey;
   }
}

tenwit · **Joined:** Thu Dec 23, 2004 9:08 pm **Posts:** 2008

You have inverse="true". You have to choose between one of these two scenarios:

Pages exist only within lists belonging to Surveys, in which case you must remove inverse="true" and any code that saves Pages outside of a survey
Pages may exist outside of surveys, in which case you must allow PAGE_NUM to be null: pages with null page_num are not in a survey.

GDW · **Joined:** Tue Jun 20, 2006 11:51 am **Posts:** 14

That did it. Your explanation made the underlying reason seem so obviously simple.

I tried unsuccessfully to give your answer credit. I guess you can't do that if you didn't start the thread.

Thanks again,
Gary