How to insert a row in UDO of type document with document line using DIAPI

Does somebody knows if there is a way to insert a new row in a UDO of type document and document lines using the DI API?
I tried it by using <i>oCompany.UserTables.Item( "PMX_DOC" )</i> but this was not working.
Thanks for the help.
Regards,
Jeffrey

We have to insert the row in a server process so it is not possible to use the UIAPI.
I wanted to do it by writing SQL statements to do the insert/update/deletes but it seems that we are not allowed to do this if you want to certify your addon. Does somebody knows how we can get an addon certified but also updating our own UDO using DIAPI?
Thanks.
Regards,
Jeffrey

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     at weblogic.security.service.SecurityManager.runAs(SecurityManager.java:120)
     at weblogic.servlet.internal.WebAppServletContext.securedExecute(WebAppServletContext.java:2277)
     at weblogic.servlet.internal.WebAppServletContext.execute(WebAppServletContext.java:2183)
     at weblogic.servlet.internal.ServletRequestImpl.run(ServletRequestImpl.java:1454)
     at weblogic.work.ExecuteThread.execute(ExecuteThread.java:209)
     at weblogic.work.ExecuteThread.run(ExecuteThread.java:178)
while I get this from my own project based on the above one:
<UIXEditableValue> <_isBeanValidationAvailable> A Bean Validation provider is not present, therefore bean validation is disabled
<UIXRegion> <_logIllegalContextChangeMessage> ADF_FACES-10026:Durante l'elaborazione del componente area, non è stata trovata una modifica di contesto oppure la modifica trovata non corrisponde all'istanza impostata dal componente corrente. Prevista oracle.adf.view.rich.component.fragment.UIXRegion$RegionContextChange, trovata UIXCollection.CollectionComponentChange[Component class: oracle.adf.view.rich.component.rich.data.RichTreeTable, component ID: tt1].
<UIXRegion$RegionSiteImpl> <validate> Attempt to validate an already invalid RegionSite:
<UIXCollection> <processSaveState> è possibile che la chiave di riga non venga reimpostata correttamente alla fine della richiesta. ID componente: :pt1:r1:pt1:pc1:tt1, ViewId: /main.jspx
Edited by: user10047839 on 20-ott-2011 7.23

How to insert a row in database trough AappModule

HI,
I have used below code for getting details from databse. Now i have to insert a row in the database.
Kindly let me know how can i use below code for inserting a row in the database. or should i make any change in its viewObject's Query.
If there is any another way of inserting data in database through method binding please let me knoe that.
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ViewObject userInfoVO = myService.findViewObject("EmployeeViewParam1");
userInfoVO.executeQuery();
DCBindingContainer bindings = (DCBindingContainer)BindingContext.getCurrent().getCurrentBindingsEntry();
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Row[] rows1 = rowIter1.getAllRowsInRange();
for(int i=0;i<rows1.length;i++)
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HI,
i have tried belod code. for this i have made AppModuleImpl Class and wrote below method in that and called this method from my bean class with help of AppModuleImpl's Object.
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ViewObjectImpl employee=getEmployeeView1();
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row.setAttribute("Employeeid",189420);
row.setAttribute("Employeename","Test");
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How to insert Existing rows by modifying some cloumns values??

Respected experts ,
I have 72 rows in a table .I want to insert another 72 rows with one column value change and another columns value same??
Please suggest me how I do this ????
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Do you want to insert 72 rows from the same table but just changing value of the one column? If so, take a look at Scott's solution
Best Regards,Uri Dimant SQL Server MVP,
http://sqlblog.com/blogs/uri_dimant/
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hi,
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Regards Thomas

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Hi,
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If you would like to do it with SQL, this would be one of the ways to achive it:
SQL*Plus: Release 10.2.0.4.0 - Production on Fri Sep 25 10:12:59 2009
Copyright (c) 1982, 2007, Oracle. All Rights Reserved.
Connected to:
Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - Production
With the Partitioning, Data Mining and Real Application Testing options
SQL>
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Name                                      Null?    Type
FIRST_PRICE                                        NUMBER(16,6)
NEXT_PRICE                                         NUMBER(16,6)
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SQL> select * from leap;
no rows selected
SQL>
SQL>
SQL> !vi multirow_insert.sql
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insert into leap(first_price, next_price) values (1,2);
insert into leap(first_price, next_price) values (3,4);
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SQL>
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1 row created.
1 row created.
1 row created.
SQL> commit;
Commit complete.
SQL>
SQL>
SQL> select * from leap;
FIRST_PRICE NEXT_PRICE
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          3          4
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SQL>

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Please Oracle EBS queries on EBS forums.

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