How define today a date of entry to create with params

Hi Experts,

JDev 12 c,

I use CreateWithParams, and I wanted to put a Date as today's date field. So how can I do?

I tried the HrDate field mapping and set value as adf.currentDate and its type like oracle.jbo.domain.TimeStamp.

When I run the page I get the below error.

java.lang.IllegalArgumentException

at oracle.jbo.domain.Date.parseISO8601FormatString(Date.java:628)

at oracle.jbo.domain.Date.toDate(Date.java:609)

to oracle.jbo.domain.Date. < init > (Date.java:390)

to oracle.jbo.domain.Date$ 1.convert(Date.java:116)

at oracle.jbo.domain.TypeFactory.get(TypeFactory.java:870)

at oracle.jbo.domain.TypeFactory.get(TypeFactory.java:574)

at oracle.jbo.domain.TypeFactory.getInstance(TypeFactory.java:351)

at oracle.adf.model.binding.DCMethodParameterDef.resolveParameterValue(DCMethodParameterDef.java:225)

Thank you

Animesh

This marker is used as reolved.

Here is the solution

public final void insert (ActionEvent actionEvent) {}
    Add the code in the event here...
  
    OperationBinding ob = (OperationBinding) ADFUtils.findOperation ("Createwithparameters");
               ob.getParamsMap () .put ("Hrdate", new oracle.jbo.domain.Timestamp (System.currentTimeMillis ()));
               OB. Execute();

}

The TIMESTAMP has not been properly defined in OT. So now it is resolved.

<>

Name = "Hrdate".

ColumnName = "HRDATE."

SQLType = 'TIMESTAMP '.

Type = "Oracle.jbo.domain.Timestamp"

ColumnType = "TIMESTAMP".

TableName = "EMPLOYEENAME" >

NDName: Hrdate

NDType:oracle.jbo.domain.Timestamp

Tags: Java

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