Given the AX 2009 statement:

ledgerJournalTrans.Dimension    = ledgerTable.Dimension;
ledgerJournalTrans.Dimension[1] = "abc";

What is the equivalent way to that in AX 2012?

This will of cause assume the existence of a “Department” dimension as the first dimension.

My case was the “default dimensions” scenario, so I had to adapt a little. This is the method I eventually used:

On the DimensionAttribute table add a new field name Number then add this method:

public static DimensionAttribute findByNumber(DimensionOrdinal _number)
{
DimensionAttribute dimensionAttribute;
select firstonly dimensionAttribute where dimensionAttribute.Number == _number;
return dimensionAttribute;
}

This explicitly identifies the dimension with a corresponding number.

On the DimensionAttributeValueSetStorage class add the method:

public void addItemNumber(DimensionOrdinal _idx, SysDim _value)
{
DimensionAttributeValue attrValue;
DimensionAttribute attr = DimensionAttribute::findByNumber(_idx);
if (!attr)
throw error(strFmt("@SYS342559", _idx));
attrValue = DimensionAttributeValue::findByDimensionAttributeAndValue(attr, _value, false, true);
this.addItemValues(attr.RecId, attrValue.RecId, attrValue.HashKey);
}

The DimensionAttributeValueSetStorage handles “default dimensions” as described in the white paper @dlannoye mentioned.

Then the corresponding code read like this:

dimensionStorage = DimensionAttributeValueSetStorage::find(salesTable.DefaultDimension);
dimensionStorage.addItemNumber(1, "abc");
salesTable.DefaultDimension = dimensionStorage.save();