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5. Drag an attribute from the Toolbox, and drop it onto the Friends element. Change its name to PlaceID, and check that string is selected in the right-hand column (either type it or select it from the combo box).

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One of the interesting enhancements to bitmap indexes in 9i was the addition of the bitmap join index an index that could hold entries for the rows in one table, but had key values that were taken from another table (or tables). For example (see script bitmap_cost_06.sql in the online code suite): create bitmap index fct_dim_name on fact_table(dim.dim_name) from dim_table dim, fact_table fct where dim.id = fct.dim_id ; create bitmap index fct_dim_par on fact_table(dim.par_name) from dim_table dim, fact_table fct where dim.id = fct.dim_id ; These index definitions demonstrate two potential benefits: The first example gives us an index on a very large fact table that uses a long dimension name which did not, however, have to be stored millions of times in the fact table. The second example shows us an index that can access the fact table based on a query against an attribute of a dimension table that (as the column name suggests) may have far fewer distinct values than the dimension ID, and therefore may be a much smaller, more desirable, index. (We assume that this is also an attribute that the users frequently use to identify and summarize the data.) Personally, I am not convinced that the bitmap join index adds a lot of value once you have managed to use simple bitmap indexes well but I can imagine there would be a few cases where the technology can be used effectively. Having got through the details of why and how you create a bitmap join index, though, the arithmetic involved is unchanged. (Although the optimizer is still allowed to consider the option for doing a join, rather than using the index.) Consider this query: select count(fct.id) from dim_table fact_table where and ; dim.par_name = 'Parent_001' fct.dim_id = dim.id dim, fct

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How It Works The element you have added corresponds to the root <Friends> element of the XML format you are going to use for uploads: < xml version="1.0" encoding="utf-8" > <Friends xmlns="http://www.apress.com/schemas/friendsreunion" PlaceID="C9796AD1-5A7E-4d9c-9F99-0090E11E5662"> ...etc... </Friends> Your schema now specifies that the root element must have the name Friends. It also says that the Friends element can have a string attribute called PlaceID. (The attribute is not compulsory, however, because attributes are optional by default; see the Restricting Element Occurrence section later in this chapter for more information.) If you switch to the XML view, you can check out the schema markup that VS .NET has generated, based on what you added in the designer: < xml version="1.0" encoding="utf-8" > <xs:schema id="Friends" targetNamespace="http://www.apress.com/schemas/friendsreunion" elementFormDefault="qualified" xmlns="http://www.apress.com/schemas/friendsreunion" xmlns:mstns="http://www.apress.com/schemas/friendsreunion" xmlns:xs="http://www.w3.org/2001/XMLSchema"> <xs:element name="Friends"> <xs:complexType> <xs:sequence /> <xs:attribute name="PlaceID" type="xs:string" /> </xs:complexType> </xs:element> </xs:schema> Notice that all of the new elements use the xs prefix. The xs prefix is associated with the XML Schema namespace: <xs:schema id="Friends" ... xmlns:xs="http://www.w3.org/2001/XMLSchema"> This marks these elements as XML Schema elements, and they must therefore obey the rules of validity for XML Schemas. This means that elements must follow the required order and have valid values and attributes, such as the type attribute on the <xs:attribute> element. According to the validity rules for an XSD, an <xs:element> that is a direct child of the <xs:schema> describes the root element of any document that conforms to this schema. The name attribute of the <xs:element> element here defines the name that the root element must have, so it is <Friends>.

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Let s take a step backward from the complexity of the previous example, and focus instead on just the requirement to list all the employees from a specific group of departments. I d like to write this (see script semi_01.sql in the online code suite) as follows: Select from where emp.* emp emp.dept_no in select from where )

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