1. Management of XML and Semistructured Data
Lecture 15: Publishing XML Data From Relations
Wednesday, May 16th, 2001

2. Overview XML Publishing Example XML Publishing Languages
Virtual XML Publishing
Materialized XML Publishing (next time)

3. XML Publishing Today: Legacy data XML data
fragmented into many flat relations
3rd normal form
proprietary
XML data
nested
un-normalized
public (450 schemas at

4. XML Publishing: an Example
Legacy data in E/R:
name
country
name
url
euSid
usSid
Eu-Stores
US-Stores
date
tax
Eu-Sales
US-Sales
date
Products
pid
name
priceUSD

5. XML Publishing: an Example
XML view
<allsales>
<country> <name> France </name>
<store> <name> Nicolas </name>
<product> <name> Blanc de Blanc </name>
<sold> 10/10/2000 </sold>
<sold> 12/10/2000 </sold> …
</product>
<product>…</product>…
</store>….
</country> …
</allsales>
In summary: group by country store product

6. allsales Output “schema”: * country * name store ? * name product url
PCDATA *
name
product
url *
PCDATA
PCDATA
name
sold ?
PCDATA
date
tax
PCDATA
PCDATA

7. XML Publishing Need a language for specifying
the Relational  XML mapping
SilkRoute:
a SQL/XML-QL blend
IBM (formerly Experanto project)
extension of SQL
SQL Server:
“FOR XML” – and extension of SQL
XDR’s

8. XML Publishing: SilkRoute
In SilkRoute [Fernandez, Suciu, Tan ’00]
{ FROM EuStores $S, EuSales $L, Products $P
WHERE $S.euSid = $L.euSid AND $L.pid = $P.pid
CONSTRUCT
<allsales()>
<country($S.country)>
<name> $S.country </name>
<store($S.euSid)>
<name> $S.name </name>
<product($P.pid)>
<name> $P.name </name>
<price> $P.priceUSD </price>
</product>
</store>
</country>
<allsales>
} /* union….. */

9. XML Publishing : SilkRoute
…. /* union */
{ FROM USStores $S, EuSales $L, Products $P
WHERE $S.usSid = $L.euSid AND $L.pid = $P.pid
CONSTRUCT
<allsales()>
<country(“USA”)>
<name> USA </name>
<store($S.euSid)>
<name> $S.name </name>
<url> $S.url </url>
<product($P.pid)>
<name> $P.name </name>
<price> $P.priceUSD </price>
<tax> $L.tax </tax>
</product>
</store>
</country>
<allsales> }

10. Internal Representation
View Tree:
Non-recursive datalog
(SELECT DISTINCT … )
allsales()
allsales():- *
country(c) :-EuStores(x,_,c), EuSales(x,y,_), Products(y,_,_)
country(“USA”) :-
country(c) *
store(c,x) :- EuStores(x,_,c), EuSales(x,y,_), Products(y,_,_)
store(c,x) :- USStores(x,_,_), USSales(x,y,_), Products(y,_,_), c=“USA”
name(c)
store(c,x) c * ?
name(n)
product(c,x,y)
url(c,x,u)
url(c,x,u):-USStores(x,_,u), USSales(x,y,_),Products(y,_,_) n u *
name(n)
sold(c,x,y,d) n
date(c,x,y,d)
Tax(c,x,y,d,t) d t

11. XML Publishing : IBM
XPERANTO: Publishing Object-Relational Data as XML, Carey, Florescu, Ives, Lu, Shanmugasundaram, Shekita, Subramanian, WebDB’2000
Efficiently Publishing Relational Data as XML Documents, Shanmugasundaram, Shekita, Barr, Carey, Lindsay, Pirahesh, Reinwald, VLDB’2000

12. XML Publishing : IBM SQL + User defined functions
(Select S.name, STORE(S.euSid, S.name,
(Select XMLAGG(PRODUCT(P.pid, P.name, P.priceUSD))
From EuSales L, Products P
Where S.euSid = L.euSid AND L.pid = P.pid))
From EuStores S)
Union All
Define XML Constructor STORE(storeID: integer,
name: varchar(20),
prodList: xml) AS {
<store id=$storeID>
<name> $name </name>
$prodList
</store> }
SQL +
User defined functions
Define XML Constructor PRODUCT( ...) AS {
. . . }

13. XML Publishing : SQL Server
Three modes
RAW mode
Auto Mode
Explicit Mode

14. XML Publishing : SQL Server, RAW Mode
Select S.euSid, P.name, P.price
From Stores S, EuSales L, Products P
Where S.euSid = L.euSid AND L.pid = P.pid
For XML Raw
<row euSid = “SLKDJFS”, name = “Saint Emilion”, price=“23.99”>
<row euSid = “DRJLKSD”, name = “Loire”, price=“12.99”>
flat XML
default tag and attribute names

15. XML Publishing : SQL Server Auto Mode
Select S.euSid, P.name, P.price
From Stores S, EuSales L, Products P
Where S.euSid = L.euSid AND L.pid = P.pid
For XML Auto
<Stores euSid = “SLKDJFS”>
<Product name = “Saint Emilion”, price=“23.99”/>
<Product name = “Loire”, price=“12.99”/>
</Stores>
<Stores euSid = “FGJISOD”>
. . .
nested XML
default tag and attribute names

16. XML Publishing : SQL Server Explicit Mode
Nested XML
User defined tags and attributes
Idea: write SQL queries with complex column names
Ad-hoc, order dependent semantics

17. XML Publishing : SQL Server Explicit Mode
(Select as Tag,
null as Parent,
S.euSid as [Store!1!id],
S.name as [Store!1!name!element],
null as [Product!2!name!element],
null as [Product!2!price!element]
From Stores S)
Union All
(Select as Tag,
as Parent,
S.euSid as [Store!1!id],
null as [Store!1!name!element],
P.name as [Product!2!name!element],
P.price as [Product!2!name!element]
From Stores S, EuSales L, Products P
Where S.euSid = L.euSid AND L.pid = P.pid)
Order By [Store!1!id]

18. XML Publishing : SQL Server Explicit Mode
All column names are legal SQL names
Special form: [tagname!k!something],
Or [tagname!k!something!element]
Or other variations...
Hence everything is legal SQL
But what does it mean ?
Construct the universal table first
Then process the table sequentially

19. XML Publishing : SQL Server Explicit Mode
Universal table:
Tag
Parent
Store!1!id
Store!1!name!element
Product!2!name!element
Product!2!price!element 1
ABCDE
Nicolas 2
Saint Emilion
23.99
Loire
12.99
FDKLS
FNAC
Databases
49.99
. . .

20. XML Publishing : SQL Server Explicit Mode
Converting universal table to XML:
scan sequentially each two
let Tag=k
look up only columns with that tag
all are called [tagname!k!something] with the same tagname
What happens if one has a different tagname ?
Create an element called tagname
Output <tagname>
Columns become its children
Either subelements or attributes
if Parent is specified, last element with that tag is the parent
otherwise it is a root element

21. XML Publishing : SQL Server Explicit Mode
<Store id=“ABCDE”>
<name> Nicolas </name>
<Product> <name> Saint Emilion </name>
<price> </price>
</Product>
<Product> <name> Loire </name>
<price> </price>
</Price>
</Store>
<Store id=“FDKLS”>
<name> FNAC </name>
. . .

22. XML Publishing : SQL Server Explicit Mode
Seems complex, but also powerful
Can construct arbitrarily deeply nested hierarchies
How ?
However, they are very, very limited
Why ?

23. XML Publishing: Virtual View
Don’t compute the XML data yet
Users ask XML queries
System composes with the view, sends to the RDBMS
Main issue: compose queries

24. XML Publishing: Virtual View in SilkRoute
find names, urls of all stores who sold on 1/1/2000 (in XML-QL / XQuery melange):
WHERE <allsales/country/store>
<product/sold/date> 1/1/2000 </>
<name> $X </>
<url> $Y </>
</>
RETURN $X , $Y

25. Query Composition View Tree XML-QL Query Pattern
date(c,x,y,d)
allsales()
country(c)
store(c,x)
name(n)
product(c,x,y)
sold(c,x,y,d)
url(c,x,u)
allsales
$n1
country
$n2
name(c)
$n3
store c
product
$n4
url
name n u $Y
name(n) $X
sold n
$n5
Tax(c,x,y,d,t)
date $Z d t
1/1/2000
“Evaluate” the XML pattern(s) on the view tree, combine all datalog rules

26. Query Composition Result (in theory…): ( SELECT S.name, S.url
FROM USStores S, USSales L, Products P
WHERE S.usSid=L.usSid AND L.pid=P.pid AND L.date=‘1/1/2000’)
UNION
( SELECT S2.name, S2.url
FROM EUStores S1, EUSales L1, Products P1
USStores S2, USSales L2, Products P2,
WHERE S1.usSid=L1.usSid AND L1.pid=P1.pid AND L1.date=‘1/1/2000’
AND S2.usSid=L2.usSid AND L2.pid=P1.pid
AND S1.country=“USA” AND S1.euSid = S2.usSid)

27. Complexity of XML Publishing
But in practice: 5-7 times more joins !
Need query minimization
Could this be avoided ?
We thought hard and couldn’t find a better way
It is NP-hard !

28. XML Publishing Is NP-Hard
View Tree:
customer ? ?
order():- Q1
order
complaint
complaint():- Q2
PCDATA
PCDATA
XML query:
WHERE <customer> <order> $x </> <complaint> $y </> </> RETURN ( )
Reduction from the clique problem.
Q1 describes a graph with named nodes: E.g. V1(x1), V2(x2), V3(x3), V4(x4), E(x1,x2), E(x4,x2),…
Q2 describes a k-clique: e.g. E(y1,y2), E(y1,y3), …, E(yk-1,yk)
If the graph has a k-clique, then (Q1 join Q2) = Q1
Conversely, if the minimized query of Q1 join Q2 is a graph isomorphic to the original graph, then the graph has a k-clique. It is easy to check isomorphism, since all nodes are named (i.e. there is at most one homeomorphism, and that can be found in polynomial time).
The composed SQL query is : Minimizing it is NP hard ! (can be shown…)
Q1 JOIN Q2