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Greg Pike Piocon Technologies

Aces in the Hole: Learning Advanced SQL Techniques from the OTN Forum Pros. Greg Pike Piocon Technologies. Greg Pike Managing Principal Piocon Technologies. History: 15+ year Oracle solution provider Home: ChicagoLand

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Greg Pike Piocon Technologies

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  1. Aces in the Hole: Learning Advanced SQL Techniques from the OTN Forum Pros Greg PikePiocon Technologies

  2. Greg PikeManaging PrincipalPiocon Technologies • History: 15+ year Oracle solution provider • Home: ChicagoLand • Focus: Business Intelligence, Data Warehousing, Portal, web applications, devious SQL • Blog: www.SingleQuery.com • Client list includes

  3. The Oracle Discussion Forums An interactive community for sharing information, questions, and comments about Oracle products and related technologies Most forums are moderated by product managers, and all of them are frequently visited by knowledgeable users from the community Anyone can read messages but must be a registered member to post a question or response Posts to the SQL and PL/SQL Forum are often answered in minutes Source: http://www.oracle.com/technology/forums/faq.html

  4. Who Can Learn from the OTN SQL and PL/SQL Forum? Beginners: Got a question…its probably already been answered 100 times. Please search for a solution first, but don’t be shy to ask IntermediateLearners: The SQL and PL/SQL Forum has a wealth of knowledge on every imaginable topic Experts: You may not feel like an expert after a visit to this Forum UniqueProblem: This is your real-time resource for solutions Contributor: Get involved and share your expertise

  5. Who Contributes? Introducing the Experts and Recent Forum Post Counts: • Justin Cave – 18,700+ • Billy Verreynne – 5,500+ • APC – 9,500+ • BluShadow – 6,200+ • William Robertson – 4,500+ • Volder – 950+ • Yingkuan – 7000+ • Kamal Kishore – 7,300+ …and 100’s of people worldwide with a passion for problem solving Warren Tolentino – 4,400+ Devmiral – 2,200+ Nicolas Gasparotto – 14,900+ Marias – 1,300+ The Flying Spontinalli – 700+ Rob van Wijk – 3,900+ Michaels – 3,400+ John Spencer – 3,700+

  6. Q:”If the advice is free, how good can it be?”A: “The Program. Enough said.” • The Oracle ACE program formally recognizes advocates of Oracle technology with strong credentials as evangelists and educators • Oracle ACE recipients are chosen based on their significant contributions and activity in the Oracle technical community with candidates nominated by anyone in the Oracle Technology and Applications communities. “ Source: http://www.oracle.com/technology/community/oracle_ace/index.html

  7. The Oracle Aces “Oracle ACEs are known for their strong credentials as Oracle community enthusiasts and advocates with candidates nominated by anyone in the Oracle Technology and Applications communities.” • Technical proficiency • Oracle-related blog • Oracle discussion forum activity • Published white paper(s) and/or article(s) • Presentation experience • Beta program participant • Oracle user group member • Oracle certification

  8. Just a Few of the 188 Oracle Aces Cary Milsap (speaking today) Peter Koletzke (speaking today) Dan Norris (Piocon) Paul Dorsey Steven Feuerstein Ken Jacobs Tom Kyte Mark Rittman Laurent Schneider

  9. Case Study: Examining a Recent Post • Warren Tolentino • Devang Bhatt (devmiral) • Nicolas Gasparotto • marias • The Flying Spontinalli • Rob van Wijk • michaels The following post is from May 28, 2007 and elicited 33 replies from long-time Forum contributors including Oracle Aces http://forums.oracle.com/forums/thread.jspa?messageID=1864354 Only a portion of the solutions are presented here and the supporting commentary is omitted. Some queries have been altered to fit on the page but retain their basic logic Thanks to the following OTN experts who contributed to this thread (OTN handle shown):

  10. The Original Question: ”I need to get the count on how many customers answered for each set of questions. For the above data, this should be the output:” Questions IDsCustomer Count 1 1 1,2,4 2 3,4 2 2,3 1 2 1 “The Question IDs can have as many as 10 questions.” “Thanks in advance.” “I need your help to generate a report. I'm using Oracle 9i database. This is our data:” Customer_IDQuestion_ID 10001 1 10002 1 10002 2 10002 4 10003 3 10003 4 10004 2 10004 3 10005 1 10005 2 10005 4 10006 3 10006 4 10007 2

  11. Examining a Few of the Posted Solutions • Method 1: Hierarchical Query • Method 2: COLLECT • Method 3: XML • Method 4: MODEL • Method 5: Pipelined Funtions

  12. Method 1: Hierarchical Query (Warren) For querying datasets that contain a parent-child relationship between rows Recursively walks through the tree of relationships from the top-down or bottom-up Includes tools for determining location in the hierarchy and the nature of individual nodes

  13. Method 1: Hierarchical Query SELECT customer_id, question_id, ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY customer_id, question_id) rn, COUNT(*) OVER (PARTITION BY customer_id,question_id) cnt FROM customer_inquiry ) ci1 DTSRT WITH ci1.rn = 1 CONNECT BY ci1.rn = PRIOR ci1.rn + 1 AND PRIOR ci1.customer_id = ci1.customer_id GROUP BY ci1.customer_id, ci1.cnt ) ci2 GROUP BY questions; SELECT ci2.questions, COUNT(ci2.cnt) “Customer Count" FROM (SELECT ci1.customer_id, SUBSTR(MAX(SUBSTR(SYS_CONNECT_BY_PATH (ci1.question_id,','),2)),1,40) questions, ci1.cnt FROM (

  14. Step 1: The pseudo-parent-child relationship Customer_IDQuestion_ID 10001 1 10002 1 10002 2 10002 4 10003 3 10003 4 10004 2 10004 3 10005 1 10005 2 10005 4 10006 3 10006 4 10007 2 Customer_IDQuestion_IDRNCNT 10001 1 1 1 10002 1 1 1 10002 2 2 1 10002 4 3 1 10003 3 1 1 10003 4 2 1 10004 2 1 1 10004 3 2 1 10005 1 1 1 10005 2 2 1 10005 4 3 1 10006 3 1 1 10006 4 2 1 10007 2 1 1 SELECT customer_id, question_id, ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY customer_id, question_id) rn, COUNT(*) OVER (PARTITION BY customer_id,question_id) cnt FROM customer_inquiry

  15. Step 2: Employ SYS_CONNECT_BY_PATH SELECT customer_id, question_id, ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY customer_id, question_id) rn, COUNT(*) OVER (PARTITION BY customer_id,question_id) cnt FROM customer_inquiry ) ci1 START WITH ci1.rn = 1 CONNECT BY ci1.rn = PRIOR ci1.rn + 1 AND PRIOR ci1.customer_id = ci1.customer_id GROUP BY ci1.customer_id, ci1.cnt ) ci2 GROUP BY questions; SYS_CONNECT_BY_PATH returns the path of a column value from root to node, with column values separated by char for each row returned by CONNECT BY condition. SELECT ci2.questions, COUNT(ci2.cnt) “Customer Count" FROM (SELECT ci1.customer_id, SUBSTR(MAX(SUBSTR(SYS_CONNECT_BY_PATH (ci1.question_id,','),2)),1,40) questions, ci1.cnt FROM (

  16. Step 2: Employ SYS_CONNECT_BY_PATH Questions IDsCustomer Count 1 1 1,2,4 2 3,4 2 2,3 1 2 1 SYS_CONNECT_BY_PATH returns the path of a column value from root to node, with column values separated by char for each row returned by CONNECT BY condition. SELECT ci2.questions, COUNT(ci2.cnt) “Customer Count" FROM (SELECT ci1.customer_id, SUBSTR(MAX(SUBSTR(SYS_CONNECT_BY_PATH (ci1.question_id,','),2)),1,40) questions, ci1.cnt FROM (

  17. Method 2: COLLECT (Greg Pike) • COLLECT takes as its argument a column of any type and creates a nested table of the input type out of the rows selected • GROUP BY can determine how the rows are COLLECTED • The database creates its own collection object to hold the data • For the required output, CAST the results into a more usable form • Get the data out of the collection with a function

  18. Method 2: The Query COLLECT(question_id) SELECT SUBSTR(col,1,10) questions, COUNT(*) FROM ( SELECT customer_id, tab_to_string( CAST( AS t_number_tab ) ) col FROM answers GROUP BY customer_id ) GROUP BY col;

  19. Step 1: COLLECT Customer_IDQuestion_ID 10001 1 10002 1 10002 2 10002 4 10003 3 10003 4 10004 2 10004 3 10005 1 10005 2 10005 4 10006 3 10006 4 10007 2 CUST_ID COLLECT(QUESTION_ID) ------- --------------------------------- 10001 SYSTP3f5GcsxkSvyjQSG5pDbjhA==(1) 10002 SYSTP3f5GcsxkSvyjQSG5pDbjhA==(1, 2, 4) 10003 SYSTP3f5GcsxkSvyjQSG5pDbjhA==(3, 4) 10004 SYSTP3f5GcsxkSvyjQSG5pDbjhA==(2, 3) 10005 SYSTP3f5GcsxkSvyjQSG5pDbjhA==(1, 2, 4) 10006 SYSTP3f5GcsxkSvyjQSG5pDbjhA==(3, 4) 10007 SYSTP3f5GcsxkSvyjQSG5pDbjhA==(2) COLLECT takes as its argument a column of any type and creates a nested table of the input type out of the rows selected. SELECT customer_id cust_id, COLLECT(question_id) FROM customer_inquiry GROUP BY customer_id;

  20. Step 2: CAST(COLLECT) Customer_IDQuestion_ID 10001 1 10002 1 10002 2 10002 4 10003 3 10003 4 10004 2 10004 3 10005 1 10005 2 10005 4 10006 3 10006 4 10007 2 CUST_ID COLLECT(QUESTION_ID) ------- --------------------- 10001 T_NUMBER_TAB(1) 10002 T_NUMBER_TAB(1, 2, 4) 10003 T_NUMBER_TAB(3, 4) 10004 T_NUMBER_TAB(2, 3) 10005 T_NUMBER_TAB(1, 2, 4) 10006 T_NUMBER_TAB(3, 4) 10007 T_NUMBER_TAB(2) CAST converts one built-in data type or collection-typed value into another built-in data type or collection-typed value. CREATE OR REPLACE TYPE t_number_tab AS TABLE OF NUMBER; SELECT customer_id, CAST(COLLECT(question_id) AS t_number_tab) col FROM customer_inquiry GROUP BY customer_id;

  21. Step 3: Simple tab_to_string Function CREATE OR REPLACE TYPE t_number_tab AS TABLE OF NUMBER; CREATE OR REPLACE FUNCTION tab_to_string ( p_number_tab IN t_number_tab, p_delimiter IN VARCHAR2 DEFAULT ',‘ ) RETURN VARCHAR2 IS l_string VARCHAR2(32767); BEGIN FOR i IN p_number_tab.FIRST .. p_number_tab.LAST LOOP IF i != p_number_tab.FIRST THEN l_string := l_string || p_delimiter; END IF; l_string := l_string || to_char(p_number_tab(i)); END LOOP; RETURN l_string; END tab_to_string;

  22. Step 3: tab_to_string(CAST(COLLECT)) Customer_IDQuestion_ID 10001 1 10002 1 10002 2 10002 4 10003 3 10003 4 10004 2 10004 3 10005 1 10005 2 10005 4 10006 3 10006 4 10007 2 CUST_ID COLLECT(QUESTION_ID) ------- --------------------- 10001 1 10002 1,2,4 10003 3,4 10004 2,3 10005 1,2,4 10006 3,4 10007 2 The tab_to_string procedure converts a table of numbers to a comma-separated VARCHAR2. SELECT customer_id, tab_to_string(CAST(COLLECT(question_id) AS t_number_tab)) col FROM customer_inquiry GROUP BY customer_id;

  23. Method 2: Putting it all together tab_to_string( CAST( COLLECT(question_id) AS t_number_tab ) ) col Questions IDsCustomer Count 1 1 1,2,4 2 3,4 2 2,3 1 2 1 SELECT SUBSTR(col,1,10) questions, COUNT(*) FROM ( SELECT customer_id, FROM answers GROUP BY customer_id ) GROUP BY col;

  24. Method 3: XML (The Flying Spontinalli) • XML functions: • Convert traditional table data into XML • Inquire upon XML data and fragments • Operate on XML data and fragments • Convert XML data back to character data type • CURSOR function: Converts a sub-query into a REF CURSOR

  25. Method 3: The Query EXTRACT ( SYS_XMLGEN(seq) ,'/SEQ/XMLTYPE/ROW/Q' ).getstringval() questions FROM ( SELECT customer_id, XMLSEQUENCE ( CURSOR ( SELECT question_id Q FROM TEST t2 WHERE customer_id = t.customer_id ) ) seq FROM TEST t ) ) GROUP BY questions; SELECT REPLACE(REPLACE(REPLACE( questions,'</Q>'||CHR(10)||'<Q>',',' ),'<Q>',NULL ),'</Q>',NULL ) questions, COUNT(*) the_count FROM ( SELECT DISTINCT customer_id,

  26. Step 1: CURSOR and XMLSEQUENCE CUST SEQ ----- ------------------------------ 10001 XMLSEQUENCETYPE(XMLTYPE( <ROW> <Q>1</Q> </ROW> )) 10002 XMLSEQUENCETYPE(XMLTYPE( <ROW> <Q>1</Q> </ROW> ), XMLTYPE( <ROW> <Q>2</Q> </ROW> ), XMLTYPE( <ROW> <Q>4</Q> </ROW> The CURSOR function converts a sub-query into a REF CURSOR. In this case, XMLSEQUENCE requires a REF CURSOR. The XMLSEQUENCE operator is used to split multi-value results from XMLTYPE queries into multiple rows. SELECT customer_id, XMLSEQUENCE( CURSOR( SELECT question_id Q FROM TEST t2 WHERE customer_id = t.customer_id ) ) seq FROM TEST t;

  27. Step 2: SYS_XMLGen and EXTRACT The SYS_XMLGen function takes an expression that evaluates to a particular row and column of the database, and returns an instance of type XMLType containing an XML document Applying EXTRACT to an XMLType value extracts the node or a set of nodes from the document identified by the XPath expression. The method getStringVal() retrieves the text from the XMLType instance SELECT DISTINCT customer_id, EXTRACT( SYS_XMLGEN(seq),'/SEQ/XMLTYPE/ROW/Q' ).getstringval () questions FROM ( SELECT customer_id, XMLSEQUENCE( CURSOR( SELECT question_id Q FROM TEST t2 WHERE customer_id = t.customer_id ) ) seq FROM TEST t )

  28. Step 2: SYS_XMLGen and EXTRACT CUST SEQ ----- ------------------------------ 10002 XMLSEQUENCETYPE(XMLTYPE( <ROW> <Q>1</Q> </ROW> ), XMLTYPE( <ROW> <Q>2</Q> </ROW> ), XMLTYPE( <ROW> <Q>4</Q> </ROW> CUST QUESTIONS ----- ------------------------- 10001 <Q>1</Q> 10002 <Q>1</Q><Q>2</Q><Q>4</Q> 10003 <Q>3</Q><Q>4</Q> 10004 <Q>2</Q><Q>3</Q> 10005 <Q>1</Q><Q>2</Q><Q>4</Q> 10006 <Q>3</Q><Q>4</Q> 10007 <Q>2</Q> The SYS_XMLGen function takes an expression that evaluates to a particular row and column of the database, and returns an instance of type XMLType containing an XML document Applying EXTRACT to an XMLType value extracts the node or a set of nodes from the document identified by the XPath expression. The method getStringVal() retrieves the text from the XMLType instance SELECT DISTINCT customer_id, EXTRACT( SYS_XMLGEN(seq),'/SEQ/XMLTYPE/ROW/Q' ).getstringval () questions

  29. Method 3: Back to the Query Questions IDsCustCnt 1 1 1,2,4 2 3,4 2 2,3 1 2 1 EXTRACT ( SYS_XMLGEN(seq) ,'/SEQ/XMLTYPE/ROW/Q' ).getstringval() questions FROM ( SELECT customer_id, XMLSEQUENCE ( CURSOR ( SELECT question_id Q FROM TEST t2 WHERE customer_id = t.customer_id ) ) seq FROM TEST t ) ) GROUP BY questions; SELECT REPLACE(REPLACE(REPLACE( questions,'</Q>'||CHR(10)||'<Q>',',' ),'<Q>',NULL ),'</Q>',NULL ) questions, COUNT(*) the_count FROM ( SELECT DISTINCT customer_id,

  30. Method 4: MODEL (Rob van Wijk) MODEL PARTITION BY (cust_id) DIMENSION BY (ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY quest_id DESC) rn) MEASURES (CAST(quest_id AS varchar2(20)) q) RULES ( q[any] ORDER BY rn DESC = q[cv()+1] || ',' || q[cv()] ) SELECT q "Questions", COUNT(*) "Customer Count" FROM ( SELECT SUBSTR(q,2) q, rn FROM a ) WHERE rn = 1 GROUP BY q;

  31. Method 4: The MODEL Portion of the Query The MODEL clause enables you to create a multidimensional array by mapping the columns of a query into three groups: partitioning, dimension, and measure columns MODEL PARTITION BY (cust_id) DIMENSION BY (ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY quest_id DESC) rn) MEASURES (CAST(quest_id AS varchar2(20)) q) RULES ( q[any] ORDER BY rn DESC = q[cv()+1] || ',' || q[cv()] ) * All text on this page from Oracle® Database Data Warehousing Guide

  32. Method 4: Dissecting the MODEL Clause The MODEL clause enables you to create a multidimensional array by mapping the columns of a query into three groups: partitioning, dimension, and measure columns PARTITION columns define the logical blocks of the result set in a way similar to the partitions of the analytical functions. Rules in the MODEL clause are applied to each partition independent of other partitions MODEL PARTITION BY (cust_id) DIMENSION BY (ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY quest_id DESC) rn) MEASURES (CAST(quest_id AS varchar2(20)) q) RULES ( q[any] ORDER BY rn DESC = q[cv()+1] || ',' || q[cv()] ) * All text on this page from Oracle® Database Data Warehousing Guide

  33. Method 4: Dissecting the MODEL Clause The MODEL clause enables you to create a multidimensional array by mapping the columns of a query into three groups: partitioning, dimension, and measure columns PARTITION columns define the logical blocks of the result set in a way similar to the partitions of the analytical functions. Rules in the MODEL clause are applied to each partition independent of other partitions DIMENSION columns define the multi-dimensional array and are used to identify cells within a partition. By default, a full combination of dimensions should identify just one cell in a partition MODEL PARTITION BY (cust_id) DIMENSION BY (ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY quest_id DESC) rn) MEASURES (CAST(quest_id AS varchar2(20)) q) RULES ( q[any] ORDER BY rn DESC = q[cv()+1] || ',' || q[cv()] ) * All text on this page from Oracle® Database Data Warehousing Guide

  34. Method 4: Dissecting the MODEL Clause The MODEL clause enables you to create a multidimensional array by mapping the columns of a query into three groups: partitioning, dimension, and measure columns PARTITION columns define the logical blocks of the result set in a way similar to the partitions of the analytical functions. Rules in the MODEL clause are applied to each partition independent of other partitions DIMENSION columns define the multi-dimensional array and are used to identify cells within a partition. By default, a full combination of dimensions should identify just one cell in a partition MEASURES are equivalent to the measures of a fact table in a star schema. MODEL PARTITION BY (cust_id) DIMENSION BY (ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY quest_id DESC) rn) MEASURES (CAST(quest_id AS varchar2(20)) q) RULES ( q[any] ORDER BY rn DESC = q[cv()+1] || ',' || q[cv()] ) * All text on this page from Oracle® Database Data Warehousing Guide

  35. Method 4: Dissecting the MODEL Clause The MODEL clause enables you to create a multidimensional array by mapping the columns of a query into three groups: partitioning, dimension, and measure columns PARTITION columns define the logical blocks of the result set in a way similar to the partitions of the analytical functions. Rules in the MODEL clause are applied to each partition independent of other partitions DIMENSION columns define the multi-dimensional array and are used to identify cells within a partition. By default, a full combination of dimensions should identify just one cell in a partition MEASURES are equivalent to the measures of a fact table in a star schema RULES are used to manipulate the measure values of the cells in the multi-dimensional array defined by partition and dimension columns MODEL PARTITION BY (cust_id) DIMENSION BY (ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY quest_id DESC) rn) MEASURES (CAST(quest_id AS varchar2(20)) q) RULES ( q[any] ORDER BY rn DESC = q[cv()+1] || ',' || q[cv()] ) * All text on this page from Oracle® Database Data Warehousing Guide

  36. Method 4: Dissecting the MODEL Clause PARTITION BY (cust_id) DIMENSION BY (ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY quest_id DESC) rn) MEASURES (CAST(quest_id AS varchar2(20)) q) RULES ( q[any] ORDER BY rn DESC = q[cv(rn)+1] || ',' ||q[cv(rn)] ) Example: q[2] = q[3]||’,’||q[2] = NULL,3 q[1] = q[2]||’,’||q[1] = ,3,4 quest_id ROW_NUMBER Cust ID CV() (rn) Q ------- ---- ------- ------ 10001 1 1 ,1 10002 1 3 ,1 10002 2 2 ,1,2 10002 4 1 ,1,2,4 10003 3 2 ,3 10003 4 1 ,3,4 10004 2 2 ,2 10004 3 1 ,2,3 10005 1 3 1 10005 2 2 ,1,2 10005 4 1 ,1,2,4 10006 3 2 ,3 10006 4 1 ,3,4 10007 2 1 ,2

  37. Method 4: Back to the Query Questions IDsCustomer Count 1 1 1,2,4 2 3,4 2 2,3 1 2 1 MODEL PARTITION BY (cust_id) DIMENSION BY (ROW_NUMBER() OVER (PARTITION BY cust_id ORDER BY quest_id DESC) rn) MEASURES (CAST(quest_id AS varchar2(20)) q) RULES ( q[any] ORDER BY rn DESC = q[cv()+1] || ',' || q[cv()] ) SELECT q "Questions", COUNT(*) "Customer Count" FROM ( SELECT SUBSTR(q,2) q, rn FROM a ) WHERE rn = 1 GROUP BY q;

  38. Method 5: The Pipelined Function (michaels) • Oracle Table Functions produce a collection of rows that can be consumed by a query much like a table. • Rows from a collection returned by a table function can also be PIPELINED or returned as they are produced instead of in a complete set upon function completion. • Two supported approached for Pipelining: • Interface method • PL/SQL method

  39. Method 5: The Query Oracle Table Functions produce a collection of rows that can be consumed by a query much like a table. SELECT COLUMN_VALUE "Questions", COUNT (*) "Customer count" FROM TABLE (f ()) GROUP BY COLUMN_VALUE;

  40. Method 5: The Pipelined Function CREATE OR REPLACE FUNCTION f RETURN SYS.dbms_debug_vc2coll PIPELINED AS l_c1 VARCHAR2 (20); l_c2 PLS_INTEGER; BEGIN FOR c IN (SELECT a.*, COUNT (*) OVER (PARTITION BY customer_id) cnt, ROW_NUMBER () OVER (PARTITION BY customer_id ORDER BY question_id) rn FROM a ORDER BY customer_id, question_id) LOOP…

  41. Method 5: The Pipelined Function continued LOOP IF l_c2 = c.customer_id OR l_c2 IS NULL THEN l_c1 := l_c1 || c.question_id || ','; l_c2 := c.customer_id; IF c.cnt = c.rn THEN PIPE ROW (RTRIM (l_c1, ',')); l_c1 := NULL; l_c2 := NULL; END IF; END IF; END LOOP; RETURN; END f;

  42. Method 5: The Pipelined Function Questions IDsCustomer Count 1 1 1,2,4 2 3,4 2 2,3 1 2 1 SELECT COLUMN_VALUE "Questions", COUNT (*) "Customer count" FROM TABLE (f ()) GROUP BY COLUMN_VALUE;

  43. So many choices, so little time… • If all these queries provide the same results, which one should be chosen? • Query cost information was added to this post - yet another topic! • The results: • Michaels PIPELINED Function • Rob's MODEL • Nicolas' HIERARCHY 1 • Nicolas' HIERARCHY 2 • Greg's COLLECT • Warren/devmiral's HIERARCHY • Michaels'/Greg‘s XML

  44. Not done yet! Even More Solutions… XML combined with COLLECT Bit Pattern User-defined Aggregate Functions (ODCIAggregate Interface) Old-school Oracle 7 solution

  45. Wow, All of These Topics in a Single Thread! • Hierarchical Queries • SYS_CONNECT_BY_ROOT • Analytic Functions • COLLECT • CAST • CURSOR • XML Functions • MODEL Clause • Pipelined Functions • Bit Pattern Techniques • User-defined Aggregates • Explain Plans/Query Costs BUT…Never underestimate the value of using widely-understood and accepted Oracle database concepts The experts here used this thread to demonstrate alternate technologies only and never advocated these rather esoteric solutions

  46. Jumping in the Pool: “Forum-Decorum” • Don’t demand urgent help. In fact, don’t demand anything from this all-volunteer community • Search for an answer first • Use the proper code notation tags – reformatting queries is a pain [pre]This is code text[/pre] • Don’t flame the Gurus…you will likely get scorched! • If you use information from the Forums on your Blog or anywhere else, please reference your source • Don’t be intimidated. If you don’t understand an answer, request clarification • Thank your responders

  47. Wrap-Up • The Oracle SQL and PL/SQL Forum has 64K+ topics with almost 350k posts • Other active Forums include: • Database-General • App Server • J-Developer • Forms • OWB • More! • Take advantage of the experts including many Oracle Aces! • Learn something new • Become a regular contributor!

  48. Acknowledgements • Oracle® Database Data Warehousing Guide 10gR2 • Oracle® Database SQL Reference 10gR2 • Oracle® XML DB Developer's Guide 11gR1 • OTN - Getting into SQL/XML - Tim Quinlan • http://www.oracle-base.com • The OTN SQL and PL/SQL Forum

  49. Thank You for attending! Thank You for attending! Greg Pike gpike@piocon.com Piocon Technologies 1420 Kensington Rd. Suite 106 Oak Brook, IL 60523 630-579-0800 Blog: www.singlequery.com Thanks to numerous contributors: • The Oracle Forum experts

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