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22 Advanced Querying and Information Retrieval
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The output is the same as in the version of the query without grouping, but with three extra columns called item-name- ag, color- ag, and size- ag In each tuple, the value of a ag eld is 1 if the corresponding eld is a null representing all Instead of using tags to indicate nulls that represent all, we can replace the null value by a value of our choice: decode(grouping(item-name), 1, all , item-name) This expression returns the value all if the value of item-name is a null corresponding to all, and returns the actual value of item-name otherwise This expression can be used in place of item-name in the select clause to get all in the output of the query, in place of nulls representing all Neither the rollup nor the cube clause gives complete control on the groupings that are generated For instance, we cannot use them to specify that we want only groupings {(color, size), (size, item-name)} Such restricted groupings can be generated by using the grouping construct in the having clause; we leave the details as an exercise for you
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Finding the position of a value in a larger set is a common operation For instance, we may wish to assign students a rank in class based on their total marks, with the rank 1 going to the student with the highest marks, the rank 2 to the student with the next highest marks, and so on While such queries can be expressed in SQL-92, they are dif cult to express and inef cient to evaluate Programmers often resort to writing the query partly in SQL and partly in a programming language A related type of query is to nd the percentile in which a value in a (multi)set belongs, for example, the bottom third, middle third, or top third We study SQL:1999 support for these types of queries here Ranking is done in conjunction with an order by speci cation Suppose we are given a relation student-marks(student-id, marks) which stores the marks obtained by each student The following query gives the rank of each student select student-id, rank() over (order by (marks) desc) as s-rank from student-marks Note that the order of tuples in the output is not de ned, so they may not be sorted by rank An extra order by clause is needed to get them in sorted order, as shown below select student-id, rank () over (order by (marks) desc) as s-rank from student-marks order by s-rank A basic issue with ranking is how to deal with the case of multiple tuples that are the same on the ordering attribute(s) In our example, this means deciding what to do if there are two students with the same marks The rank function gives the same
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Silberschatz Korth Sudarshan: Database System Concepts, Fourth Edition
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22 Advanced Querying and Information Retrieval
The McGraw Hill Companies, 2001
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rank to all tuples that are equal on the order by attributes For instance, if the highest mark is shared by two students, both would get rank 1 The next rank given would be 3, not 2, so if three students get the next highest mark, they would all get rank 3, and the next student(s) would get rank 5, and so on There is also a dense rank function that does not create gaps in the ordering In the above example, the tuples with the second highest value all get rank 2, and tuples with the third highest value get rank 3, and so on Ranking can be done within partitions of the data For instance, suppose we have an additional relation student-section(student-id, section) that stores for each student the section in which the student studies The following query then gives the rank of students within each section select student-id, section, rank () over (partition by section order by marks desc) as sec-rank from student-marks, student-section where student-marksstudent-id = student-sectionstudent-id order by section, sec-rank The outer order by clause orders the result tuples by section, and within each section by the rank Multiple rank expressions can be used within a single select statement; thus we can obtain the overall rank and the rank within the section by using two rank expressions in the same select clause An interesting question is what happens when ranking (possibly with partitioning) occurs along with a group by clause In this case, the group by clause is applied rst, and partitioning and ranking are done on the results of the group by Thus aggregate values can then be used for ranking For example, suppose we had marks for each student for each of several subjects To rank students by the sum of their marks in different subjects, we can use a group by clause to compute the aggregate marks for each student, and then rank students by the aggregate sum We leave details as an exercise for you The ranking functions can be used to nd the top n tuples by embedding a ranking query within an outer-level query; we leave details as an exercise Note that bottom n is simply the same as top n with a reverse sorting order Several database systems provide nonstandard SQL extensions to specify directly that only the top n results are required; such extensions do not require the rank function, and simplify the job of the optimizer, but are (currently) not as general since they do not support partitioning SQL:1999 also speci es several other functions that can be used in place of rank For instance, percent rank of a tuple gives the rank of the tuple as a fraction If there are n tuples in the partition3 and the rank of the tuple is r, then its percent rank is de ned as (r 1)/(n 1) (and as null if there is only one tuple in the partition) The function cume dist, short for cumulative distribution, for a tuple is de ned as p/n where p is the number of tuples in the partition with ordering values preceding or equal to the ordering value of the tuple, and n is the number of tuples in the parti3 The entire set is treated as a single partition if no explicit partition is used
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