c# print barcode zebra FIGURE 6-14 Query plan for islands, solution 1 in C#

Draw QR in C# FIGURE 6-14 Query plan for islands, solution 1

FIGURE 6-14 Query plan for islands, solution 1
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Subqueries, Table Expressions, and Ranking Functions
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The plan shows two merge joins, each between the results of two ordered scans of the index on seqval. Each such merge join is used to process a logical anti-semi join that lters points before or after gaps. Each such merge join lters as many rows as the number of islands (10,000 in our case). Finally, another merge join is used to pair starting and ending points. Even though the last merge is many-to-many and can potentially be slow, it s pretty fast because it handles only a small number of islands in our case. This solution ran on my system for 17 seconds and incurred 64,492 logical reads. To apply the solution to a temporal sequence, simply use the DATEADD function as usual to add an interval to the sequence value:
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WITH StartingPoints AS ( SELECT seqval, ROW_NUMBER() OVER(ORDER BY seqval) AS rownum FROM dbo.TempSeq AS A WHERE NOT EXISTS (SELECT * FROM dbo.TempSeq AS B WHERE B.seqval = DATEADD(hour, -4, A.seqval)) ), EndingPoints AS ( SELECT seqval, ROW_NUMBER() OVER(ORDER BY seqval) AS rownum FROM dbo.TempSeq AS A WHERE NOT EXISTS (SELECT * FROM dbo.TempSeq AS B WHERE B.seqval = DATEADD(hour, 4, A.seqval)) ) SELECT S.seqval AS start_range, E.seqval AS end_range FROM StartingPoints AS S JOIN EndingPoints AS E ON E.rownum = S.rownum;
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To apply the solution to a sequence with duplicates, query a derived table with the distinct values:
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WITH StartingPoints AS ( SELECT seqval, ROW_NUMBER() OVER(ORDER BY seqval) FROM (SELECT DISTINCT seqval FROM dbo.NumSeqDups) WHERE NOT EXISTS (SELECT * FROM dbo.NumSeqDups AS B WHERE B.seqval = A.seqval - 1) ), EndingPoints AS ( SELECT seqval, ROW_NUMBER() OVER(ORDER BY seqval) FROM (SELECT DISTINCT seqval FROM dbo.NumSeqDups) WHERE NOT EXISTS (SELECT * FROM dbo.NumSeqDups AS B WHERE B.seqval = A.seqval + 1) )
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AS rownum AS A
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AS rownum AS A
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Inside Microsoft SQL Server 2008: T-SQL Querying
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SELECT S.seqval AS start_range, E.seqval AS end_range FROM StartingPoints AS S JOIN EndingPoints AS E ON E.rownum = S.rownum;
Islands, Solution 2: Using Group Identi er Based on Subqueries
The second solution to the islands problem involves a concept I haven t discussed yet a grouping factor, or group identi er. You basically need to group data by a factor that does not exist in the data as a base attribute. In our case, you need to calculate some x value for all members of the rst subset of consecutive values {2, 3}, some y value for the second {11, 12, 13}, some z value for the third {27}, and so on. When you have this grouping factor available, you can group the data by this factor and return the minimum and maximum col1 values in each group. One approach to calculating this grouping factor brings me to another technique: calculating the min or max value of a group of consecutive values. Take the group {11, 12, 13} as an example. If you can manage to calculate for each of the members the max value in the group (13), you can use it as your grouping factor. The logic behind the technique to calculating the maximum within a group of consecutive values is: return the minimum value that is greater than or equal to the current, after which there s a gap. Here s the translation to T-SQL:
SELECT seqval, (SELECT MIN(B.seqval) FROM dbo.NumSeq AS B WHERE B.seqval >= A.seqval AND NOT EXISTS (SELECT * FROM dbo.NumSeq AS C WHERE C.seqval = B.seqval + 1)) AS grp FROM dbo.NumSeq AS A;
This code generates the following output:
seqval ----------2 3 11 12 13 27 33 34 35 42 grp ----------3 3 13 13 13 27 35 35 35 42
The rest is really easy: create a CTE table out of the previous step s query, group the data by the grouping factor, and return the minimum and maximum values for each group:
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