qr code c#.net generator sdk Inside Microsoft SQL Server 2008: T-SQL Querying in Visual C#.NET

Creation QR Code in Visual C#.NET Inside Microsoft SQL Server 2008: T-SQL Querying

Inside Microsoft SQL Server 2008: T-SQL Querying
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This query generates the following output, shown here in abbreviated form:
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custid ----------1 2 3 4 5 6 7 8 9 10 11 12 ... empid ----------4 3 3 4 3 9 4 4 4 3 6 8 cnt ----------2 2 3 4 6 3 3 2 4 4 2 2
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If you want to return all ties, simply use the RANK function instead of ROW_NUMBER and calculate it based on count ordering alone (without the employee ID tiebreaker), like so:
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WITH C AS ( SELECT custid, empid, COUNT(*) AS cnt, RANK() OVER(PARTITION BY custid ORDER BY COUNT(*) DESC) AS rn FROM Sales.Orders GROUP BY custid, empid ) SELECT custid, empid, cnt FROM C WHERE rn = 1;
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This time, as you can see in the following output, ties are returned:
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custid ----------1 1 2 3 4 5 6 7 8 9 10 11 11 11 12 ... empid ----------1 4 3 3 4 3 9 4 4 4 3 6 4 3 8 cnt ----------2 2 2 3 4 6 3 3 2 4 4 2 2 2 2
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In case you do want to apply a tiebreaker, you can use another solution that is very ef cient. It is based on the concatenation technique that I presented earlier in the chapter. Write a query that groups the data by customer ID and employee ID, and for each group, concatenate the count of rows and the employee ID to a single value (call it binval). De ne a table expression based on this query. Have the outer query group the data by customer ID and calculate for each customer the maximum binval. This maximum value contains the max count and within it the maximum employee ID. What s left is to extract the count and employee ID from the binary value by using the SUBSTRING function and convert the values to the original types. Here s the complete solution query:
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SELECT custid, CAST(SUBSTRING(MAX(binval), 5, 4) AS INT) AS empid, CAST(SUBSTRING(MAX(binval), 1, 4) AS INT) AS cnt FROM (SELECT custid, CAST(COUNT(*) AS BINARY(4)) + CAST(empid AS BINARY(4)) AS binval FROM Sales.Orders GROUP BY custid, empid) AS D GROUP BY custid;
As an exercise, you can test the solutions against a table with a large number of rows. You will see that this solution is very fast.
Histograms
Histograms are powerful analytical tools that express the distribution of items. For example, suppose you need to gure out from the order information in the Sales.OrderValues view how many small, medium, and large orders you have, based on the order values. In other words, you need a histogram with three steps. The extreme values (the minimum and maximum values) are what de nes values as small, medium, or large. Suppose for the sake of simplicity that the minimum order value is 10 and the maximum is 40. Take the difference between the two extremes (40 10 = 30) and divide it by the number of steps (3) to get the step size. In this case, it s 30 divided by 3, which is 10. So the boundaries of step 1 (small) would be 10 and 20; for step 2 (medium), they would be 20 and 30; and for step 3 (large), they would be 30 and 40. To generalize this, let mn = MIN(val) and mx = MAX(val) and let stepsize = (mx mn) / @numsteps. Given a step number n, the lower bound of the step (lb) is mn + (n 1) * stepsize and the higher bound (hb) is mn + n * stepsize. Something is tricky here. What predicate do you use to bracket the elements that belong in a speci c step You can t use val BETWEEN lb and hb because a value that is equal to hb appears in this step and also in the next step, where it equals the lower bound. Remember that the same calculation yielded the higher bound of one step and the lower bound of the next step. One approach to deal with this problem is to increase each of the lower bounds besides the rst by one so that they exceed the previous step s higher bounds. With integers, this is a ne solution, but with another data type (such as NUMERIC in our case) it doesn t work because there are potential values between adjacent steps but not within either one between the cracks, so to speak.
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