As of Swift 4.2 (Xcode 10) you can declare conformance to the CaseIterable protocol, this works for all enumerations without associated values: enum Stuff: CaseIterable { case first case second case third case forth } The number of cases is now simply obtained with print(Stuff.allCases.count) // 4 For more information, see SE-0194 Derived Collection of … Read more
Use the HAVING clause and GROUP By the fields that make the row unique The below will find all users that have more than one payment per day with the same account number SELECT user_id , COUNT(*) count FROM PAYMENT GROUP BY account, user_id , date HAVING COUNT(*) > 1 Update If you want to … Read more
You can use this: SELECT COUNT(*) FROM (SELECT DISTINCT column_name FROM table_name) AS temp; This is much faster than: COUNT(DISTINCT column_name)
You have multiple options. Two options are the following. boolarr.sum() numpy.count_nonzero(boolarr) Here’s an example: >>> import numpy as np >>> boolarr = np.array([[0, 0, 1], [1, 0, 1], [1, 0, 1]], dtype=np.bool) >>> boolarr array([[False, False, True], [ True, False, True], [ True, False, True]], dtype=bool) >>> boolarr.sum() 5 Of course, that is a bool-specific … Read more
In 2.7 and 3.1, there is the special Counter (dict subclass) for this purpose. >>> from collections import Counter >>> Counter([‘apple’,’red’,’apple’,’red’,’red’,’pear’]) Counter({‘red’: 3, ‘apple’: 2, ‘pear’: 1})
SELECT category, COUNT(*) AS `num` FROM posts GROUP BY category
For older versions of Hibernate (<5.2): Assuming the class name is Book: return (Number) session.createCriteria(“Book”) .setProjection(Projections.rowCount()) .uniqueResult(); It is at least a Number, most likely a Long.
Bottom Line Use either COUNT(field) or COUNT(*), and stick with it consistently, and if your database allows COUNT(tableHere) or COUNT(tableHere.*), use that. In short, don’t use COUNT(1) for anything. It’s a one-trick pony, which rarely does what you want, and in those rare cases is equivalent to count(*) Use count(*) for counting Use * for … Read more
SELECT ( SELECT COUNT(*) FROM tab1 ) AS count1, ( SELECT COUNT(*) FROM tab2 ) AS count2 FROM dual
This will do what you want (list of towns, with the number of users in each): SELECT `town`, COUNT(`town`) FROM `user` GROUP BY `town`; You can use most aggregate functions when using a GROUP BY statement (COUNT, MAX, COUNT DISTINCT etc.) Update: You can declare a variable for the number of users and save the … Read more