If you’re using Java and Jena’s ARQ, you can use ARQ’s extensions for aggregates. Your query would look something like: SELECT ?tag (count(distinct ?tag) as ?count) WHERE { ?r ns9:taggedWithTag ?tagresource. ?tagresource ns9:name ?tag } LIMIT 5000 The original SPARQL specification from 2008 didn’t include aggregates, but the current version, 1.1, from 2013 does.
int divisor = AllMyControls.Where(p => p.IsActiveUserControlChecked).Count() or simply int divisor = AllMyControls.Count(p => p.IsActiveUserControlChecked); Since you are a beginner, it would be worthwhile to take a look at Enumerable documentation
SELECT Country, COUNT(*) AS Total, COUNT(CASE WHEN Reconnect = true THEN 1 END) AS With_Reconnect FROM PlayerSession S LEFT JOIN Player P ON P.id = S.player_id GROUP BY country
Here’s one option using a subquery with DISTINCT: SELECT COUNT(*) gender_count, SUM(IF(gender=”male”,1,0)) male_count, SUM(IF(gender=”female”,1,0)) female_count FROM ( SELECT DISTINCT tel, gender FROM example_dataset ) t SQL Fiddle Demo This will also work if you don’t want to use a subquery: SELECT COUNT(DISTINCT tel) gender_count, COUNT(DISTINCT CASE WHEN gender=”male” THEN tel END) male_count, COUNT(DISTINCT CASE WHEN … Read more
count returns how many times an object occurs in a list, so if you count occurrences of ” you get 6 because the empty string is at the beginning, end, and in between each letter. Use the len function to find the length of a string.
It’s hard to know how to help you without understanding the context / structure of your data, but I believe this might help you: SELECT SUM(CASE WHEN column1 IS NOT NULL THEN 1 ELSE 0 END) AS column1_count ,SUM(CASE WHEN column2 IS NOT NULL THEN 1 ELSE 0 END) AS column2_count ,SUM(CASE WHEN column3 IS … Read more
SELECT (CHAR_LENGTH(str) – CHAR_LENGTH(REPLACE(str, substr, ”))) / CHAR_LENGTH(substr) AS cnt … ORDER BY cnt DESC Yep, looks bloated but afaik there is no any other possible solution. mysql> select (CHAR_LENGTH(‘asd’) – CHAR_LENGTH(REPLACE(‘asd’, ‘s’, ”))) / CHAR_LENGTH(‘s’); +—————————————————————–+ | (CHAR_LENGTH(‘asd’) – CHAR_LENGTH(REPLACE(‘asd’, ‘s’, ”))) / CHAR_LENGTH(‘s’) | +—————————————————————–+ | 1.0000 | +—————————————————————–+ 1 row in set … Read more
On systems that have pgrep available, the -c option returns a count of the number of processes that match the given name pgrep -c command_name Note that this is a grep-style match, not an exact match, so e.g. pgrep sh will also match bash processes. If you want an exact match, also use the -x … Read more
This would work even with multiple spaces and leading and/or trailing spaces and blank lines: String trim = s.trim(); if (trim.isEmpty()) return 0; return trim.split(“\\s+”).length; // separate string around spaces More info about split here.