[Solved] Sql Server sorting query result [closed]
[ad_1] add an order by clause –at the end of the query order by ChatDate, ChatTime 0 [ad_2] solved Sql Server sorting query result [closed]
sql-order-by
[Solved] SELECT * FROM people WHERE user_id=’$user_id’ ORDER BY time GROUP BY surname [closed]
[ad_1] The ORDER BY clause should be the last, but you have to specify fields to be aggregated. Such as: SELECT surname, count(*) FROM people WHERE user_token=’$user_token’ GROUP BY surname ORDER BY surname 1 [ad_2] solved SELECT * FROM people WHERE user_id=’$user_id’ ORDER BY time GROUP BY surname [closed]
[Solved] How to assign rank to students to where they share the highest rank when their scores equal?
[ad_1] You need two counters Absolute Counter (always 1, 2, 3, 4, 5, etc). Rank Counter – it counts from 1 but if the scores are same, it does not update. As soon as scores are different, it updates with the absolute counter. Sample Code $counter = 1; // init absolute counter $rank = 1; … Read more
[Solved] How to order by multiple columns with same datatype [closed]
[ad_1] Your sample data was ambiguous and misleading. If you really just want to order by whatever column isn’t ”, and given your assurance that no columns are NULL and there are no ‘ ‘ etc., you can use CONCAT, which will simply munge whatever value is not empty with two empty strings: SELECT FirstName, … Read more
[Solved] mySQL Largest number by group
[ad_1] In general ORDER BY in a sub-query makes no sense. (It only does when combined with FETCH FIRST/LIMIT/TOP etc.) The solution is to use a correlated sub-query to find the heaviest fish for the “main query”‘s current row’s username, location, species combination. If it’s a tie, both rows will be returned. SELECT * FROM … Read more