![]() ![]() For example, for there were 20 payments so, the percentage of successful payments = 50%, failed = 25% and refunded = 25%. The query successfully returns how many payments were processed on each day and the status:, successful, 10īut I don't know how to make it calculate the percentage for each payment for each day. My current query is this: select pdate, status, count(*) as total Here are three suggestions: select coalesce (p.psarea,p1.psarea) as psarea, Male, Female from ( select psarea, count () as Male from personal where psgender 'M' group by psarea ) as p full outer join ( select psarea, count () as Female from personal where psgender 'F' group by psarea ) as p1 on p1.psarea p. ![]() I need to write a query to get, for each day how many payments were successful, how many failed and how many were refunded, with the percentage for that day. ![]() HAVING 100*count(p.id)/(SELECT count(p.I have a payments table with the following fields: Demo Database Below is a selection from the 'Products' table in the Northwind sample database: COUNT () Example The following SQL statement finds the number of products: Example SELECT COUNT(ProductID) FROM Products Try it Yourself Note: NULL values are not counted. 1 SELECT COUNT(), HOUR(timestampcolumnname) 2 FROM tablename 3 GROUP BY HOUR(timestampcolumnname) Note: This solution works both for TIMESTAMP and DATETIME type. HAVING 100*count(p.id)/(SELECT count(p.id) FROM participants as p JOIN participants_events as pe On p.id = pe.participant_id) as percentage , 100*count(p.id)/(SELECT count(p.id) FROM participants as p JOIN participants_events as pe So i.signori answer will be : SELECT count(p.id) as count CPUCreditUsage: CPU Credit Usage (Count) (T2 instances) The number of CPU credits spent by the instance for CPU utilization. CPUUtilization: CPU Utilization (Percent) The percentage of CPU utilization. Example Code: SELECT RepresentativeName, Sale, round( ( (Sale 100) / temp. TO SELECT count(p.id) as count, round(100*count(p.id)/(SELECT count(p.id) FROM participants as p JOIN participants_events as pe on p.id = pe.participant_id),1) as percentage, p.Nationality Connection Attempts (Count) The number of attempts to connect to an instance, whether successful or not. To find the percentage of the Sale field, we can CROSS JOIN the SUM () function of the Sale attribute with the original relation (table). 1) COUNT 2) SUM 3) AVG 4) MIN 5) MAX In this MySQL Aggregate Functions tutorial, you will learn: Why Use Aggregate Functions COUNT function DISTINCT Keyword MIN function MAX function SUM function AVG function Why Use Aggregate Functions From a business perspective, different organization levels have different information requirements. So modify your initial query from : SELECT count(p.id) as count, round(100*count(p.id)/(SELECT count(p.id) FROM participants as p),1) as percentage, p.Nationality Please check in your initial query to use the same sets: Perhaps the next option will be more to your liking. Unfortunately, that's not easily done using the over () clause. Name Result a 100 b 130.45 c 182.96 d 65. Here's the formula to obtain a percentage: count () 100.0 / sum (count ()) over () Adding the above SQL to our original query produces the following results: Looks good, but some rounding wouldn't hurt. I have a table named PERSON (In the live system I expect several hundred thousand records or more). HAVING round(100*count(p.id)/(SELECT count(p.id) FROM participants as p),1) <= 1 I am using MySQL and I need help using COUNT() for a range of values within a table. HAVING round(100*count(p.id)/(SELECT count(p.id) FROM participants as p),1) > 1įor the second part, try to combine the results of the first with those of the sum of percentages less than 1. resolution'Cancelled' is an expression returning 0 for false, or 1 for true. ![]() , round(100*count(p.id)/(SELECT count(p.id) FROM participants as p),1) as percentage 1 Answer Sorted by: 1 To get the percentage you need to divide the sum of the value, with the total and multiply by 100 to get a percentage. Try using Having clause: SELECT count(p.id) as count So with the new query the results will be: +-+-+-+ count (distinct col1, col2) : distinct (col1, col2) values combinations. count (distinct col2) : distinct col2 values. I want to modify the query in order to don't show any rows with percentage less than 1% but summarize their number and add is as 'Other' with the new percentage. There are several things you can count with COUNT () function: count () : rows count (col1) : rows where col1 is not null count (col2) : rows where col2 is not null count (distinct col1) : distinct col1 values. JOIN participants_events as pe on p.id = pe.participant_id This is the query I am using to get the percentage per country for participants in events: SELECT count(p.id) as count, round(100*count(p.id)/(SELECT count(p.id) FROM participants as p),1) as percentage, p.Nationality ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |