Order by COUNT per value

问题

I have a table which stores IDs and the city where the store is located.

I want to list all the stores starting with the stores that are in the city where there are the most stores.

TABLE

ID CITY
1  NYC
2  BOS
3  BOS
4  NYC
5  NYC

The output I want is the following since I have the most stores in NYC, I want all the NYC location to be listed first.

1  NYC
4  NYC
5  NYC
2  BOS
3  BOS

回答1:

SELECT count(City), City
FROM table
GROUP BY City
ORDER BY count(City);

SELECT count(City) as count, City
FROM table
GROUP BY City
ORDER BY count;

Ahh, sorry, I was misinterpreting your question. I believe Peter Langs answer was the correct one.

回答2:

This one calculates the count in a separate query, joins it and orders by that count (SQL-Fiddle):

SELECT c.id, c.city
FROM cities c
JOIN ( SELECT city, COUNT(*) AS cnt
       FROM cities
       GROUP BY city
     ) c2 ON ( c2.city = c.city )
ORDER BY c2.cnt DESC;

回答3:

This solution is not a very optimal one so if your table is very large it will take some time to execute but it does what you are asking.

 select c.city, c.id, 
      (select count(*) as cnt from city c2 
       where c2.city = c.city) as order_col
 from city c
 order by order_col desc

That is, for each city that you come across you are counting the number of times that that city occurs in the database.

Disclaimer: This gives what you are asking for but I would not recommend it for production environments where the number of rows will grow too large.

回答4:

SELECT `FirstAddressLine4`, count(*) AS `Count` 
FROM `leads` 
WHERE `Status`='Yes'
AND `broker_id`='0'
GROUPBY `FirstAddressLine4` 
ORDERBY `Count` DESC 
LIMIT 0, 8

来源：https://stackoverflow.com/questions/2283305/order-by-count-per-value

标签

mysql

sql

count

sql-order-by