A query without COUNT returns 3 records, with only 1.
SELECT `blog_cate` . * , COUNT( blogi.blog_cate ) AS num
FROM (
`blog_cate`
)
JOIN `blogi` ON `blogi`.`blog_cate` = `blog_cate`.`blogi_cate_url`
results:
+----+------------------+----------------+-----+
| id | blogi_cate_title | blogi_cate_url | num |
+----+------------------+----------------+-----+
| 1 | Базы данных | batabase | 3 |
+----+------------------+----------------+-----+
And the same query, but without a COUNT:
SELECT `blog_cate` . *
FROM (
`blog_cate`
)
JOIN `blogi` ON `blogi`.`blog_cate` = `blog_cate`.`blogi_cate_url`
That returns me 3 records:
+----+------------------+----------------+
| id | blogi_cate_title | blogi_cate_url |
+----+------------------+----------------+
| 1 | Базы данных | batabase |
| 1 | Базы данных | batabase |
| 3 | Разработка | razrabotka |
+----+------------------+----------------+
Is it possible to use a COUNT and have a normal results?
p.s. tables:
+----+------------+
| id | blog_cate |
+----+------------+
| 1 | batabase |
| 2 | batabase |
| 3 | razrabotka |
+----+------------+
+----+------------------+----------------+
| id | blogi_cate_title | blogi_cate_url |
+----+------------------+----------------+
| 1 | Базы данных | batabase |
| 2 | PHP | php |
| 3 | Разработка | razrabotka |
+----+------------------+----------------+
COUNT() with out a group by will group all records and produce a count of them. Adding more fields to the select will only show the details of the first record
You could build one query to get the three rows and one query to get the count result and join them via cross join to combine every detail row with the count row.
Related
I am trying to get the sum of a single column from all rows that share similar data. For example, given the following data:
|ppID | cID | Count | NameSpace|
|-----|-------|-------|----------|
|586 | 18281 | 1 | LAB |
|587 | 18269 | 1 | LAB |
|588 | 18281 | 1 | LAB |
|589 | 17823 | 1 | IPB |
|590 | 18281 | 1 | LAB |
|591 | 18256 | 1 | LAB |
|592 | 18256 | 1 | LAB |
|593 | 18269 | 1 | IPB |
|-----|-------|-------|----------|
I'm hoping to get:
|ppID | cID | Count | NameSpace|
|-----|-------|-------|----------|
|586 | 18281 | 3 | LAB |
|587 | 18269 | 1 | LAB |
|589 | 17823 | 1 | IPB |
|591 | 18256 | 2 | LAB |
|593 | 18269 | 1 | IPB |
|-----|-------|-------|----------|
I've pieced together a couple of different things and come up with `
SELECT * FROM PopularPages
WHERE cID IN (SELECT cID FROM PopularPages
GROUP BY cID
HAVING COUNT(cID) > 1)
ORDER BY cID, Namespace
which will list out each of the rows but without counting up the sum of the Count column. Any help would be appreciated.
is this what you want ?
SELECT
MIN(ppID) as ppID,
cID,
SUM(`Count`) as COUNT,
NameSpace
FROM PopularPages
GROUP BY cID
HAVING Count > 1;
I want to find the average of the following data via mysql query (assume these are 719 rows).
| 1 |
| 3 |
| 1 |
| 2 |
| 2 |
| 1 |
| 1 |
| 2 |
| 1 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 1 |
+----------+
719 rows in set (2.43 sec)
SELECT COUNT(*) FROM osdial_agent_log WHERE DATE(event_time)='2015-11-01' GROUP BY lead_id;
I ran this query to get that data
Can someone help me to find the average for the above data.
Use
SELECT AVG(total)
FROM (SELECT COUNT(*) AS total
FROM osdial_agent_log
WHERE DATE(event_time)='2015-11-01'
GROUP BY lead_id) t
I've got a table that looks like:
Table 1 ->
+----+--------+--------+
| id | name | author |
+----+--------+--------+
| 1 | First | Me |
| 2 | Second | You |
+----+--------+--------+
Table 2 ->
+-----+------------+-----------+------------+
| mid | table1_id | key | value |
+-----+------------+-----------+------------+
| 1 | 1 | desc | hello |
| 2 | 1 | begin_day | monday |
| 3 | 1 | end_day | tuesday |
| 4 | 2 | desc | goodbye |
| 5 | 2 | begin_day | wednesday |
| 6 | 2 | end_day | friday |
+-----+------------+-----------+------------+
The relationship here is that the id in table 1 corresponds to the table1_id in table 2.
The output that I'm trying to get is...
+----+---------+---------+-------------+-----------+-----------+
| id | name | author | desc | begin_day | end_day |
+----+---------+---------+-------------+-----------+-----------+
| 1 | First | Me | hello | monday | tuesday |
| 1 | Second | You | goodbye | wednesday | friday |
+----+---------+---------+-------------+-----------+-----------+
I've tried several different join statements -- all a variation of the below. I'm not that well versed in MySQL queries, however.
SELECT * FROM table_1 LEFT JOIN table_2 on table_1.id = table_2.table1_id
Which produces...
+----+----------+----------+----------+------------+-----------+
| id | mid | name | author | key | value |
+----+----------+----------+----------+------------+-----------+
| 1 | 1 | First | Me | desc | hello |
| 1 | 2 | First | Me | begin_day | monday |
| 1 | 3 | First | Me | end_day | tuesday |
| 2 | 4 | Second | You | desc | goodbye |
| 2 | 5 | Second | You | begin_day | wednesday|
| 2 | 6 | Second | You | end_day | friday |
Obviously, iterating over this join statement produces 6 results, 1 for each row in table 2 that matches the id in table 1. How can I avoid this with a proper query statement?
Thank you in advance.
You can use a case statement if you know all of the columns you will be getting, as follows:
Select distinct table_1.*,
case when table_2.key='desc' then value end as desc,
case when table_2.key='begin_day' then value end as begin_day,
case when table_2.key='end_day' then value end as end_day
FROM table_1 LEFT JOIN table_2 on table_1.id = table_2.table1_id
Hope this helps!
SELECT
table_1.*,
MAX(IF(key='desc', value, NULL)) AS 'desc',
MAX(IF(key='begin_day', value, NULL)) AS begin_day,
MAX(IF(key='end_day', value, NULL)) AS end_day
FROM table_1
LEFT JOIN table_2 ON (id = table1_id)
GROUP BY id;
I have a query to get the top 'n' users who commented on a specific keyword,
SELECT `user` , COUNT( * ) AS magnitude
FROM `results`
WHERE `keyword` = "economy"
GROUP BY `user`
ORDER BY magnitude DESC
LIMIT 5
I have approx 6000 keywords, and would like to run this query to get me the top 'n' users for each and every keyword we have data for. Assistance appreciated.
Since you haven't given the schema for results, I'll assume it's this or very similar (maybe extra columns):
create table results (
id int primary key,
user int,
foreign key (user) references <some_other_table>(id),
keyword varchar(<30>)
);
Step 1: aggregate by keyword/user as in your example query, but for all keywords:
create view user_keyword as (
select
keyword,
user,
count(*) as magnitude
from results
group by keyword, user
);
Step 2: rank each user within each keyword group (note the use of the subquery to rank the rows):
create view keyword_user_ranked as (
select
keyword,
user,
magnitude,
(select count(*)
from user_keyword
where l.keyword = keyword and magnitude >= l.magnitude
) as rank
from
user_keyword l
);
Step 3: select only the rows where the rank is less than some number:
select *
from keyword_user_ranked
where rank <= 3;
Example:
Base data used:
mysql> select * from results;
+----+------+---------+
| id | user | keyword |
+----+------+---------+
| 1 | 1 | mysql |
| 2 | 1 | mysql |
| 3 | 2 | mysql |
| 4 | 1 | query |
| 5 | 2 | query |
| 6 | 2 | query |
| 7 | 2 | query |
| 8 | 1 | table |
| 9 | 2 | table |
| 10 | 1 | table |
| 11 | 3 | table |
| 12 | 3 | mysql |
| 13 | 3 | query |
| 14 | 2 | mysql |
| 15 | 1 | mysql |
| 16 | 1 | mysql |
| 17 | 3 | query |
| 18 | 4 | mysql |
| 19 | 4 | mysql |
| 20 | 5 | mysql |
+----+------+---------+
Grouped by keyword and user:
mysql> select * from user_keyword order by keyword, magnitude desc;
+---------+------+-----------+
| keyword | user | magnitude |
+---------+------+-----------+
| mysql | 1 | 4 |
| mysql | 2 | 2 |
| mysql | 4 | 2 |
| mysql | 3 | 1 |
| mysql | 5 | 1 |
| query | 2 | 3 |
| query | 3 | 2 |
| query | 1 | 1 |
| table | 1 | 2 |
| table | 2 | 1 |
| table | 3 | 1 |
+---------+------+-----------+
Users ranked within keywords:
mysql> select * from keyword_user_ranked order by keyword, rank asc;
+---------+------+-----------+------+
| keyword | user | magnitude | rank |
+---------+------+-----------+------+
| mysql | 1 | 4 | 1 |
| mysql | 2 | 2 | 3 |
| mysql | 4 | 2 | 3 |
| mysql | 3 | 1 | 5 |
| mysql | 5 | 1 | 5 |
| query | 2 | 3 | 1 |
| query | 3 | 2 | 2 |
| query | 1 | 1 | 3 |
| table | 1 | 2 | 1 |
| table | 3 | 1 | 3 |
| table | 2 | 1 | 3 |
+---------+------+-----------+------+
Only top 2 from each keyword:
mysql> select * from keyword_user_ranked where rank <= 2 order by keyword, rank asc;
+---------+------+-----------+------+
| keyword | user | magnitude | rank |
+---------+------+-----------+------+
| mysql | 1 | 4 | 1 |
| query | 2 | 3 | 1 |
| query | 3 | 2 | 2 |
| table | 1 | 2 | 1 |
+---------+------+-----------+------+
Note that when there are ties -- see users 2 and 4 for keyword "mysql" in the examples -- all parties in the tie get the "last" rank, i.e. if the 2nd and 3rd are tied, both are assigned rank 3.
Performance: adding an index to the keyword and user columns will help. I have a table being queried in a similar way with 4000 and 1300 distinct values for the two columns (in a 600000-row table). You can add the index like this:
alter table results add index keyword_user (keyword, user);
In my case, query time dropped from about 6 seconds to about 2 seconds.
You can use a pattern like this (from Within-group quotas (Top N per group)):
SELECT tmp.ID, tmp.entrydate
FROM (
SELECT
ID, entrydate,
IF( #prev <> ID, #rownum := 1, #rownum := #rownum+1 ) AS rank,
#prev := ID
FROM test t
JOIN (SELECT #rownum := NULL, #prev := 0) AS r
ORDER BY t.ID
) AS tmp
WHERE tmp.rank <= 2
ORDER BY ID, entrydate;
+------+------------+
| ID | entrydate |
+------+------------+
| 1 | 2007-05-01 |
| 1 | 2007-05-02 |
| 2 | 2007-06-03 |
| 2 | 2007-06-04 |
| 3 | 2007-07-01 |
| 3 | 2007-07-02 |
+------+------------+
hi how can i count integers in specific column of mysql table?
+------------+--------+
| product_id | stores |
+------------+--------+
| 371374 | 1 |
| 283994 | 1 |
| 232191 | 2 |
| 131127 | 1 |
| 284000 | 1 |
| 371383 | 1 |
| 83 | 3 |
| 131156 | 1 |
| 371385 | 1 |
| 284004 | 1 |
+------------+--------+
i want to count all stores like 1 + 1 + 2 + 1 + ...
SELECT SUM(stores) FROM tablename
Try:
SELECT SUM(stores)
FROM mytable
try this query
select sum(stores) from <tablename>;
SELECT SUM( stores ) AS sum
FROM `table_name`