Select count of distinct column in table - mysql

Lets say I have a sql table in the format:
id INT PRIMARY KEY AUTO_INCREMENT,
action TINYINT NOT NULL,
user_id INT NOT NULL
I would like to build a query that produces the same result as the following:
SELECT
(SELECT COUNT(*), action FROM table WHERE action=1 AND user_id=1),
(SELECT COUNT(*), action FROM table WHERE action=2 AND user_id=1),
(SELECT COUNT(*), action FROM table WHERE action=3 AND user_id=1),
...
(SELECT COUNT(*), action FROM table WHERE action=n AND user_id=1)
(1<->n is simply the range of action)
But without all the redundancy and inefficiency.

If I understand the question correctly, this is what you're looking for:
SELECT action, COUNT(*)
FROM your_table
WHERE user_id = 1
GROUP BY action

You can use GROUP BY:
SELECT
COUNT(*) as Count,
action as Action
FROM table1
WHERE user_id = 1
GROUP BY action
This will produce a row for each unique action, and in that row have a column Count being the count of rows with that unique action.
In the example above, all user_id conditions are =1, so if this is case you can include it as in the query, otherwise you can form your own condition based on the needed values.

SELECT COUNT(id), action FROM table WHERE user_id=1 GROUP BY action
HTH, though I am a bit confused if this is what you are looking for.

Related

How to count occurrences with derived tables in SQL?

I have this very simple table:
CREATE TABLE MyTable
(
Id INT(6) PRIMARY KEY,
Name VARCHAR(200) /* NOT UNIQUE */
);
If I want the Name(s) that is(are) the most frequent and the corresponding count(s), I can neither do this
SELECT Name, total
FROM table2
WHERE total = (SELECT MAX(total) FROM (SELECT Name, COUNT(*) AS total
FROM MyTable GROUP BY Name) table2);
nor this
SELECT Name, total
FROM (SELECT Name, COUNT(*) AS total FROM MyTable GROUP BY Name) table1
WHERE total = (SELECT MAX(total) FROM table1);
Also, (let's say the maximum count is 4) in the second proposition, if I replace the third line by
WHERE total = 4;
it works.
Why is that so?
Thanks a lot
You can try the following:
WITH stats as
(
SELECT Name
,COUNT(id) as count_ids
FROM MyTable
GROUP BY Name
)
SELECT Name
,count_ids
FROM
(
SELECT Name
,count_ids
,RANK() OVER(ORDER BY count_ids DESC) as rank_ -- this ranks all names
FROM stats
) s
WHERE rank_ = 1 -- the most popular ```
This should work in TSQL.
Your queries can't be executed because "total" is no column in your table. It's not sufficient to have it within a sub query, you also have to make sure the sub query will be executed, produces the desired result and then you can use this.
You should also consider to use a window function like proposed in Dimi's answer.
The advantage of such a function is that it can be much easier to read.
But you need to be careful since such functions often differ depending on the DB type.
If you want to go your way with a sub query, you can do something like this:
SELECT name, COUNT(name) AS total FROM myTable
GROUP BY name
HAVING COUNT(name) =
(SELECT MAX(sub.total) AS highestCount FROM
(SELECT Name, COUNT(*) AS total
FROM MyTable GROUP BY Name) sub);
I created a fiddle example which shows both queries mentioned here will produce the same and correct result:
db<>fiddle

I want to pull distinct values for 2 columns in the same table and associated unique value columns for these unique values in SQL

The table is something like:
User_id Bidid timestamp ... (about 25-30 more columns)
How do I pull all the distinct user ids with all the distinct Bidid's associated with them and also the timestamp which is unique to each Bidid?
select distinct comes to mind:
select distinct User_id, Bidid, timestamp
from t;
I think this could be idea you are looking for, it depends of the version SQL/environment/etc:
SELECT USER_ID, BIDID, TIMESTAMP, ...
FROM THE_TABLE
WHERE BIDID=(SELECT DISTINCT BIDIT FROM THE_TABLE)
or
SELECT USER_ID, BIDID, TIMESTAMP,..
FROM THE_TABLE
WHERE BIDID IN (SELECT DISTINCT BIDIT FROM THE_TABLE)

Insert empty row after group

I have a table that contains transaction data. The rows with the same 'group_id' are a part of the same transaction. I am running the following SQL query to show all the transactions:
SELECT * FROM transactions
When I run this query I get as expected a list of all the transactions. But this large list makes it difficult to seperate the data with a different group_id from the other data.
For that reason I want to add an empty row at the end of the group_id, so I get:
1
1
(empty row)
2
2
2
instead of:
1
1
2
2
2
Can someone help me with this?
Here is my database:
http://sqlfiddle.com/#!9/b9bf79/1
I do not suggest you do this at all but if you just want to separate two groups you could do this:
SELECT * FROM transactions WHERE group_id = 1
UNION ALL
(SELECT '','','','','','')
UNION ALL
SELECT * FROM transactions WHERE group_id = 2
Obviously this can added to if there are more group ids in the future but it is not a general purpose solution you are really better off dealing with appearance issues like this in application code.
you can use (abuse) rollup.
SELECT *
FROM transactions
group by group_id, id
with rollup
having group_id is not null
this will insert a row with id set to null after each group_id.
mysql will also sort by group_id because of the group by.
The group by id` makes sure that all rows are shown (your schema does not show it, but I assume id is unique? Otherwise you need to add other fields)
However only id will be null in the extra rows. The other columns repeat the value above.
You can filter them like this:
SELECT
id,
case id is not null when true then date else null end as date,
case id is not null when true then group_id else null end as group_id
-- ....
FROM transactions
group by group_id, id
with rollup
having group_id is not null
Alternatively:
select * from
(SELECT *
FROM transactions
union all
select distinct null, null, group_id, null, null,null from transactions
) as t
order by 3,1
but null values are sorted first, so the "gap" is in front of each section

Order a Table in MySql without sendtime

How can I query to find the most recent message in a table, when the values in the column sendTime are all null?
I have tried.
SELECT `from`
,MAX(column) AS most_recent_message
FROM table
GROUP BY `from`
if you have an auto increment primary key, then you can find the maximum primary key belongs to which column.
SELECT * FROM your_table WHERE id = (SELECT MAX(id) FROM your_table);

PHP SQL - Advanced delete query

I have a table with 3 columns: id, date and name. What I am looking for is to delete the records that have a duplicate name. The rule should be to keep the record that has the oldest date. For instance in the example below, there is 3 records with the name Paul. So I would like to keep the one that has the oldest date (id=1) and remove all the others (id = 4 and 6). I know how to make insert, update, etc queries, but here I do not see how to make the trick work.
id, date, name
1, 2012-03-10, Paul
2, 2012-03-10, James
4, 2012-03-12, Paul
5, 2012-03-11, Ricardo
6, 2012-03-13, Paul
mysql_query(?);
The best suggestion I can give you is create a unique index on name and avoid all the trouble.
Follow the steps as Peter Kiss said from 2 to 3. Then do this
ALTER Table tablename ADD UNIQUE INDEX name (name)
Then Follow 4 Insert everything from the temporary table to the original.
All the new duplicate rows, will be omitted
Select all the records what you want to keep
Insert them to a temporary table
Delete everything from the original table
Insert everything from the temporary table to the original
Like Matt, but without the join:
DELETE FROM `table` WHERE `id` NOT IN (
SELECT `id` FROM (
SELECT `id` FROM `table` GROUP BY `name` ORDER BY `date`
) as A
)
Without the first SELECT you will get "You can't specify target table 'table' for update in FROM clause"
Something like this would work:
DELETE FROM tablename WHERE id NOT IN (
SELECT tablename.id FROM (
SELECT MIN(date) as dateCol, name FROM tablename GROUP BY name /*select the minimum date and name, for each name*/
) as MyInnerQuery
INNER JOIN tablename on MyInnerQuery.dateCol = tablename.date
and MyInnerQuery.name = tablename.name /*select the id joined on the minimum date and the name*/
) /*Delete everything which isn't in the list of ids which are the minimum date fore each name*/
DELETE t
FROM tableX AS t
LEFT JOIN
( SELECT name
, MIN(date) AS first_date
FROM tableX
GROUP BY name
) AS grp
ON grp.name = t.name
AND grp.first_date = t.date
WHERE
grp.name IS NULL
DELETE FROM thetable tt
WHERE EXISTS (
SELECT *
FROM thetable tx
WHERE tx.thename = tt.thename
AND tx.thedate > tt. thedate
);
(note that "date" is a reserver word (type) in SQL, "and" name is a reserved word in some SQL implementations)