how to find the number of consecutive repeats - mysql

So I'm trying to write a mysql script to find the number of consecutive repeats in 'value' column of this table.
id value result
-- ----- ------
1 1 0
2 1 1
3 2 0
4 3 0
5 3 1
So in this case I want get the value 2

Get the next value using user variables,
GROUP so consecutive values more than 2 are not counted again,put all in a subquery,and use a simple CASE to increment the value you need in case value=next value.Add salt and pepper.
SELECT SUM(CASE WHEN y.value=y.next_value THEN #var+1 ELSE #var END) consecIds
FROM
(SELECT t.id, t.value, next_id, n.value next_value
FROM
(
SELECT t.id, t.value,
(
SELECT id
FROM table1
WHERE id > t.id
ORDER BY id
LIMIT 1
) next_id
FROM table1 t,(SELECT #var:=0)x
) t LEFT JOIN table1 n
ON t.next_id = n.id
GROUP BY t.value,n.value)y
FIDDLE

SELECT COUNT(DISTINCT column_name) FROM table_name;
DISTINCT will erase duplicated repetitions from specified column in result.
COUNT will count the rows in result.
The COUNT(DISTINCT column_name) function returns the number of distinct values of the specified column.

Related

Add Additional Index to Select Query & Fetch result using greater than Index ID

I have a table in which right now I am fetching result by adding a additional index but I want to fetch result using greater than operator to Additional index id
This query gives me this
SET #a:=0;
SELECT #a:=#a+1 additional_id, output.*
FROM (SELECT sum(item_sale + item_viewed) as totalSum ,item_id FROM items WHERE item_active='1' GROUP BY item_id order by totalSum desc ) output
additional_id
item_id
totalSum
1
3
17
2
1
5
3
2
2
But i want to use greater than operator and want result like this if additional_id > 1 then find only 2 result
additional_id
item_id
totalSum
2
1
5
3
2
2
How could i achieve this ?
SELECT *
FROM (
-- your query
SELECT #a:=#a+1 additional_id, output.*
FROM (SELECT sum(item_sale + item_viewed) as totalSum ,item_id FROM items WHERE item_active='1' GROUP BY item_id order by totalSum desc ) output
-- end of your query
CROSS JOIN (SELECT #a:=0) init_variable
) AS subquery
WHERE additional_id > 1

MYSQL - Select Only Records Where Previous Record Column Data Differs

Hoping this is possible with just sql. I have a query that returns a data set with time_stamp and hash_index columns. Basically something to the effect of:
1 1583365548 6ff11ad5536f28d66098f6d74f97d877
2 1583365554 6ff11ad5536f28d66098f6d74f97d877
3 1583365556 6ff11ad5536f28d66098f6d74f97d877
4 1583365562 a2e99acb2540d49955ef93fb2684ac25
5 1583365571 a2e99acb2540d49955ef93fb2684ac25
6 1583365572 a2e99acb2540d49955ef93fb2684ac25
7 1583365574 a2e99acb2540d49955ef93fb2684ac25
8 1583365578 a2e99acb2540d49955ef93fb2684ac25
9 1583365580 a2e99acb2540d49955ef93fb2684ac25
What I want to do is further filter this query to only include the record if the hash_index differs from the previous record. Is this something I can do without having to dump it into PHP and loop through it?
My current query is below:
SELECT
(#cnt:=#cnt + 1) AS row_number,
time_stamp,
MD5(GROUP_CONCAT(CONCAT(user_state_name,
option_id,
option_code,
item_id,
item_code))) AS hash_index
FROM
user_state
WHERE
user_id = 2
GROUP BY
time_stamp;
What is with a query like this:
SELECT t1.*
FROM user_state t1
LEFT JOIN user_state t2 ON t1.id-1 = t2.id
WHERE t1.hash_index <> t2.hash_index;
If you want to filter out adjacent duplicates, I would just use lag() and dispense with hashing and aggregation:
SELECT us.*
FROM (SELECT us.*,
LAG(time_stamp) OVER (PARTITION BY user_id ORDER BY time_stamp) as prev_ts,
LAG(time_stamp) OVER (PARTITION BY user_id, user_state_nae, option_id, option_code, item_id, item_code ORDER BY time_stamp) as prev_ts_values
FROM user_state us
WHERE user_id = 2
) t
WHERE prev_ts_values is null or prev_ts_values <> prev_ts;
You can select whichever rows you want.

How can I get all affected row ids in mysql group by sentence?

I am doing a selection using GROUP BY, so I get a lot of different rows. What I need is to UPDATE the status field of each row affected by the SELECT query, but it is grouped... so I don't know all the ids, just the MAX(id).
This is the select query:
SELECT d.*, n.* FROM
(SELECT MAX(id) as id, MAX(datetime) as datetime, COUNT(DISTINCT content) as total FROM user_notifications
WHERE id_user
GROUP BY id_ref, type
ORDER BY datetime DESC) d
JOIN user_notifications n USING (id)
Edit:
Simple example, this is the table
id user content status
1 1 aaa 0
2 1 aaa 0
3 1 bbb 0
4 2 aaa 0
5 3 bbb 0
this is the query
select max(id), user, content from table where user=1 group by content
this is the result
id user content
2 1 aaa
3 1 bbb
in this query in fact sql internally select all user=1 (id= 1, 2, 3) and then the query is grouped, showing just two rows.
So, I want to update all ids involved in the query (id= 1 , 2 and 3)
Based on your example, you need the result exact to this query:
select id from `table` where user = 1;
But by using the query you provided which groups record by content
select max(id), user, content from `table` where user = 1 group by content
So, in such case you can join same table with derived table:
select id from `table` as t
join (select max(id), user, content from `table` where user = 1 group by content) as dt
on dt.content = t.content and dt.user = t.user;
This query is now equivalent to first straight forward query.

How do I find duplicate values across multiple columns in Mysql?

I have a table like this
I want to check the all rows in Column A with column B and get the count of duplicates.
For example, I want to get the
count of 12 as 3(2 times in A+1 time in B)
count of 11 as 2(2 times in A+0 time in B)
count of 13 as 2(1 time in A+0 time in B)
How can I acheive it?
You can calculate the total occurrences from a union all. A where clause can show only the values that occur in the A column:
select nr
, count(*)
from (
select A as nr
from YourTable
union all
select B
from YourTable
) sub
where nr in -- only values that occur at least once in the A column
(
select A
from YourTable
)
group by
nr
having count(*) > 1 -- show only duplicates
You can combine all values in A and B then do the group by.
Then only select those values found in column A.
Select A, count(A) as cnt
From (
Select A
from yourTable
Union All
Select B
from yourTable) t
Where t.A in
(select distinct A from yourTable)
Group by t.A
Order by t.A;
Result:
A cnt
11 2
12 3
13 1
See demo: http://sqlfiddle.com/#!9/9fcfe9/3

MySql Sum and Count for simple table

Could you help me with simple table SUM and COUNT calculating?
I've simple table 'test'
id name value
1 a 4
2 a 5
3 b 3
4 b 7
5 b 1
I need calculate SUM and Count for "a" and "b". I try this sql request:
SELECT name, SUM( value ) AS val, COUNT( * ) AS count FROM `test`
result:
name val count
a 20 5
But should be
name val count
a 9 2
b 11 3
Could you help me with correct sql request?
Add GROUP BY. That will cause the query to return a count and sum per group you defined (in this case, per name).
Without GROUP BY you just get the totals and any of the names (in your case 'a', but if could just as well have been 'b').
SELECT name, SUM( value ) AS val, COUNT( * ) AS count
FROM `test`
GROUP BY name
You need group by
select
name,
sum(value) as value,
count(*) as `count`
from test group by name ;