MySQL SELECT query -> UPDATE (numbering grouped by rows) - mysql

This query selects the desired result as it should: it puts row numbers for each id, grouped by id, starting from 1 again at the each id:
SELECT `id_product`,
`lowest_price`,
IF(#CG=`id_product`, #I:=#I+1, #I:=1) AS rownr,
#CG:=`id_product` AS run_order
FROM `table`
INNER JOIN (SELECT #I:=0, #CG:=-1) x
ORDER BY id_product ASC, lowest_price ASC
I am having trouble updating the table:
UPDATE table SET row_number = .....
Solved it with temporary table, but is there any more neat solution?

Related

My sql select that has multiple row in the same criteria

I have mysql table like this
I want to get row that has minimum 2 or more than 2 (multiple) row only from this table, so the result would be like this
What do i do?
thank you
Use GROUP BY and HAVING clauses
SELECT t.* FROM my_table t
JOIN (
SELECT cust_id, MIN(transaction_no) AS transaction_no
FROM my_table
GROUP BY cust_id
HAVING COUNT(cust_id) > 1
) agg ON t.transaction_no = agg.transaction_no

run a query if a value on the last row is correct

Id like to run a query only if a value in the last row is correct. In my exemple if the value in ColumnA is 1 on the last row then i want to run MyQuery. But if the value is not 1 stop there and do nothing.
i've try with case and count(*) and also with If exists. but i keep getting error
SELECT CASE WHEN ((SELECT COUNT(*) FROM
(Select a.* from table as a order by a.index desc limit 1) as b
where b.ColumnA = 1)) > 0 )
THEN (MyQuery)
END
i've also try with if exists but it doesn'work either
if exists Select b.* from (Select a.* from table as a order by a.index desc limit 1) where b.ColumnA = 1
begin
(MyQuery)
end
can you point me what wrong in those query or maybee there's a better way to achive this.
EDIT. This query will be run on a trigger after each insert in that table the goal is to avoid running MyQuery on row that dont required it. MyQuery is a bit slow and most row dont required it to run.
I think we can rephrase your logic here to make it work as you want:
WITH cte AS (
SELECT ColumnA, ROW_NUMBER() OVER (ORDER BY index DESC) rn
FROM yourTable
)
(your query here)
WHERE (SELECT ColumnA FROM cte WHERE rn = 1) = 1;
The WHERE clause above would return either true or false, and would apply to all records in the potential result set from your query. That is, if the ColumnA value from the "last" record were 1, then you would get back the entire result set, otherwise it would be empty set.
Assuming your version of MariaDB supports neither ROW_NUMBER nor CTEs, then use:
(your query here)
WHERE (SELECT ColumnA FROM yourTable ORDER BY index DESC LIMIT 1) = 1;
It depends on what your query is.
INSERT ...
SELECT ... WHERE ... -- this could lead to zero rows being inserted
DELETE ...
WHERE NOT EXISTS ( SELECT ... ) -- this could lead to zero rows being deleted
UPDATE t1 JOIN t2 ... -- the JOIN may cause no rows to be updated
Note:
(Select a.* from table as a order by a.index desc limit 1) as b
where b.ColumnA = 1)) > 0 )
can be simplified (and sped up) to
( ( SELECT ColumnA FROM table ORDER BY index DESC LIMIT 1 ) = 1 )
Note that that is a true/false "expression", so it can be used in various places.

Query to fetch second highest salary from table [duplicate]

What is the simplest SQL query to find the second largest integer value in a specific column?
There are maybe duplicate values in the column.
SELECT MAX( col )
FROM table
WHERE col < ( SELECT MAX( col )
FROM table )
SELECT MAX(col)
FROM table
WHERE col NOT IN ( SELECT MAX(col)
FROM table
);
In T-Sql there are two ways:
--filter out the max
select max( col )
from [table]
where col < (
select max( col )
from [table] )
--sort top two then bottom one
select top 1 col
from (
select top 2 col
from [table]
order by col) topTwo
order by col desc
In Microsoft SQL the first way is twice as fast as the second, even if the column in question is clustered.
This is because the sort operation is relatively slow compared to the table or index scan that the max aggregation uses.
Alternatively, in Microsoft SQL 2005 and above you can use the ROW_NUMBER() function:
select col
from (
select ROW_NUMBER() over (order by col asc) as 'rowNum', col
from [table] ) withRowNum
where rowNum = 2
I see both some SQL Server specific and some MySQL specific solutions here, so you might want to clarify which database you need. Though if I had to guess I'd say SQL Server since this is trivial in MySQL.
I also see some solutions that won't work because they fail to take into account the possibility for duplicates, so be careful which ones you accept. Finally, I see a few that will work but that will make two complete scans of the table. You want to make sure the 2nd scan is only looking at 2 values.
SQL Server (pre-2012):
SELECT MIN([column]) AS [column]
FROM (
SELECT TOP 2 [column]
FROM [Table]
GROUP BY [column]
ORDER BY [column] DESC
) a
MySQL:
SELECT `column`
FROM `table`
GROUP BY `column`
ORDER BY `column` DESC
LIMIT 1,1
Update:
SQL Server 2012 now supports a much cleaner (and standard) OFFSET/FETCH syntax:
SELECT [column]
FROM [Table]
GROUP BY [column]
ORDER BY [column] DESC
OFFSET 1 ROWS
FETCH NEXT 1 ROWS ONLY;
I suppose you can do something like:
SELECT *
FROM Table
ORDER BY NumericalColumn DESC
LIMIT 1 OFFSET 1
or
SELECT *
FROM Table ORDER BY NumericalColumn DESC
LIMIT (1, 1)
depending on your database server. Hint: SQL Server doesn't do LIMIT.
The easiest would be to get the second value from this result set in the application:
SELECT DISTINCT value
FROM Table
ORDER BY value DESC
LIMIT 2
But if you must select the second value using SQL, how about:
SELECT MIN(value)
FROM ( SELECT DISTINCT value
FROM Table
ORDER BY value DESC
LIMIT 2
) AS t
you can find the second largest value of column by using the following query
SELECT *
FROM TableName a
WHERE
2 = (SELECT count(DISTINCT(b.ColumnName))
FROM TableName b WHERE
a.ColumnName <= b.ColumnName);
you can find more details on the following link
http://www.abhishekbpatel.com/2012/12/how-to-get-nth-maximum-and-minimun.html
MSSQL
SELECT *
FROM [Users]
order by UserId desc OFFSET 1 ROW
FETCH NEXT 1 ROW ONLY;
MySQL
SELECT *
FROM Users
order by UserId desc LIMIT 1 OFFSET 1
No need of sub queries ... just skip one row and select second rows after order by descending
A very simple query to find the second largest value
SELECT `Column`
FROM `Table`
ORDER BY `Column` DESC
LIMIT 1,1;
SELECT MAX(Salary)
FROM Employee
WHERE Salary NOT IN ( SELECT MAX(Salary)
FROM Employee
)
This query will return the maximum salary, from the result - which not contains maximum salary from overall table.
Old question I know, but this gave me a better exec plan:
SELECT TOP 1 LEAD(MAX (column)) OVER (ORDER BY column desc)
FROM TABLE
GROUP BY column
This is very simple code, you can try this :-
ex :
Table name = test
salary
1000
1500
1450
7500
MSSQL Code to get 2nd largest value
select salary from test order by salary desc offset 1 rows fetch next 1 rows only;
here 'offset 1 rows' means 2nd row of table and 'fetch next 1 rows only' is for show only that 1 row. if you dont use 'fetch next 1 rows only' then it shows all the rows from the second row.
Simplest of all
select sal
from salary
order by sal desc
limit 1 offset 1
select * from (select ROW_NUMBER() over (Order by Col_x desc) as Row, Col_1
from table_1)as table_new tn inner join table_1 t1
on tn.col_1 = t1.col_1
where row = 2
Hope this help to get the value for any row.....
Use this query.
SELECT MAX( colname )
FROM Tablename
where colname < (
SELECT MAX( colname )
FROM Tablename)
select min(sal) from emp where sal in
(select TOP 2 (sal) from emp order by sal desc)
Note
sal is col name
emp is table name
select col_name
from (
select dense_rank() over (order by col_name desc) as 'rank', col_name
from table_name ) withrank
where rank = 2
SELECT
*
FROM
table
WHERE
column < (SELECT max(columnq) FROM table)
ORDER BY
column DESC LIMIT 1
It is the most esiest way:
SELECT
Column name
FROM
Table name
ORDER BY
Column name DESC
LIMIT 1,1
As you mentioned duplicate values . In such case you may use DISTINCT and GROUP BY to find out second highest value
Here is a table
salary
:
GROUP BY
SELECT amount FROM salary
GROUP by amount
ORDER BY amount DESC
LIMIT 1 , 1
DISTINCT
SELECT DISTINCT amount
FROM salary
ORDER BY amount DESC
LIMIT 1 , 1
First portion of LIMIT = starting index
Second portion of LIMIT = how many value
Tom, believe this will fail when there is more than one value returned in select max([COLUMN_NAME]) from [TABLE_NAME] section. i.e. where there are more than 2 values in the data set.
Slight modification to your query will work -
select max([COLUMN_NAME])
from [TABLE_NAME]
where [COLUMN_NAME] IN ( select max([COLUMN_NAME])
from [TABLE_NAME]
)
select max(COL_NAME)
from TABLE_NAME
where COL_NAME in ( select COL_NAME
from TABLE_NAME
where COL_NAME < ( select max(COL_NAME)
from TABLE_NAME
)
);
subquery returns all values other than the largest.
select the max value from the returned list.
This is an another way to find the second largest value of a column.Consider the table 'Student' and column 'Age'.Then the query is,
select top 1 Age
from Student
where Age in ( select distinct top 2 Age
from Student order by Age desc
) order by Age asc
select age
from student
group by id having age< ( select max(age)
from student
)
order by age
limit 1
SELECT MAX(sal)
FROM emp
WHERE sal NOT IN ( SELECT top 3 sal
FROM emp order by sal desc
)
this will return the third highest sal of emp table
select max(column_name)
from table_name
where column_name not in ( select max(column_name)
from table_name
);
not in is a condition that exclude the highest value of column_name.
Reference : programmer interview
Something like this? I haven't tested it, though:
select top 1 x
from (
select top 2 distinct x
from y
order by x desc
) z
order by x
See How to select the nth row in a SQL database table?.
Sybase SQL Anywhere supports:
SELECT TOP 1 START AT 2 value from table ORDER BY value
Using a correlated query:
Select * from x x1 where 1 = (select count(*) from x where x1.a < a)
select * from emp e where 3>=(select count(distinct salary)
from emp where s.salary<=salary)
This query selects the maximum three salaries. If two emp get the same salary this does not affect the query.

MYSQL: Return results ONLY for the first true encountered SELECT expression

Desired result:
Return results ONLY for the first true encountered SELECT expression.
Explanation:
So, I have three different SELECT expresions:
SELECT * FROM table WHERE column1 = 'sometext' AND column2='1'
SELECT * FROM table WHERE column1 = 'someothertext' AND column2='2'
SELECT * FROM table WHERE column1 = 'somethirdtext' AND column2='3'
I want to have the results from 1. If 1. is returning NULL, I would like result from select number 2. If Select number 2. is returning NULL, I would like to use select number 3. and so on.
Please note that I am expecting more than one row to be returned for each condition that is true - and I only want the result from either SELECT 1) 2) or 3) (in that order)
It is important to only return results from the one single SELECT expression, so even if 2. and 3. would return something, I would only like results from 1.
The code I have right now is following that expected logic BUT when a I have more than one rows being returned by some of the below SELECTS, it gives me error:
1242 - Subquery returns more than 1 row
The code right now:
SELECT IFNULL( (SELECT * FROM table WHERE column = 'sometext'), IFNULL( (SELECT * FROM table WHERE column = 'someothertext'), IFNULL( (SELECT * FROM table WHERE column = 'somethirdtext'), 0 ) ) )
You're looking for COALESCE function.
SELECT COALESCE(
(SELECT col FROM t WHERE `column` = 'sometext'),
(SELECT col FROM t WHERE `column` = 'someothertext'),
(SELECT col FROM t WHERE `column` = 'somethirdtext')
);
-please, note that subquery should not return more than 1 row/column.
I would approach this slightly differently, since you can only return one row per condition anyway, I would use the following to limit the number of selects done:
SELECT *
FROM table
WHERE column IN ('sometext', 'someothertext', 'somethirdtext')
ORDER BY CASE column
WHEN 'sometext' THEN 1
WHEN 'someothertext' THEN 2
WHEN 'somethirdtext' THEN 3
END
LIMIT 1;
As pointed out in the comments, you can use FIELD for the sort too:
SELECT *
FROM table
WHERE column IN ('sometext', 'someothertext', 'somethirdtext')
ORDER BY FIELD(column, 'sometext', 'someothertext', 'somethirdtext')
LIMIT 1;
I think you can get multiple rows per condition using the following:
SELECT T.*
FROM Table T
INNER JOIN
( SELECT Column
FROM Table
WHERE column IN ('sometext', 'someothertext', 'somethirdtext')
ORDER BY FIELD(column, 'sometext', 'someothertext', 'somethirdtext')
LIMIT 1
) MinT
ON MinT.Column = T.Column;
Basically the subquery MinT does the same as before, ordering by whichever condition matches. Then gets the value for the column of the first match and limits the whole table to this value.
Example on SQL Fiddle
SELECT t.*
FROM
( SELECT o.column1, o.column2
FROM
( SELECT 1 AS ord, 'sometext' AS column1, '1' AS column2 UNION ALL
SELECT 2, 'someothertext', '2' UNION ALL
SELECT 3, 'somethirdtext', '3'
) AS o
WHERE EXISTS
( SELECT 1
FROM table AS td
WHERE td.column1 = o.column1
AND td.column2 = o.column2
)
ORDER BY o.ord
LIMIT 1
) AS d
JOIN
table AS t
ON t.column1 = d.column1
AND t.column2 = d.column2 ;
MySQL isn't my daily db so I might be off on this, but can't you just use LIMIT 1 on your subqueries?

MySQL: check that a set of queries returns the same row count : : but I don't know what the count is

We read values from a set of sensors, occasionally a reading or two is lost for a particular sensor , so now and again I run a query to see if all sensors have the same record count.
GROUP BY sensor_id HAVING COUNT(*) != xxx;
So I run a query once to visually get a value of xxx and then run it again to see if any vary.
But is there any clever way of doing this automatically in a single query?
You could do:
HAVING COUNT(*) != (SELECT MAX(count) FROM (
SELECT COUNT(*) AS count FROM my_table GROUP BY sensor_id
) t)
Or else group again by the count in each group (and ignore the first result):
SELECT count, GROUP_CONCAT(sensor_id) AS sensors
FROM (
SELECT sensor_id, COUNT(*) AS count FROM my_table GROUP BY sensor_id
) t
GROUP BY count
ORDER BY count DESC
LIMIT 1, 18446744073709551615
SELECT sensor_id,COUNT(*) AS count
FROM table
GROUP BY sensor_id
ORDER BY count
Will show a list of the sensor_id along with a count of all the records it has, you can then manually check to see if any vary.
SELECT * FROM (
SELECT sensor_id,COUNT(*) AS count
FROM table
GROUP BY sensor_id
) AS t1
GROUP BY count
Will show all the counts that vary, but the group by will lose information about which sensor_ids have which counts.
---EDIT---
Taken a bit from both mine and eggyal's answer and created this, for the count that is most frequent I call the id default, and then for any values that stand out I have given them separate rows. This way you maintain the readability of a table if you have many results Multi Row, but also have a simple one row column if all counts are the same One Row. If however you are happy with the concocted strings then go with eggyal's answer.
Might be a bit over the top but here goes:
select 'default' as id,t5.c1 as count from(
select id,count(*) as c1 from your_table group by id having count(*)=
(select t4.count from
(
select max(t3.count2) as max,t3.count as count from
(
select count(*) as count2,t2.count from
(
SELECT id,COUNT(*) AS count
FROM your_table
GROUP BY id
) as t2
GROUP BY count
) as t3
) as t4)) as t5 group by count
union all
select t5.id as id,t5.c1 as count from(
select id,count(*) as c1 from your_table group by id having count(*)<>
(select t4.count from
(
select max(t3.count2) as max,t3.count as count from
(
select count(*) as count2,t2.count from
(
SELECT id,COUNT(*) AS count
FROM your_table
GROUP BY id
) as t2
GROUP BY count
) as t3
) as t4)) as t5