What I want to do with SQL is find repeating values in a column until there's an other value sorted by date.
This is my table. I use "CURRENT_TIMESTAMP" for the date column.
Name |date
-------------------
Bart |12-12-2014
Bart |23-12-2014
Joost |24-12-2014
Bart |25-12-2014
Bart |26-12-2014
Bart |27-12-2014
So in this example I want the number "3" returned and the last known value of name, so in this case "Bart".
I hope I made myself clear, sorry for the unclear title!
Hmmm . . . Here is one method where the logic is placed in the where clause:
select count(*), max(lastname.name)
from tablename t cross join
(select t2.name from tablename t2 order by date desc limit 1) as lastname
where t.name = lastname.name and
t.date > (select max(t2.date) from tablename t2 where t2.name <> lastname.name);
You can use the following query:
SELECT Name, COUNT(*) AS cnt
FROM (
SELECT Name, [date],
ROW_NUMBER() OVER (ORDER BY [date]) -
ROW_NUMBER() OVER (PARTITION BY Name ORDER BY [date]) AS grp
FROM mytable ) AS t
GROUP BY Name, grp
ORDER BY COUNT(*) DESC
This query tries to identify islands of records, i.e. consecutive rows having the same Name: grp calculated field does exactly this.
If you want the Name having the largest number of consecutive records, just use TOP 1 in the above query.
Demo here
Related
There is a table with fields - id, student_name, score.
How do a write a mysql/php query that will get the position
of a student based on the score field.
For example if we have scores: 60,57,80,90,85; student with score 90 will
have 1st position, 85 2nd position etc.
Two scores if equal will have same position nth, but the next score will
have n+2th position.
What does the SQL query look like?
You can use rank():
select t.*, rank() over (order by score desc) as rnk
from t;
If you want this for a particular student, use a subquery:
select t.*
from (select t.*, rank() over (order by score desc) as rnk
from t
) t
where studentid = ?
Dummy table:
id FileName DateLastSaved
1 Marium.doc 2015-01-01
2 Amna.doc 2016-01-01
3 Marium.doc 2016-01-01
I want the query to return such rows where FileName is unique in the whole table. Rows should be returned for particular date range.
Suppose date ranges are of 2016 only, so third row should not be returned as FileName is not unique.
The query that I have created is:
$presentquery="SELECT * FROM InitialLog i WHERE MDid='$MDid' AND
(DateLastSaved>='$firstdate' AND DateLastSaved<='$presentdate') AND NOT
EXISTS (SELECT id FROM InitialLog i2 WHERE i2.id<i.id AND i.FileName=i2.FileName )";
(Where $firstdate and $presentdate are 2 dates for date ranges)
The query is returning the accurate results but it's taking time to execute. Is there any other way that I can rewrite this query??
(I have table with many rows)
I put this query together and it returns the results very quickly.
Select *
FROM foo
Where (`datelastsaved` > '2015-12-31' && `datelastsaved` < '2017-01-01')
AND `filename` NOT IN (
Select `filename`
FROM foo
GROUP BY `filename`
HAVING COUNT(*) > 1);
The first part is your normal select statement with the where clauses to filter on the dates.
The second part is the NOT IN where the select statement finds all of the ones with duplicate filenames.
Select `filename` FROM foo GROUP BY `filename` HAVING COUNT(*) > 1)
You can get the same logic using a LEFT JOIN and looking for nulls, that is,
$presentquery = "SELECT DISTINCT i.* FROM InitialLog i
LEFT JOIN InitialLog i2 ON i2.id<i.id AND i.FileName=i2.FileName
WHERE i.MDid='$MDid'
AND i.DateLastSaved>='$firstdate'
AND i.DateLastSaved<='$presentdate'
AND i2.id IS NULL";
This way you are doing a single join rather than subquerying against each value in i.
It looks like you are trying to get the data associated with the first occurrence of each file name, this should work:
SELECT *
FROM InitialLog i
WHERE MDid='$MDid'
AND DateLastSaved>='$firstdate'
AND DateLastSaved<='$presentdate'
AND id IN (SELECT MIN(id) FROM InitialLog GROUP BY FileName)
;
Alternatively, you can do a JOIN with the same subquery instead:
SELECT i.*
FROM InitialLog AS i
INNER JOIN (SELECT MIN(id) AS id
FROM InitialLog
GROUP BY FileName
) AS firsts USING (id)
WHERE i.MDid='$MDid'
AND i.DateLastSaved>='$firstdate'
AND i.DateLastSaved<='$presentdate'
;
I have Problems with a select statement, as a little help here are the important columns:
Table1
ID NAME
TABLE 2
ID U_ID COUNTER
The ID of Table 1 Matches the U_ID of Table 2. Table 2 contains many entries for the same u_id.
What I want to do is to get the Name of the "user" (table 1) who has in sum the max. counter.
What I got since now is the join of the tables (Where clause depends on other rows which are not important for the problem).
Can anyone help me on this issue?
So what you need is an aggregate of an aggregate (max of sum of column). The easiest will be to create a view providing the sum and u_id end then select the max of it:
create view table2sums
as
select u_id, sum(counter) as total
from table2
group by u_id;
and then
select t1.name
from table1 t1, table2sums t2
where t1.id = t2.u_id
and t2.total >= all (
select total
from table2sums
)
In this special case you can also do it directly:
select t1.name
from table1 t1, table2 t2
where t1.id = t2.u_id
group by t1.name
having sum(t2.counter) >= all (
select sum(counter)
from table2
group by t2.u_id
)
NOTE: The other proposed solutions will show a better performance. My solution only selects the name (which is what you said you wanted) and works in any RDBMS.
There exist RDBMS without the LIMIT possibility.
In the end, I'd say: regard my solution as educational, the others as practical
SELECT name,
SUM(counter) as counter
FROM table1
JOIN table2
ON table1.id = table2.u_id
GROUP BY u_id
ORDER BY counter DESC
LIMIT 1
You can try this:
SELECT name, SUM(counter) as total_counter
FROM table1
JOIN table2
ON table1.id = table2.u_id
GROUP BY u_id
ORDER BY total_counter DESC
LIMIT 1
Working Demo: http://sqlfiddle.com/#!2/45419/4
I have a query that gets me a users rank in a table of scores.
SELECT
*
FROM
(SELECT
*, #rank:=#rank + 1 rank
FROM
(SELECT
user_id, SUM(round_total) TotalPoints
FROM
sx14sp_mem_picks
GROUP BY user_id) s, (SELECT #rank:=0) init
ORDER BY TotalPoints DESC) r
WHERE
user_id = 22234
There is a problem with ties. I have a table field "pick_date" that i would like to use to break ties with. The user who made his picks first beats the tie.
Any ideas?
If sx14sp_mem_picks.pickdate is the field to break ties then in the order by sx14sp_mem_picks subquery, add
min( pickdate) asc
This will put the earliest pickdate first - you have to use MIN() bc you need to use an aggregate function given the use of "group by".
You need to order by the pick date in addition to the total points. However, you are talking about multiple rows per user. So, let's take the last pick date:
SELECT *
FROM (SELECT *, (#rank:=#rank + 1) as rank
FROM (SELECT user_id, SUM(round_total) as TotalPoints, max(pick_date) as max_pick_date
FROM sx14sp_mem_picks
GROUP BY user_id
) s CROSS JOIN
(SELECT #rank := 0) init
ORDER BY TotalPoints DESC, max_pick_date asc
) r
WHERE user_id = 22234;
I have table
**id name status date**
1 john 2 01.01.2010
2 mike 5 04.01.2010
3 john 2 06.01.2010
4 sam 1 08.01.2010
john has status 2 twice and i need to select john,mike from this table where status = 2 but i need to show latest record.
I cannot use order by i use it already for something else.
You can use order by for multiple criteria like this:
ORDER BY date desc, status desc
You need to use a correlated subquery such as this:
select *
from table t1
where t1.date = ( select max( t2.date )
from table t2
where t1.name = t2.name
and t1.status = t2.status )
The query would go much faster if you didn't need the ID field:
SELECT t.name, t.status, max(t.date) date
FROM table t
GROUP BY t.name, t.status
ORDER BY [whatever]
If you DID need id, AND the ID is guarenteed to be larger on the record with the newer date, you could just add max(t.id) id to the field list.
SELECT *
FROM table t
WHERE status = 2
AND date = (SELECT MAX(date) FROM table tmp WHERE tmp.name = t.name GROUP BY name)