I have a database with 2 columns (ex: no_1, no_2) and then 5000 rows of data, numbers are between 1 - 20 , I need to find a pairs of numbers which where repeating the most of the times?
Any help please ?
Something like this maybe:-
SELECT no_1, no_2, COUNT(*) AS numbercount
FROM SomeTable
GROUP BY no_1, no_2
ORDER BY numbercount DESC
Related
I am trying to find the count of how many times a location is used in my table of search results, but I want to group together cases where the same post code area start is used. The following groups purely on the text used:
SELECT count(*),
search_browse_log.postcode_start
FROM search_browse_log
GROUP BY postcode_start
ORDER BY count(*) DESC
But in the data, I have for example CR0, CR1, CR2 (postcode starts). I want to group them all together so I have a count of 3 for "CR", rather than 1 each of CR0, CR1 and CR2.
Thanks in advance if you can help!
Use a conditional in the GROUP BY clause to get either 1 or 2 characters, depending on whether the postcode starts with 1 or 2 letters.
GROUP BY IF(postcode_start REGEXP '^[A-Z][A-Z]',
LEFT(postcode_start, 2),
LEFT(postcode_start, 1))
If it's only the first 2 characters of the postcode, you could use the LEFT function in the group by:
SELECT count(*),
LEFT(postcode_start,2) as `postcode_start`
FROM search_browse_log
GROUP BY LEFT(postcode_start,2)
ORDER BY count() DESC
To group by the non-numeric characters at the start of the string:
SELECT count(*),
IF(postcode_start REGEX '^[a-ZA-Z][0-9]',LEFT(postcode_start,1),LEFT(postcode_start,2)) as `postcode_start`
FROM search_browse_log
GROUP BY IF(postcode_start REGEX '^[a-ZA-Z][0-9]',LEFT(postcode_start,1),LEFT(postcode_start,2))
ORDER BY count() DESC
For reference, see https://dev.mysql.com/doc/refman/5.7/en/regexp.html#operator_regexp
I have a database that lists a few area codes, area code + office codes and some whole numbers and a action. I want it to return a result by the digits given but I am not sure how to accomplish it. I have some MySQL knowledge but its not very deep.
Here is a example:
match | action
_____________________
234 | goto 1
333743 | goto 2
8005551212| goto 3
234843 | goto 4
I need to query the database with a full 10 digit number -
query 8005551212 gives "goto 3"
query 2345551212 gives "goto 1"
query 3337431212 gives "goto 2"
query 2348431212 gives "goto 4"
This would be similar to the LIKE selection, but I need to match against the database value instead of the query value. Matching the full number is easy,
SELECT * FROM database WHERE `match` = 8005551212;
First the number to query will always be 10 digits, so I am not sure how to format the SELECT statement to differentiate the match of 234XXXXXXX and 234843XXXX, as I can only have one match return. Basically if it does not match the 10 digits, then it checks 6 digits, then it will check the 3 digits.
I hope this makes sense, I do not have any other way to format the number and it has to be accomplished with just a single SQL query and return over a ODCB connection in Asterisk.
Try this
SELECT match, action FROM mytable WHERE '8005551212' like concat(match,'%')
The issue is that you will get two rows in one case .. given your data..
SELECT action
FROM mytable
WHERE '8005551212' like concat(match,'%')
order by length(match) desc limit 1
That should get the row that had the most digits matched..
try this:
SELECT * FROM (
SELECT 3 AS score,r.* FROM mytable r WHERE match LIKE CONCAT(SUBSTRING('1234567890',1,3),'%')
UNION ALL
SELECT 6 AS score,r.* FROM mytable r WHERE match LIKE CONCAT(SUBSTRING('1234567890',1,6),'%')
UNION ALL
SELECT 10 AS score,r.* FROM mytable r WHERE match LIKE CONCAT(SUBSTRING('1234567890',1,10),'%')
) AS tmp
ORDER BY score DESC
LIMIT 1;
What ended up working -
SELECT `function`,`destination`
FROM reroute
WHERE `group` = '${ARG2}'
AND `name` = 0
AND '${ARG1}' LIKE concat(`match`,'%')
ORDER BY length(`match`) DESC LIMIT 1
I have table rozpis_riesitelov which contains columns :
id_rozpisu_riesit, id_zam, id_projektu, id_ulohy.
I made query :
select id_zam, id_ulohy, count(*) as counted
from rozpis_riesitelov
group by id_zam
having id_ulohy in (1,2,8)
which shows me id of employee (id_zam) and how many times He was in project (id_ulohy is irrevelant but I had to select it beacuse of having clause). It shows me everyone in db but I am looking for employee with ID of 4 who is in 6 projects (Yes, I could do order by but I want to see max). When I do max of this query like this:
select max(counted)
from (select id_zam, id_ulohy, count(id_zam) as counted
from rozpis_riesitelov
group by id_zam
having id_ulohy in (1,2,8)) as riesitel
which shows me number 149 instead of 6.
So basically I only need to find employee that occurs in the most of the projects.
What's wrong with sorting by the COUNT() value, and limiting to one result?
SELECT `id_zam`,
`id_ulohy`,
COUNT(*) AS `counted`
FROM `rozpis_riesitelov `
WHERE `id_ulohy` IN ( 1, 2, 8 )
GROUP BY `id_zam`
ORDER BY `counted` DESC
LIMIT 1
Not sure exactly what you are trying to accomplish but you only use HAVING to filter on your aggregate like this:
HAVING COUNT(*) > 1
you should be able to move the condition to a WHERE clause and get correct max returned:
select max(counted)
from (select id_zam, count(id_zam) as counted
from rozpis_riesitelov
where id_ulohy in (1,2,8)
group by id_zam) as riesitel
I want to count from the row with the least value to the row with a specific value.
For example,
Name / Point
--------------------
Pikachu / 7
Voltorb / 1
Abra / 4
Sunflora / 3
Squirtle / 8
Snorlax / 12
I want to count to the 7, so I get the returned result of '4' (counting the rows with values 1, 3, 4, 7)
I know I should use count() or mysql_num_rows() but I can't think of the specifics.
Thanks.
I think you want this :
select count(*) from mytable where Point<=7;
Count(*) counts all rows in a set.
If you're working with MySQL, then you could ORDER BY Point:
SELECT count(*) FROM table WHERE Point < 7 ORDER BY Point ASC
If you want to know all about ORDER BY, check out the w3schools page: http://www.w3schools.com/sql/sql_orderby.asp
Just in case you want to only count the rows based on the Point values:
SELECT count(*) FROM table WHERE Point < 7 GROUP BY Point
This may help you to get rows falling between range of values :
select count(*) from table where Point >= least_value and Point<= max_value
Im trying to get a result set that contains the 10 values that are closest to, in this case, the number 3.
I have a database that has values in a column named rated which can be 1,2,3,4 or 5. What im trying to do is query the database and return the first 10 rows that have the values closest to 3. The values can be above 3 or below 3. I should note that these values in the rated column are floats.
I then need to sort these rows in order so that rows with value of 3 are first and then the row with lowest offset (+/-) from 3.
Is there any SQL query that can return atleast the result set of values closest to 3 ? or am i going to have to return the whole db and sort it myself?
To get the first 10 rows with highest value down i used the statement
SELECT * FROM tabs ORDER BY 5 DESC LIMIT 10";
5 refers to the column rated
Is there some way to modify this to do what i want ?
Thanks
If I understand your problem correctly, this should do the trick:
select *
from tabs
order by abs(`rated` - 3) asc
limit 10
Note that it sorts by the difference in ascending order, so those with a difference of 0 will come first.
SELECT * FROM tabs ORDER BY ABS(3 - Rate) ASC LIMIT 10
If I got right what you need try:
select *
from (
select
case when -(3-rated) > 0 then -(3-rated) else (3-rated) end as distance,
tabs.*
from tabs
) subsel
order by distance
limit 10