id qid answer date answer_userKey
72 2 2 2012-07-30 00:00:00 1
71 1 4 2012-07-30 00:00:00 1
70 2 2 2012-07-30 00:00:00 2
69 1 4 2012-07-30 00:00:00 2
68 2 2 2012-07-30 00:00:00 3
67 1 3 2012-07-30 00:00:00 3
66 2 2 2012-07-31 00:00:00 4
65 1 4 2012-07-31 00:00:00 4
64 2 2 2012-07-31 00:00:00 5
Here's my sample table, I need to get all data + all distinct answer_userKeys for every date like this
date DISTINCT(answer_userKey)
2012-07-30 3
2012-07-31 2
As well as all the single values in a normal associative array like you get when you do
SELECT * FROM tbl_data
WHERE date BETWEEN '2012-07-29' AND '2012-08-01'
Tried everything here :(
Try this:
SELECT DATE(date) dte, COUNT(DISTINCT answer_userKey) cnt
FROM tbl_data
WHERE DATE(date) BETWEEN '2012-07-29' AND '2012-08-01'
GROUP BY dte;
i hope this is your requirement..this one works in oracle
select to_char(date,'yyyy-mm-dd') date
,count(distinct answer_userKey) DISTINCT(answer_userKey)
from table_name
group by to_char(date,'yyyy-mm-dd')
If you need both the detail and aggregate data, then you will likely either need to perform two queries to get these two different sets of data, or simply query the details and build the aggregation within whatever language you are using (perhaps in a multidimensional array). For example in pseudo-code:
Query: SELECT * FROM tbl_data WHERE date BETWEEN '?' AND '?'
array; // multidimensional associative array to be populated
while ([fetch next row from result set as row] ) {
array[row['date']][row['answer_userKey']] = row
}
Related
I have a table where each quiz ID is repeated multiple times. there is a date in front of each quiz id in each row. I want to select entire row for each quiz ID where date is latest with user. The date format is mm/dd/YYYY.
Sample -
USER_ID Quiz_id Name Date Marks .. .. ..
1 2 poly 4/3/2020 27
1 2 poly 4/3/2019 98
1 4 moro 4/3/2020 09
2 5 cat 4/12/2015 87
2 4 moro 4/3/2009 56
2 6 PP 4/3/2011 76
3 2 poly 4/3/2020 12
3 2 poly 5/3/2020 09
3 7 dog 4/3/2011 23
I want result look like this:Result
USER_ID Quiz_id Name Date Marks .. .. ..
1 2 poly 4/3/2020 27
1 4 moro 4/3/2020 09
2 5 cat 4/12/2015 87
2 4 moro 4/3/2009 56
2 6 PP 4/3/2011 76
3 2 poly 5/3/2020 09
3 7 dog 4/3/2011 23
You can use rank function to get the desired result:
Demo
SELECT A.* FROM (
SELECT A.*, RANK() OVER(PARTITION BY USER_ID,QUIZ_ID, NAME ORDER BY DATE DESC) RN FROM
Table1 A ORDER BY USER_ID) A WHERE RN = 1 ORDER BY USER_ID, QUIZ_ID;
I don't have MySQL installed so you will need to test and report back. The general idea is to identify the row of interest using max and a group by (table t). As the Date column appears to be text column (MySQL uses the format YYYY-MM-DD for dates) you will need to convert it to a date with str_to_date() so you can use the max() aggregate function. Finally, join with the original table (here table t2 to do the date conversion), as only the aggregate column(s) and columns named in the group by are well defined (in table t1), i.e.:
select USER_ID, Quiz_id, Date, Marks from (
select USER_ID, Quiz_id, max(str_to_date(Date, '%m/%d/%Y')) as Date2 from quiz group by 1, 2
) as t natural join (
select *, str_to_date(Date, '%m/%d/%Y') Date2 from Quiz
) as t2;
I don't recall off-hand but Date might be reserved word, in which case you will need to quote the column name, or ideally rename said column to use a better name.
Also, the original table is not in 3rd normal form as Quiz_id depends on Name. Quiz_id, as implied, should be a foreign key to a lookup table that holds the Name.
I´m trying to do some analysis in the following data
WeekDay Date Count
5 06/09/2018 20
6 07/09/2018 Null
7 08/09/2018 19
1 09/09/2018 16
2 10/09/2018 17
3 11/09/2018 24
4 12/09/2018 25
5 13/09/2018 24
6 14/09/2018 23
7 15/09/2018 23
1 16/09/2018 9
2 17/09/2018 23
3 18/09/2018 33
4 19/09/2018 22
5 20/09/2018 31
6 21/09/2018 17
7 22/09/2018 10
1 23/09/2018 12
2 24/09/2018 26
3 25/09/2018 29
4 26/09/2018 27
5 27/09/2018 24
6 28/09/2018 29
7 29/09/2018 27
1 30/09/2018 19
2 01/10/2018 26
3 02/10/2018 39
4 03/10/2018 32
5 04/10/2018 37
6 05/10/2018 Null
7 06/10/2018 26
1 07/10/2018 11
2 08/10/2018 32
3 09/10/2018 41
4 10/10/2018 37
5 11/10/2018 25
6 12/10/2018 20
The problem that I want to solve is: I want to create a table with the average of the 3 last same weekdays related to the day. But, when there is a NULL in the weekday, I want to ignore and do the average only with the remain numbers, not count NULL as an 0. I will give you an example here:
The date in this table is day/month/year :)
Ex: On day 12/10/2018, I need the average from
the days 05/10/2018; 28/09/2018; 21/09/2018. These are the last 3 same weekday(six) as 12/10/2018.
. Their values are Null; 29; 17. Then the result of this average must be 23, because I need to ignore the NULL, and not be 15,333.
How can I do this?
The count() function ignores nulls (i.e. does NOT increment if it encounters null) so I suggest you simply count the values then may contain the nulls you wish to ignore.
dow datecol value
6 21/09/2018 17
6 28/09/2018 29
6 05/10/2018 Null
e.g. sum(value) above = 46, and the count(value) = 2 so the average is 23.0 (and avg(value) will also return 23.0 as it also ignores nulls)
select
weekday
, `date`
, `count`
, (select (sum(`count`) * 1.0) / (count(`count`) * 1.0)
from atable as t2
where t2.weekday = t1.weekday
and t2.`date` < t1.`date
order by t2.`date` DESC
limit 3
) as average
from atable as t1
You could just use avg(count) in the query above, and get the same result.
ps. I do hope you do NOT use count as a column name! I also would suggest you do NOT use date as a column name either. i.e. Avoid using SQL terms as names.
SELECT WeekDay, AVG(Count)
FROM myTable
WHERE Count IS NOT NULL
GROUP BY WeekDay
Use IsNULL(Count,0) in your Select
SELECT WeekDay, AVG(IsNULL(Count,0))
FROM myTable
GROUP BY WeekDay
First off, you need to get the number of instances of that weekday in the data since you just need the last 3 same week days
create table table2
as
select
row_number() over(partition by weekday order by date desc) as rn
,weekday
,date
,count
from table
From here, you can get what you want. With you explanation, you don't need to filter out the NULL values for count. Just doing the avg() aggregation will simply ignore it.
select
weekday
,avg(count)
from table2
where rn in (1,2,3)
group by weekday
I have to create a reports in one currency. I need to do query in MySQL without using PHP process. but unable to figure it out.
There is a table called currency_exchange_rate table as follows, (exchange rate in LKR to other currency).this table is updating like one record for each currency in LKR in every month
exchange_rates
id currency_id start_date exchange_rate
1 5 2017-01-2 155
2 4 2017-01-3 25
3 6 2017-01-3 53
4 5 2017-02-1 156
5 4 2017-02-1 24
6 6 2017-02-1 54
There is a project table as follows
pro_id name value currency_id status_id owner_id date
1 studio1 500 5 1 44 2017-01-20
2 lotus 120 5 1 42 2017-01-21
3 auro 300 4 2 45 2017-01-21
4 studio2 400 6 1 44 2017-01-22
5 holland 450 4 3 46 2017-02-05
6 studio3 120 4 3 47 2017-02-06
7 studio4 400 6 3 48 2017-02-06
how to generate reports in one currency(DKK but exchange rate in LKR) like status wise,monthly total, total by owner, etc..
and we have to consider currency id,currency to be convert and exchange rate for the month for those currency types to get relevant value for project row.
hope you are clear about my scenario. your help is much appreciated.
I don't need every report. just want a sql for convert values in project table using exchange rates table or status wise report as follows
status_id value_in_one_currency
1 xxxx
2 xxxx
3 xxxx
Try this:
SELECT A.status_id, A.`value` * B.exchange_rate `value_in_one_currency`
FROM project A JOIN exchange_rates B
ON A.currency_id=C.currency_id
AND DATE_FORMAT(A.`date`,'%m-%Y')=DATE_FORMAT(B.`start_date`,'%m-%Y');
See MySQL Join Made Easy for some insight.
This is what I finalize:
I took currency_id=5 as the final currency to be converted
SELECT A.*,C.exchange_rate AS DKK,D.exchange_rate AS LKR, (order_value * D.exchange_rate /C.exchange_rate ) AS `converted_value`
FROM projects A
LEFT JOIN exchange_rates C ON (DATE_FORMAT(C.start_date,'%Y-%m')=DATE_FORMAT(A.`date`,'%Y-%m') AND C.currency_id=5)
LEFT JOIN exchange_rates D ON DATE_FORMAT(D.start_date,'%Y-%m')=DATE_FORMAT(A.`date`,'%Y-%m') AND D.currency_id=A.currency_id
I need to extract the required fields from a table along with relevant time stamp
SELECT * FROM Glm_Test.LicenseUsage where FeatureId='2';
Output :
VendorId,FeatureId,Total_Lic_Installed,Total_Lic_Used,Reserved,CurrentTime
1 2 106 19 67 2015-12-15 15:00:05
1 2 106 19 67 2015-12-15 15:02:02
1 2 106 19 69 2015-12-15 15:04:02
1 2 106 19 67 2015-12-15 15:06:01
1 2 106 20 67 2015-12-15 15:08:02
select VendorId,FeatureId,Total_Lic_Installed,Max(Total_Lic_Used),Reserved,CurrentTime from Glm_Test.LicenseUsage where FeatureId= '2' group by VendorId,FeatureId;
output:
1 2 106 20 69 2015-12-15 15:00:05
In the above 2 queries
1st query lists all entries from the table
and i want second query to return time stamp for the MAX value of column Total_Lic_Used but somehow it is returning me only timestamp of the first entry.
Help is much appreciated.
Selecting the columns which are not part of an aggregation function like count/max/min/sum... or not in group by clause will give unexpected results:
Other RBBMS wont allow these statements(gives error like):
sql server ==> the select list because it is not contained in either
an aggregate function or the GROUP BY clause
Oracle ==>not a GROUP BY expression
You can do this by a sub query and join
select
a.VendorId,
a.FeatureId,
a.Total_Lic_Installed,
b.max_Total_Lic_Used,
a.Reserved,
a.CurrentTime
from Glm_Test.LicenseUsage a
join (
select
VendorId,
FeatureId,
Max(Total_Lic_Used) max_Total_Lic_Used
from Glm_Test.LicenseUsage
where FeatureId = '2'
group by VendorId, FeatureId
) b
on a.VendorId = b.VendorId and
a.FeatureId = b.FeatureId and
a.Total_Lic_Used = b.max_Total_Lic_Used
sql fiddle demo
You can try this also;
select
`VendorId`,
`FeatureId`,
`Total_Lic_Installed`,
`Total_Lic_Used`,
`Reserved`,
`CurrentTime`
from Glm_Test.LicenseUsage
order by Total_Lic_Used desc
limit 1
demo
Fixture Table
uid home_uid away_uid winner date season_division_uid
1 26 6 6 2013-07-30 18
2 8 21 8 2013-06-30 18
3 6 8 8 2013-06-29 18
4 21 26 21 2013-05-20 18
5 6 26 6 2013-04-19 18
This table contains hundreds of rows.
Currently I have a query to select all the teams in a division, i.e.
SELECT team_uid
FROM Season_Division_Team
WHERE season_division_uid='18'
which lists the rows of team uid's i.e. [6,26,8,21,26].
Now for each of the unique team ids, I would like to return the last 3 winner values, ordered by the date column, that they were involved in (they could be an away_uid or home_uid).
So the returned value example would be:
team_id winner date
6 6 2013-07-30
6 8 2013-06-29
6 26 2013-04-19
26 6 2013-07-30
26 21 2013-05-20
26 6 2013-04-19
Any ideas? Thank you
Im not sure how to get it direct, a query like
select * from Season_division_Team where
`date >= (select min(`date`) from
(select `date` from season_division_team order by date desc limit 3))
and (home_uid = 6 or away_uid = 6)
Thats not going to be a good query. But only way i can think of currently
Its hard to get the 3rd largest value from SQL Example
the sub query is trying to get the date where the last win occured, and then getting all dates after that where the team played.
EDIT:
SELECT * FROM Season_Division_Team WHERE winner = 6 ORDER BY `date` DESC LIMIT 3
that sounds more like your latter comment