I have a table 'exam_table' containing : User_ID, Exam_date, Exam_status.
Exam_status = ['Success' or 'Fail']
The question is :
Based on the above data, propose an SQL
query to finds the 5 candidates with the most failures. In case
of equality, we wish to obtain first the students whose date of first exam is the most distant in time.
I found the 5 candidates with the most failures but I still don't know how to sort them according to exam_date in case of equality.
Do you have any suggestions? Thank you in advance for helping !
Your order by is a clause which has ordering criteria separated by ,. So you can easily add another criteria, like below:
SELECT User_ID, count(exam_status) as nb_Failures
FROM exam_table
GROUP BY User_ID
ORDER BY nb_Failures, min(exam_date)
LIMIT 5;
UPDATED:
corrected by the date of the first exam:
SELECT
user_id,
MIN (exam_date) AS first_exam_date,
SUM (
CASE exam_status
WHEN 'Failed' THEN 1
ELSE 0
END
) AS nb_failures
FROM exam_table
GROUP BY user_id
ORDER BY nb_failures DESC, first_exam_date ASC
LIMIT 5;
or like this:
SELECT
user_id,
MIN (exam_date) AS first_exam_date,
COUNT(exam_status) AS nb_failures
FROM exam_table
WHERE exam_status = 'Failed'
GROUP BY user_id
ORDER BY nb_failures DESC, first_exam_date ASC
LIMIT 5;
PS: aggregate functions must also be applied to the date
PPS: but the first and second queries have different results. In the first, the date of the first exam is selected, in principle, it does not matter if it is successful or not. The second selects only the date of the first failed exam.
Related
DO we have a way to get first record considering the time.
example
get first record today, get first record yesterday, get first record day before yesterday ...
Note: I want to get all records considering the time
sample expected output should be
first_record_today,
first_record_yesterday,..
As I understand the question, the "first" record per day is the earliest one.
For that, we can use RANK and do the PARTITION BY the day only, truncating the time.
In the ORDER BY clause, we will sort by the time:
SELECT sub.yourdate FROM (
SELECT yourdate,
RANK() OVER
(PARTITION BY DATE_TRUNC('DAY',yourdate)
ORDER BY DATE_TRUNC('SECOND',yourdate)) rk
FROM yourtable
) AS sub
WHERE sub.rk = 1
ORDER BY sub.yourdate DESC;
In the main query, we will sort the data beginning with the latest date, meaning today's one, if available.
We can try out here: db<>fiddle
If this understanding of the question is incorrect, please let us know what to change by editing your question.
A note: Using a window function is not necessary according to your description. A shorter GROUP BY like shown in the other answer can produce the correct result, too and might be absolutely fine. I like the window function approach because this makes it easy to add further conditions or change conditions which might not be usable in a simple GROUP BY, therefore I chose this way.
EDIT because the question's author provided further information:
Here the query fetching also the first message:
SELECT sub.yourdate, sub.message FROM (
SELECT yourdate, message,
RANK() OVER (PARTITION BY DATE_TRUNC('DAY',yourdate)
ORDER BY DATE_TRUNC('SECOND',yourdate)) rk
FROM yourtable
) AS sub
WHERE sub.rk = 1
ORDER BY sub.yourdate DESC;
Or if only the message without the date should be selected:
SELECT sub.message FROM (
SELECT yourdate, message,
RANK() OVER (PARTITION BY DATE_TRUNC('DAY',yourdate)
ORDER BY DATE_TRUNC('SECOND',yourdate)) rk
FROM yourtable
) AS sub
WHERE sub.rk = 1
ORDER BY sub.yourdate DESC;
Updated fiddle here: db<>fiddle
I have a question regarding the max-statement in a select -
Without the MAX-statemen i have this select:
SELECT stockID, DATE, close, symbol
FROM ta_stockprice JOIN ta_stock ON ta_stock.id = ta_stockprice.stockID
WHERE stockid = 8648
ORDER BY close
At the end i only want to have the max row for the close-column so i tried:
Why i didn´t get date = "2021-07-02" as output?
(i saw that i allways get "2021-07-01" as output - no matter if i use MAX / MIN / AVG...)
The MAX() turns the query into an aggregation query. With no GROUP BY, it returns one row. But the query is syntactically incorrect, because it mixes aggregated and unaggregated columns.
Once upon a time, MySQL allowed such syntax in violation of the SQL Standard but returned values from arbitrary rows for the unaggreged columns.
Use ORDER BY to do what you want:
SELECT stockID, DATE, close, symbol
FROM ta_stockprice JOIN ta_stock ON ta_stock.id = ta_stockprice.stockID
WHERE stockid = 8648
ORDER BY close DESC
LIMIT 1;
I have been trying to write a query to perfect this instance but cant seem to do the trick because I am still receiving duplicated. Hoping I can get help how to fix this issue.
SELECT DISTINCT
1.Client
1.ID
1.Thing
1.Status
MIN(1.StatusDate) as 'statdate'
FROM
SAMPLE 1
WHERE
[]
GROUP BY
1.Client
1.ID
1.Thing
1.status
My output is as follows
Client Id Thing Status Statdate
CompanyA 123 Thing1 Approved 12/9/2019
CompanyA 123 Thing1 Denied 12/6/2019
So although the query is doing what I asked and showing the mininmum status date per status, I want only the first status date. I have about 30k rows to filter through so whatever does not run overload the query and have it not run. Any help would be appreciated
Use window functions:
SELECT s.*
FROM (SELECT s.*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY statdate) as seqnum
FROM SAMPLE s
WHERE []
) s
WHERE seqnum = 1;
This returns the first row for each id.
Use whichever of these you feel more comfortable with/understand:
SELECT
*
FROM
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY statusdate) as rn
FROM sample
WHERE ...
) x
WHERE rn = 1
The way that one works is to number all rows sequentially in order of StatusDate, restarting the numbering from 1 every time ID changes. If you thus collect all the number 1's togetyher you have your set of "first records"
Or can coordinate a MIN:
SELECT
*
FROM
sample s
INNER JOIN
(SELECT ID, MIN(statusDate) as minDate FROM sample WHERE ... GROUP BY ID) mins
ON s.ID = mins.ID and s.StatusDate = mins.MinDate
WHERE
...
This one prepares a list of all the ID and the min date, then joins it back to the main table. You thus get all the data back that was lost during the grouping operation; you cannot simultaneously "keep data" and "throw away data" during a group; if you group by more than just ID, you get more groups (as you have found). If you only group by ID you lose the other columns. There isn't any way to say "GROUP BY id, AND take the MIN date, AND also take all the other data from the same row as the min date" without doing a "group by id, take min date, then join this data set back to the main dataset to get the other data for that min date". If you try and do it all in a single grouping you'll fail because you either have to group by more columns, or use aggregating functions for the other data in the SELECT, which mixes your data up; when groups are done, the concept of "other data from the same row" is gone
Be aware that this can return duplicate rows if two records have identical min dates. The ROW_NUMBER form doesn't return duplicated records but if two records have the same minimum StatusDate then which one you'll get is random. To force a specific one, ORDER BY more stuff so you can be sure which will end up with 1
I have "tasks" table with 3 fields:
date
priority (0,1,2)
done (0,1)
What I am trying to achieve is with the whole table sorted by done flag, tasks that are not done should be sorted by priority, while tasks that are done should be sorted by date:
Select * from tasks order by done asc
If done=0 additionally order by priority desc
If done=1 additionally order by date desc
Is it possible to do this in MySQL without unions?
Thanks.
You could try ORDER BY (done asc, aux desc) where aux is computed with a CASE to yield either priority or date based on the value of done (you may have to cast them to the same type to fit in the same expression, e.g. cast the date to a suitable integer day number).
For example:
SELECT * FROM tab
ORDER BY done desc,
case done
when 0 then prio
else to_days(thedate)
end desc;
Taken from Alex Martelli just shortened a bit with IF() and fixed the ASC/DESC ordering
SELECT * FROM tab
ORDER BY done ASC, IF(done, to_days(thedate), prio) DESC;
I have a bunch of tasks in a MySQL database, and one of the fields is "deadline date". Not every task has to have to a deadline date.
I'd like to use SQL to sort the tasks by deadline date, but put the ones without a deadline date in the back of the result set. As it is now, the null dates show up first, then the rest are sorted by deadline date earliest to latest.
Any ideas on how to do this with SQL alone? (I can do it with PHP if needed, but an SQL-only solution would be great.)
Thanks!
Here's a solution using only standard SQL, not ISNULL(). That function is not standard SQL, and may not work on other brands of RDBMS.
SELECT * FROM myTable
WHERE ...
ORDER BY CASE WHEN myDate IS NULL THEN 1 ELSE 0 END, myDate;
SELECT * FROM myTable
WHERE ...
ORDER BY ISNULL(myDate), myDate
SELECT foo, bar, due_date FROM tablename
ORDER BY CASE ISNULL(due_date, 0)
WHEN 0 THEN 1 ELSE 0 END, due_date
So you have 2 order by clauses. The first puts all non-nulls in front, then sorts by due date after that
The easiest way is using the minus operator with DESC.
SELECT * FROM request ORDER BY -date DESC
In MySQL, NULL values are considered lower in order than any non-NULL value, so sorting in ascending (ASC) order NULLs are listed first, and if descending (DESC) they are listed last.
When a - (minus) sign is added before the column name, NULL become -NULL.
Since -NULL == NULL, adding DESC make all the rows sort by date in ascending order followed by NULLs at last.