SELECT start FROM scheduling
WHERE residentid='2'
ORDER BY DATE_FORMAT(start,'%H:%i:%s') desc;
------- start is type VARCHAR(10)--------------
'02:19PM'
'9:25PM'
'4:45PM'
NULL
'02:19PM'
''
'02:25AM'
I don't understand why my dates aren't being ordered correctly, I made everything as per what I found, but I still cant get the start times to be in descending order. If you see there is a 9:25PM then there is a 4:45PM., but that does not make sense, since 9:35PM comes after 4:45PM but also I treid asc,(ascending) still does not work
SELECT * FROM scheduling
ORDER BY STR_TO_DATE(REPLACE(REPLACE(start, 'AM', ' AM'), 'PM', ' PM'),'%l:%i %p') desc;
Updated answer.
sqlfiddle.com/#!9/f49d7b/18/0
Try this:
SELECT * FROM scheduling
WHERE residentid='2'
ORDER BY DATE_FORMAT(start,'%H:%i:%s') desc;
Related
I have a requirement to display spend estimation for last 30 days. SpendEstimation is calculated multiple times a day. This can be achieved using simple SQL query:
SELECT DISTINCT ON (date) date(time) AS date, resource_id , time
FROM spend_estimation
WHERE
resource_id = '<id>'
and time > now() - interval '30 days'
ORDER BY date DESC, time DESC;
Unfortunately I can't seem to be able to do the same using SQLAlchemy. It always creates select distinct on all columns. Generated query does not contain distinct on.
query = session.query(
func.date(SpendEstimation.time).label('date'),
SpendEstimation.resource_id,
SpendEstimation.time
).distinct(
'date'
).order_by(
'date',
SpendEstimation.time
)
SELECT DISTINCT
date(time) AS date,
resource_id,
time
FROM spend
ORDER BY date, time
It is missing ON (date) bit. If I user query.group_by - then SQLAlchemy adds distinct on. Though I can't think of solution for given problem using group by.
Tried using function in distinct part and order by part as well.
query = session.query(
func.date(SpendEstimation.time).label('date'),
SpendEstimation.resource_id,
SpendEstimation.time
).distinct(
func.date(SpendEstimation.time).label('date')
).order_by(
func.date(SpendEstimation.time).label('date'),
SpendEstimation.time
)
Which resulted in this SQL:
SELECT DISTINCT
date(time) AS date,
resource_id,
time,
date(time) AS date # only difference
FROM spend
ORDER BY date, time
Which is still missing DISTINCT ON.
Your SqlAlchemy version might be the culprit.
Sqlalchemy with postgres. Try to get 'DISTINCT ON' instead of 'DISTINCT'
Links to this bug report:
https://bitbucket.org/zzzeek/sqlalchemy/issues/2142
A fix wasn't backported to 0.6, looks like it was fixed in 0.7.
Stupid question: have you tried distinct on SpendEstimation.date instead of 'date'?
EDIT: It just struck me that you're trying to use the named column from the SELECT. SQLAlchemy is not that smart. Try passing in the func expression into the distinct() call.
I am having a small difficulty understanding the below simple DISTINCT ON query:
SELECT DISTINCT
ON (bcolor) bcolor,
fcolor
FROM
t1
ORDER BY
bcolor,
fcolor;
I have this table here:
What is the order of execution of the above table and why I am getting the following result:
As I understand since ORDER BY is used it will display the table columns (both of them), in alphabetical order and since ON is used it will return the 1st matched duplicate, but I am still confused about how the resulting table is displayed.
Can somebody take me through how exactly this query is executed ?
This is an odd one since you would think that the SELECT would happen first, then the ORDER BY like any normal RDBMS, but the DISTINCT ON is special. It needs to know the order of the records in order to properly determine which records should be dropped.
So, in this case, it orders first by the bcolor, then by the fcolor. Then it determines distinct bcolors, and drops any but the first record for each distinct group.
In short, it does ORDER BY then applies the DISTINCT ON to drop the appropriate records. I think it would be most helpful to think of 'DISTINCT ON' as being special functionality that differs greatly from DISTINCT.
Added after initial post:
This could be done using window functions and a subquery as well:
SELECT
bcolor,
fcolor
FROM
(
SELECT
ROW_NUMBER() OVER (PARTITION BY bcolor ORDER BY fcolor ASC) as rownumber,
bcolor,
fcolor
FROM t1
) t2
WHERE rownumber = 1
This question is geared for those who have more SQL experience than me.
I am writing a query(that will eventually be a Stored Procedure but this should be irrelevant) where I want to select the count of rows if the most recent entry's is equivalent to the one that was just entered before. And i want to continue to do this until it hits an entry that has a different value. (Poorly explained so I will show the example)
In my table I have a column 'Product_Id' and when this query is run i want it take the product_id and compare it to the previously entered product Id, if its the same I want to add one, and I want it to keep checking the previously entered product_id until it runs into a different product_id
I'm hoping it sounds more complicated than it is, and the query would look something like
Select count(Product_ID)
FROM dbo.myTable
Where Product_Id = previous(Product_Id)
Now, i know that previous isn't a keyword in TSQL, and neither was Last, but I'm hoping of someone who knows a keyword that does what I am asking.
Edit for Sam
USE DbName;
GO
WITH OrderedCount as
(
select ROW_NUMBER() OVER (Order by dbo.Line_Production.Run_Date DESC) as RowNumber,
Line_Production.Product_ID
From dbo.Line_Production
)
Select RowNumber, COUNT(OrderedCount.Product_ID) as PalletCount
From OrderedCount
WHERE OrderedCount.RowNumber + 1 = RowNumber
and Product_ID = Product_ID
Group by RowNumber
The OrderedCount portion works, and it returns the data back how I want it, I'm now having trouble comparing the Product_ID's for different RowNumbers
my Where Clause is wrong
There's no keyword. That would be a nice magic solution, but it doesn't exist, at least in part because there is no guaranteed ordering (okay, you could have the keyword only if there is an ORDER BY...). I can write you a query, but that'll take time, so for now I'll give you a few steps and I'll come back and see if you still need help in a bit.
Figure out an ORDER BY, otherwise no order is guaranteed. If there is a time entered field, that's a good choice, or an index, that works too.
Learn to use Row_Number.
Compare the table (with Row_Number) to itself where instance1.row - 1 = instance2.row.
If product_id is an identity column, couldn't you just do product_id - 1? In other words, if it's sequential, it's the same as using ROW_NUMBER mentioned in the previous comment.
I'm new to postgres, so please take it easy on me.
I'm trying to write a query so that for any user, I can pull ALL of the log files (both their activity and the activity of others) for one minute prior and one minute after their name appears in the logs within the same batchstamp.
chat.batchstamp is varchar
chat.datetime is timestamp
chat.msg is text
chat.action is text (this is the field with the username)
Here are the separate commands I want to use, I just don't know how to put them together and if this is really the right path to go on this.
SELECT batchstamp, datetime, msg FROM chat WHERE action LIKE 'username';
Anticipated output:
batchstamp datetime msg
abc 2010-12-13 23:18:00 System logon
abc 2010-12-13 10:12:13 System logon
def 2010-12-14 11:12:18 System logon
SELECT * FROM chat WHERE datetime BETWEEN datetimefrompreviousquery - interval '1 minute' AND datetimefrompreviousquery + interval '1 minute';
Can you please help explain to me what I should do to feed data from the previous query in to the second query? I've looked at subqueries, but do I need to run two subqueries? Should I build a temporary table?
After this is all done, how do I make sure that the times the query matches are within the same batchstamp?
If you're able to point me in the right direction, that's great. If you're able to provide the query, that's even better. If my explanation doesn't make sense, maybe I've been looking at this too long.
Thanks for your time.
Based on nate c's code below, I used this:
SELECT * FROM chat,
( SELECT batchstamp, datetime FROM chat WHERE action = 'fakeuser' )
AS log WHERE chat.datetime BETWEEN log.datetime - interval '1 minute' AND log.datetime + '1 minute';
It doesn't seem to return every hit of 'fakeuser' and when it does, it pulls the logs from every 'batchstamp' instead of just the one where 'fakeuser' was found. Am I in for another nested query? What's this type of procedure called so I can further research it?
Thanks again.
You first query can go in the from clause with '(' brackets around it and 'as alias' name. After that you can reference it as you would a normal table in the rest of the query.
SELECT
*
FROM chat,
(
SELECT
batchstamp,
datetime,
msg
FROM log
WHERE action LIKE 'username'
) AS log
WHERE chat.datetime BETWEEN
log.datetime - interval '1 minute'
AND log.datetime + interval '1 minute';
That should get you started.
A colleague at work came up with the following solution which seems to provide the results I'm looking for. Thanks for everyone's help.
SELECT batchstamp, datetime, msg INTO temptable FROM chat WHERE action = 'fakeusername';
select a.batchstamp, a.action, a.datetime, a.msg
FROM chat a, temptable b
WHERE a.batchstamp = b.batchstamp
and (
a.datetime BETWEEN b.datetime - interval '1 minute'
AND b.datetime + interval '1 minute'
) and a.batchstamp = '2011-3-1 21:21:37'
group by a.batchstamp, a.action, a.datetime, a.msg
order by a.datetime;
I have a table with 3 columns.
ORDER CATEGORY
NAV_PER_SHARE
Number_OUTSTANDING_SHARES
Now:
SELECT ORDER_CATEGORY, SUM(OUTSTANDING_SHARES) GROUP BY ORDER_CATEGORY , it runs fine
But:
SELECT ORDER_CATEGORY, NAV_PER_SHARE * SUM(OUTSTANDING_SHARES) GROUP BY ORDER_CATEGORY , it says : Not a group by expression .
What am I missing?
You caNnot group because NAV_PER_SHARE is given per result. Did you mean
SUM(NAV_PER_SHARE*OUTSTANDING_SHARES)
?
From what it looks like, the solution would be to use a subquery.
SELECT ORDER_CATEGORY, NAV_PER_SHARE * SUM_OF_OS
FROM (SELECT ORDER_CATEGORY, SUM(OUTSTANDING_SHARES) SUM_OF_OS
GROUP BY ORDER_CATEGORY);
Although, I'm not sure how your one query works without a FROM keyword.