SQL Query question

[mLipok]

I have an MS SQL DB with Table which have columns :   [Tabel ID], [SIGNATURE ID ], [DATE], [TEXT]

this is an simple SQL QUERY

Quote

SELECT [SIGNATURE ID ], [DATE] FROM [TABLE]
WHERE [TEXT] like '%urz%skarb%'
ORDER BY [SIGNATURE ID ], [DATE] ASC

This query is only first step in my task.

I was trying to get unique [SIGNATURE ID] but only for the latest [DATE]

Can anybody show me the proper SQL Query ?

 

Regards,
mLipok

Remarks:
It may sound strange even stupid, because although I know ADO and other technologies well, I do not work with complicated SQL Queries, I usually have very simple cases.

 

EDIT:
I know that I can use .GetRows from this SQL Query which I post, and then loop through this table, analyze them and create new result table, but I was trying to do this with one query without any further AutoIt analysis, which should be much faster for large amount of data (about 10000 rows which I get from my first query ).

 

Edited December 20, 2018 by mLipok

[jchd]

SELECT [SIGNATURE ID ], [DATE] FROM [TABLE]
WHERE [TEXT] like '%urz%skarb%'
ORDER BY [DATE] ASC LIMIT 1

Will return one row with the requested columns for the oldest date satisfying text search.

[user4157124]

To group by [SIGNATURE ID] showing last [DATE] for each:

SELECT
    [SIGNATURE ID],
    MAX([DATE])
FROM
    [TABLE]
WHERE
    [TEXT] LIKE '%urz%skarb%'
GROUP BY
    [SIGNATURE ID]
ORDER BY
    [DATE] ASC
;

 

[mLipok]

1 hour ago, jchd said:

Will return one row with the requested columns for the oldest date satisfying text search.

My intention was to get latest [DATE] for each unique [SIGNATURE ID] ,   I mean not single row.

16 minutes ago, user4157124 said:

To group by [SIGNATURE ID] showing last [DATE] for each:

SELECT
    [SIGNATURE ID],
    MAX([DATE])
FROM
    [TABLE]
WHERE
    [TEXT] LIKE '%urz%skarb%'
GROUP BY
    [SIGNATURE ID]
ORDER BY
    [DATE] ASC
;

 

I think/hope this should be what I was looking for.
I will check it soon and back with results.

 

Edited December 20, 2018 by mLipok

[jchd]

Probably not that then: grouping is done first, which result in one (random) row being selected for each signature; then the resultset is sorted on date ascending. SQL doesn't explicitely specify which row is selected by GROUP BY, so assume the choice is random and not repeatable. Hence the resulting row may not be the oldest date for each signature.

In the OP I understood "unique signature" as "one row".

 

[mLipok]

Is there any way to make any QUERY < SUBQUERY < SUBQUERY solution ?

 

[jchd]

That can be done with a CTE:

with [sigs] as (select distinct [signature id] [s] from [table] where [TEXT] like '%urz%skarb%') 
select [s], (select min([date]) from [table] [d] where [d.sign] = [s]) from [sigs];

Edit: can as well be written as a self-join, but I find CTEs clearer.

Edited December 20, 2018 by jchd

[user4157124]

@jchd Aggregate function MAX() on [DATE] explicitly defines selected value per group. Exact record should be irrelevant ([SIGNATURE ID] is equal for each group's record; they're grouped). ORDER BY aims to sort result (not records in each group). Described behavior would indeed occur if no aggregation were applied (incorrect use of GROUP BY). Did I misunderstand something?

[jchd]

OK, you're right. I was also confused by MAX while you actually mean MIN. The order by clause is irrelevant, unless output order is meant by date or any other criterion.

I'm an SQLite fan and I know engines differ on how grouping-aggregation work: some bark other not, SQLite being overly friendly in trying to be kind. Not having ever used MS SQL I suspected it could behave differently.

Sorry for noise I should have paid more attention.

[Adisharma]

I can't understand properly because i am new in this filed to learning SQL as a beginner. I have also some query can you suggest the best forum. I follow https://hackr.io/tutorials/learn-sql to learn SQL programming. is there any other better way

[jchd]

Which engine are you using? SQL has a number of variants (even deviant) implementations.

My recommendation to learn SQL would be to start with a "simple" yet powerful engine you can play with easily: SQLite. No installation nor maintenance needed as it's an embedded engine (not client-server). You can download and play with some free third-pary SQLite DB manager (I recommend SQLite Expert).

Create some DBs and learn the basics by guided experimentation (W3school is a good reference; also sqlite.org website). Assign yourself a sensible goal: i.e. create a cute complete directory of relatives and friends (adding more details at each design step). Then build an appointments scheduler coupled with your contacts with features similar to what the apps in your smartphone offer.

Another direction: supply chain management. Build something to handle suppliers, items, buys, stock, orders, sales, deliveries, payments. You can just do with what you buy and use in your own home (real or simulated). All products come with a barcode, add designation, type; enter date bought, price, quantity, date used.  Create views to profile your consumption of various types of items: food, or wearables, or ..., etc. grouped by period, or maker, or whatever criterion of your choice.

If you're serious about learning the possibilities are endless.

Don't waste your time making programmatic applications: just enter directly data in base tables then use the built-in SQL tabs to issue SQL statements and/or use ad-hoc views to access your data. This way you can concentrate your time on manipulating SQL, accumulating working knowledge during your learning curve.

Don't forget to try large DBs (samples available on the net) and see how to optimize both the DB design and queries.

[Skysnake]

On 12/20/2018 at 1:05 PM, mLipok said:

I was trying to get unique [SIGNATURE ID] but only for the latest [DATE]

SELECT sig
        WHERE date in 
        (select max(date) from table);

Or

SELECT distinct sig, date 
    from [table]
    order by date desc

==> Should return a two column list of sigs and dates, with the top items being most recent dates. DISTINCT should give you a single instance of each sig

 

Am I too late to the party?

Is it possible to provide SQLite example? (the SQL should be very similar)

Edited March 28, 2019 by Skysnake
another option

[PercivalOgg]

SELECT distinct
    [SIGNATURE ID ]
FROM 
    [TABLE]
WHERE 
    [TEXT] like '%urz%skarb%'
    and DATE in
    (
        select 
            max([date])
        from
            [table]
        where
            [TEXT] like '%urz%skarb%'
    )
 

[mLipok]

Thank you for you answer.

And  to the AutoIt forum.

 

[Andreik]

If you use MS SQL Server a better approach would be an OUTER APPLY instead of sub-query.

[jchd]

This works with SQL[ite]:

CREATE TABLE "Signatures" (
  "Signature" CHAR, 
  "Stamp" CHAR DEFAULT (Datetime('now', 'utc')));

insert into Signatures values 
('aaa','2019-12-10 13:30:18'),
('bbb','2019-12-10 13:30:22'),
('aaa','2019-12-10 13:30:26'),
('ccc','2019-12-10 13:30:30'),
('aaa','2019-12-10 13:30:32'),
('ddd','2019-12-10 13:30:37'),
('bbb','2019-12-10 13:30:39'),
('ccc','2019-12-10 13:30:41'),
('aaa','2019-12-10 13:30:44'),
('eee','2019-12-10 13:30:47'),
('fff','2019-12-10 13:39:40'),
('ggg','2019-12-10 13:39:44'),
('bbb','2019-12-10 13:39:47'),
('bbb','2019-12-10 13:39:49'),
('ddd','2019-12-10 13:39:53'),
('aaa','2019-12-10 13:39:54'),
('hhh','2019-12-10 13:39:57');

select
  distinct Signature Who,
  stamp LastDate
from
  signatures
group by
  who
having
  stamp = max(stamp)
order by
  LastDate asc;

gives:

Who LastDate
ccc 2019-12-10 13:30:41
eee 2019-12-10 13:30:47
fff 2019-12-10 13:39:40
ggg 2019-12-10 13:39:44
bbb 2019-12-10 13:39:49
ddd 2019-12-10 13:39:53
aaa 2019-12-10 13:39:54
hhh 2019-12-10 13:39:57

 

[Andreik]

Why not simple?

select
  distinct Signature Who,
  max(stamp) LastDate
from
  signatures
group by
  who
order by
  LastDate asc;

 

[jchd]

You're right, don't know why I used a having clause here. Anyway, no need of a subquery.