LIKE '%[9-15]%' (SQL Server) - sql-server

I'm using SQL Server 2017 and I would like to ask if it's possible to use the LIKE like operator as follows:
LIKE '%Ticket [8-14]%'
Is this correct or the numbers greater than 9 (10,11, etc etc) will be identified as 1 and 0, 1 and 1, 1 and 2 etc etc.
If this way doesn't work, what can i do to select all the data that contain strings like 'Ticket 10', 'Ticket11' and so on..?
Thank you for your time

As far as I know, no. However, You can do something like:
WITH Demo AS
(
SELECT * FROM (VALUES
('Ticket 1'),
('Ticket 7'),
('Ticket 8'),
('Ticket 10'),
('Ticket 12'),
('Ticket 15')
) T(X)
)
SELECT *
FROM Demo
WHERE X LIKE '%Ticket [8-9]%' OR X LIKE '%Ticket 1[0-4]%'
Also, consider normalization - create TicketNumber column if you need to query over this value. It's much easier to concatenate Ticket and number than parse string. TicketNumber could also be easily indexed if needed.
There is also more clever idea to parse numbers:
WITH Demo AS
(
SELECT * FROM (VALUES
('Ticket 1'),
('Ticket 7'),
('My Ticket 8A'),
('Ticket 10'),
('Some Ticket 12'),
('Ticket 15 other text'),
('Ticket 135 and more')
) T(X)
)
SELECT *, CAST(CASE WHEN PATINDEX('%Ticket [0-9][0-9][0-9]%',X)!=0 THEN SUBSTRING(X, PATINDEX('%Ticket [0-9][0-9][0-9]%',X)+7, 3)
WHEN PATINDEX('%Ticket [0-9][0-9]%',X)!=0 THEN SUBSTRING(X, PATINDEX('%Ticket [0-9][0-9]%',X)+7, 2)
WHEN PATINDEX('%Ticket [0-9]%',X)!=0 THEN SUBSTRING(X, PATINDEX('%Ticket [0-9]%',X)+7, 1)
END AS int) Number
FROM Demo
Column number should now contain simple int value - ready to compare and take part in calculations.

'%Ticket [8-14]%'
The brackets are used to specify a single character, that's usually specified as a lower and upper range like "[0-9]" or "[a-z]".
Your string would match:
"Ticket 1" through "Ticket 8". The 4 would be ignored because it's already handled by the 8-1 range.
It would not match "Ticket 0" or "Ticket 9" or "Ticket 10".

A slightly different approach to Pawel's, but a very similar idea:
SELECT *
FROM YourTable YT
CROSS APPLY(VALUES('8'),('9'),('10'),('11'),('12'),('13'),('14')) V(TN)
WHERE YT.YourColumn LIKE '%Ticket ' + V.TN + '%';
If you're using this as a Stored Procedure, you could use a table-value paramter to hold the data instead. Something like:
CREATE TYPE numbers AS table (Number int);
GO
CREATE PROC YourProc #TicketNumbers numbers READONLY AS
SELECT *
FROM YourTable YT
CROSS JOIN #TicketNumbers TN
WHERE YT.YourColumn LIKE 'Ticket ' + CONVERT(varchar(3),TN.Number) + '%';
GO

Related

How can i order by column with data "01","02","03","1","2","3" and so on in sql server

I am having SQL table with data as shown below
01 Buy-1
010 Buy-10
011 Buy-11
02 Buy-2
1 Direct-1
10 Direct-10
11 Direct-11
2 Direct-2
I want to order by the data like this
01 Buy-1
02 Buy-2
010 Buy-10
011 Buy-11
1 Direct-1
2 Direct-2
10 Direct-10
11 Direct-11
This requires a little extra handling of the strings to get the right sorting criteria.
Assuming your data is consistent with your sample data with a hyphen followed by digits you can order by the data to the left of the hyphen and then by casting the first column as an int
select id, data
from t
order by Left(data, nullif(CharIndex('-',data),0)-1),
Try_Cast(id as int)
This was an intriguing problem and I had to come back and find a solution to this issue as it has been seen many times over the years. When you don't have the leading zeros it is a fairly easy solution but this is different.
This works for your sample data and even some edge cases that I tried. The ordering first effectively counts the number of number of leading zeros and moves that with the most leading zeros to the top. Then it orders by converting the value to an int. This allows you to have more than 1 leading zero and it will still sort correctly. Additionally it will work even if you have character data in the first column.
declare #Something table
(
SomeValue varchar(10)
, SomeOtherValue varchar(30)
)
insert #Something
select '01' , 'Buy-1' union all
select '010', 'Buy-10' union all
select '011', 'Buy-11' union all
select '02', 'Buy-2' union all
select '1', 'Direct-1' union all
select '10', 'Direct-10' union all
select '11', 'Direct-11' union all
select '2', 'Direct-2'
select *
from #Something s
order by len(SomeValue) - len(convert(varchar(10), try_convert(int, SomeValue))) desc
, try_convert(int, SomeValue)
I would first order by your code having a leading zero, and secondly by the code value itself.
select code, description
from MyTable
order by case when left(code,1) = '0' then 0 else 1 end,
convert(int, code)
Id: 01
_Value: Buy-1
SELECT Id, _Value from tblName order by left(_Value, CHARINDEX('-', _Value)), cast(SUBSTRING(_Value, PATINDEX('%[0-9]%', _Value),len(_Value)) as int)

Query for date value AFTER specified patindex value in TEXT column using SQL Server

I am querying for a date value that follows a specific phrase in the same text column value. There are a number of date values in this text column but I'm needing only the DATE following the phrase I've found via a patindex value. Any help/direction would be appreciated. Thanks.
Here is my SQL code:
SELECT
NotesSysID,
PATINDEX('%Demand Due Date:%', NoteText) AS [Index of DemandDueDate text],
PATINDEX('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%', NoteText) AS [Index of DemandDueDate date],
SUBSTRING(NoteText, PATINDEX('%Demand Due Date:%', NoteText), (PATINDEX('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%', NoteText)))
FROM
#temp_ExtractedNotes;
Here is a pic of my data, please note that the 2nd index is LESS than the Index of DemandDueDate text found and I'm needing the subsequent date AFTER the Index of DemandDueDate text column. Hopefully this makes sense.
I think your code will be easier for you to work with if you use a CTE and some stepwise refinement. That'll free you from trying to do everything with one hugely-nested SELECT statement.
;
WITH FirstCut as (
SELECT
NotesSysID,
LocationOfText = PATINDEX('%Demand Due Date:%', NoteText),
NoteText
FROM #temp_ExtractedNotes
),
SecondCut as (
SELECT
NotesSysID,
NoteText,
-- making assumption date will be within first 250 chars of text
DemandDueDateSection = SUBSTRING( NoteText, [LocationOfText], 250)
FROM FirstCut
),
ThirdCut as (
SELECT
NotesSysID,
NoteText,
DemandDueDateSection,
LocationOfDate = PATINDEX('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%', DemandDueDateSection)
FROM SecondCut
),
FourthCut as (
SELECT
NotesSysID,
NoteText,
DateAsText = SUBSTRING( DemandDueDateSection, LocationOfDate, 10 )
FROM ThirdCut
)
SELECT
NotesSysID,
NoteText,
DemandDueDate = CONVERT( DateTime, DateAsText)
FROM FourthCut
The issue you're having is because your second PATINDEX isn't finding the date you want; it's finding the first date in the string, which, as you noted, appears before the PATINDEX of the phrase you're searching for %Demand Due Date:%. The third parameter of SUBSTRING is LENGTH, which is to say how many characters after the second parameter you want to pull. By using that second PATINDEX value as the third parameter in your SUBSTRING you're returning a sub-string that starts where you want it to, and is of LENGTH equal to the number of characters into the string where that first date appears.
Which, of course, isn't what you want. To #ZLK's point in the comments, first, you need to do a nested PATINDEX. That's going to be pretty slow, so I'm hoping there aren't a bazillion records in your temp table.
Based on the sample image, it looks like the date you're interested in can appear a variable number of characters after %Demand Due Date:%. We'll start by adding 16 to the PATINDEX of %Demand Due Date:% (because that's how many characters are in %Demand Due Date:%, so we'll just start right after that). Then we'll pick up the next 100 characters. You can tweak that later if you need more or not that many.
So you're first SUBSTRING will look like this:
SUBSTRING(NoteText, PATINDEX('%Demand Due Date:%', NoteText) + 16, 100)
Now we have to search that sub-string for the second pattern, the one that should yield a date for you.
PATINDEX('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%', SUBSTRING(NoteText, PATINDEX('%Demand Due Date:%', NoteText) + 16, 100))
The number returned there is the point where your date value starts within the 100 characters following %Demand Due Date:%. Armed with that number, you just need to SUBSTRING out the next ten characters, and, just for fun, CAST it as a DATE. That big ugly formula will look like this:
DECLARE #test VARCHAR(200) = 'foo bar Demand Due Date: 12/21/2018 bar foo foo bar';
SELECT
CAST(
SUBSTRING(
SUBSTRING
(#test, PATINDEX('%Demand Due Date:%', #test) + 16, 100),
PATINDEX
( '%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%',
SUBSTRING
(#test, PATINDEX('%Demand Due Date:%', #test) + 16, 100)
)
,10)
AS DATE);
Result:
2018-12-21
Rextester: https://rextester.com/KCY79989
Grab a copy of PatternSplitCM.
CREATE FUNCTION dbo.PatternSplitCM
(
#List VARCHAR(8000) = NULL
,#Pattern VARCHAR(50)
) RETURNS TABLE WITH SCHEMABINDING
AS
RETURN
WITH numbers AS (
SELECT TOP(ISNULL(DATALENGTH(#List), 0))
n = ROW_NUMBER() OVER(ORDER BY (SELECT NULL))
FROM
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) d (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) e (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) f (n),
(VALUES (0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) g (n))
SELECT
ItemNumber = ROW_NUMBER() OVER(ORDER BY MIN(n)),
Item = SUBSTRING(#List,MIN(n),1+MAX(n)-MIN(n)),
Matched
FROM (
SELECT n, y.Matched, Grouper = n - ROW_NUMBER() OVER(ORDER BY y.Matched,n)
FROM numbers
CROSS APPLY (
SELECT Matched = CASE WHEN SUBSTRING(#List,n,1) LIKE #Pattern THEN 1 ELSE 0 END
) y
) d
GROUP BY Matched, Grouper;
Then it's pretty easy:
-- Sample Data
DECLARE #temp_ExtractedNotes TABLE (someID INT IDENTITY, NoteText VARCHAR(8000));
INSERT #temp_ExtractedNotes (NoteText) VALUES
('blah blah blah 2/4/2016... Demand Due Date: 1/05/2011; blah blah 12/1/2017...'),
('Yada yad..... Demand Due Date: 11/21/2016;...'),
('adas dasd asd a sd asdas Demand Due Date: 09/09/2019... 5/05/2005, 10/10/2010....'),
('nothing to see here - moving on... 01/02/2003!');
-- Solution
SELECT TOP (1) WITH TIES t.someID, ns.s, DueDate = CASE SIGN(nt.s) WHEN 1 THEN f.item END
FROM #temp_ExtractedNotes AS t
CROSS APPLY (VALUES(CHARINDEX('Demand Due Date:',t.NoteText))) AS nt(s)
CROSS APPLY (VALUES(SUBSTRING(t.noteText, nt.s+16, 8000))) AS ns(s)
CROSS APPLY dbo.patternsplitCM(ns.s,'[0-9/]') AS f
WHERE f.matched = 1
ORDER BY ROW_NUMBER() OVER (PARTITION BY t.someID ORDER BY f.itemNumber);
Results:
someID s DueDate
----------- ---------------------------------------- ------------
1 1/05/2011; blah blah 12/1/2017... 1/05/2011
2 11/21/2016;... 11/21/2016
3 09/09/2019... 5/05/2005, 10/10/2010.... 09/09/2019
4 here - moving on... 01/02/2003! NULL

How to merge multiple values from multiple columns into one row?

I have this table:
I need to convert it to (with parenthesis as well):
row_nbr - row_label - default_order
10 - TOTAL ACCOUNTABLE GROSS - (1, 3)
12 - DEDUCTIBLE TERMS - (3)
20 - TOTAL DEDUCTIBLE TERMS - (3)
34 - AMOUNT DUE (UNRECOUPED) - (4)
36 - ACCOUNTABLE GROSS - (2)
41 - TOTAL CONTINGENT COMPENSATION - (3)
I could have more than twice of the same row_nbr.
In this case the 10 is there twice, but I could have 3 10's, 4 12's, etc.
I kind of started the pivot table but honestly, even by looking at the Microsoft site, I cannot for the life of me figure this out.
select row_nbr, row_label, default_order
from #temp
pivot
(
max(row_nbr)
for default_order in (default_order)
) piv;
Anyone care to help?
Thanks.
As #Vinit says, you can use the string_agg function in 2017, but if you're at least on 2005 you can use a horrible, torturous XML generator:
SELECT row_nbr
,row_label
,default_order = '(' +
STUFF(
(SELECT ', ' + CAST(default_order AS VARCHAR(10))
FROM #temp
WHERE row_nbr = t.row_nbr
ORDER BY default_order
FOR XML PATH('') ,
ROOT('MyString'),
TYPE ).value('/MyString[1]', 'varchar(max)'), 1, 2, '')
+ ')'
FROM #temp t;
You can read more about it in this blog post
PIVOTS are definitely wonky to get a grip on. Fortunately, in this case, while you could use one as an intermediate step, it's not necessary. PIVOT will take each value and put it into a corresponding distinct column, and what you're wanting is a single column, with them all concatenated together. Like I said, you could do the pivot, then just concatenate all the generated columns together, but that's way more work than is needed.
On 2014, the easiest way to do this is using FOR XML. Russell Fox's answer pretty much covers how that technique works (although there are a few variants on how you can do that should you so choose).
If you know definitively the values are all integers, you can save a bit of typing and omit the type and value operators, as those are only necessary when you have to escape certain XML characters in string fields
select
row_nbr,
row_label,
default_order,
stuff
(
(
select concat(',', default_order)
from #temp i
where i.row_nbr = o.row_nbr
for xml path('')
), 1, 1, ''
)
from #temp o

Expression to find multiple spaces in string

We handle a lot of sensitive data and I would like to mask passenger names using only the first and last letter of each name part and join these by three asterisks (***),
For example: the name 'John Doe' will become 'J***n D***e'
For a name that consists of two parts this is doable by finding the space using the expression:
LEFT(CardHolderNameFromPurchase, 1) +
'***' +
CASE WHEN CHARINDEX(' ', PassengerName) = 0
THEN RIGHT(PassengerName, 1)
ELSE SUBSTRING(PassengerName, CHARINDEX(' ', PassengerName) -1, 1) +
' ' +
SUBSTRING(PassengerName, CHARINDEX(' ', PassengerName) +1, 1) +
'***' +
RIGHT(PassengerName, 1)
END
However, the passenger name can have more than two parts, there is no real limit to it. How should can I find the indices of all spaces within an expression? Or should I maybe tackle this problem in a different way?
Any help or pointer is much appreciated!
This solution does what you want it to, but is really the wrong approach to use when trying to hide personally identifiable data, as per Gordon's explanation in his answer.
SQL:
declare #t table(n nvarchar(20));
insert into #t values('John Doe')
,('JohnDoe')
,('John Doe Two')
,('John Doe Two Three')
,('John O''Neill');
select n
,stuff((select ' ' + left(s.item,1) + '***' + right(s.item,1)
from dbo.fn_StringSplit4k(t.n,' ',null) as s
for xml path('')
),1,1,''
) as mask
from #t as t;
Output:
+--------------------+-------------------------+
| n | mask |
+--------------------+-------------------------+
| John Doe | J***n D***e |
| JohnDoe | J***e |
| John Doe Two | J***n D***e T***o |
| John Doe Two Three | J***n D***e T***o T***e |
| John O'Neill | J***n O***l |
+--------------------+-------------------------+
String splitting function based on Jeff Moden's Tally Table approach:
create function [dbo].[fn_StringSplit4k]
(
#str nvarchar(4000) = ' ' -- String to split.
,#delimiter as nvarchar(1) = ',' -- Delimiting value to split on.
,#num as int = null -- Which value to return, null returns all.
)
returns table
as
return
-- Start tally table with 10 rows.
with n(n) as (select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1 union all select 1)
-- Select the same number of rows as characters in #str as incremental row numbers.
-- Cross joins increase exponentially to a max possible 10,000 rows to cover largest #str length.
,t(t) as (select top (select len(isnull(#str,'')) a) row_number() over (order by (select null)) from n n1,n n2,n n3,n n4)
-- Return the position of every value that follows the specified delimiter.
,s(s) as (select 1 union all select t+1 from t where substring(isnull(#str,''),t,1) = #delimiter)
-- Return the start and length of every value, to use in the SUBSTRING function.
-- ISNULL/NULLIF combo handles the last value where there is no delimiter at the end of the string.
,l(s,l) as (select s,isnull(nullif(charindex(#delimiter,isnull(#str,''),s),0)-s,4000) from s)
select rn
,item
from(select row_number() over(order by s) as rn
,substring(#str,s,l) as item
from l
) a
where rn = #num
or #num is null;
GO
If you consider PassengerName as sensitive information, then you should not be storing it in clear text in generally accessible tables. Period.
There are several different options.
One is to have reference tables for sensitive information. Any table that references this would have an id rather than the name. Viola. No sensitive information is available without access to the reference table, and that would be severely restricted.
A second method is a reversible compression algorithm. This would allow the the value to be gibberish, but with the right knowledge, it could be transformed back into a meaningful value. Typical methods for this are the public key encryption algorithms devised by Rivest, Shamir, and Adelman (RSA encoding).
If you want to do first and last letters of names, I would be really careful about Asian names. Many of them consist of two or three letters, when written in Latin script. That isn't much hiding. SQL Server does not have simple mechanisms to do this. You can write a user-defined function with a loop to manager the process. However, I view this as the least secure and least desirable approach.
This uses Jeff Moden's DelimitedSplit8K, as well as the new functionality in SQL Server 2017 STRING_AGG. As I don't know what version you're using, I've just gone "whole hog" and assumed you're using the latest version.
Jeff's function is invaluable here, as it returns the ordinal position, something which Microsoft have foolishly omitted from their own function, STRING_SPLIT (and didn't add in 2017 either). Ordinal position is key here, so we can't make use of the built in function.
WITH VTE AS(
SELECT *
FROM (VALUES ('John Doe'),('Jane Bloggs'),('Edgar Allan Poe'),('Mr George W. Bush'),('Homer J Simpson')) V(FullName)),
Masking AS (
SELECT *,
ISNULL(STUFF(Item, 2, LEN(item) -2,'***'), Item) AS MaskedPart
FROM VTE V
CROSS APPLY dbo.delimitedSplit8K(V.Fullname, ' '))
SELECT STRING_AGG(MaskedPart,' ') AS MaskedFullName
FROM Masking
GROUP BY Fullname;
Edit: Nevermind, OP has commented they are using 2008, so STRING_AGG is out of the question. #iamdave, however, has posted an answer which is very similar to my own, just do it the "old fashioned XML way".
Depending on your version of SQL Server, you may be able to use the built-in string split to rows on spaces in the name, do your string formatting, and then roll back up to name level using an XML path.
create table dataset (id int identity(1,1), name varchar(50));
insert into dataset (name) values
('John Smith'),
('Edgar Allen Poe'),
('One Two Three Four');
with split as (
select id, cs.Value as Name
from dataset
cross apply STRING_SPLIT (name, ' ') cs
),
formatted as (
select
id,
name,
left(name, 1) + '***' + right(name, 1) as out
from split
)
SELECT
id,
(SELECT ' ' + out
FROM formatted b
WHERE a.id = b.id
FOR XML PATH('')) [out_name]
FROM formatted a
GROUP BY id
Result:
id out_name
1 J***n S***h
2 E***r A***n P***e
3 O***e T***o T***e F***r
You can do that using this function.
create function [dbo].[fnMaskName] (#var_name varchar(100))
RETURNS varchar(100)
WITH EXECUTE AS CALLER
AS
BEGIN
declare #var_part varchar(100)
declare #var_return varchar(100)
declare #n_position smallint
set #var_return = ''
set #n_position = 1
WHILE #n_position<>0
BEGIN
SET #n_position = CHARINDEX(' ', #var_name)
IF #n_position = 0
SET #n_position = LEN(#var_name)
SET #var_part = SUBSTRING(#var_name, 1, #n_position)
SET #var_name = SUBSTRING(#var_name, #n_position+1, LEN(#var_name))
if #var_part<>''
SET #var_return = #var_return + stuff(#var_part, 2, len(#var_part)-2, replicate('*',len(#var_part)-2)) + ' '
END
RETURN(#var_return)
END

sort float numbers as a natural numbers in SQL Server

Well I had asked the same question for jquery on here, Now my question is same with SQL Server Query :) But this time this is not comma separated, this is separate row in Database like
I have separated rows having float numbers.
Name
K1.1
K1.10
K1.2
K3.1
K3.14
K3.5
and I want to sort this float numbers like,
Name
K1.1
K1.2
K1.10
K3.1
K3.5
K3.14
actually in my case, the numbers which are after decimals will consider as a natural numbers, so 1.2 will consider as '2' and 1.10 will consider as '10' thats why 1.2 will come first than 1.10.
You can remove 'K' because it is almost common and suggestion or example would be great for me, thanks.
You can use PARSENAME (which is more of a hack) or String functions like CHARINDEX , STUFF, LEFT etc to achieve this.
Input data
;WITH CTE AS
(
SELECT 'K1.1' Name
UNION ALL SELECT 'K1.10'
UNION ALL SELECT 'K1.2'
UNION ALL SELECT 'K3.1'
UNION ALL SELECT 'K3.14'
UNION ALL SELECT 'K3.5'
)
Using PARSENAME
SELECT Name,PARSENAME(REPLACE(Name,'K',''),2),PARSENAME(REPLACE(Name,'K',''),1)
FROM CTE
ORDER BY CONVERT(INT,PARSENAME(REPLACE(Name,'K',''),2)),
CONVERT(INT,PARSENAME(REPLACE(Name,'K',''),1))
Using String Functions
SELECT Name,LEFT(Name,CHARINDEX('.',Name) - 1), STUFF(Name,1,CHARINDEX('.',Name),'')
FROM CTE
ORDER BY CONVERT(INT,REPLACE((LEFT(Name,CHARINDEX('.',Name) - 1)),'K','')),
CONVERT(INT,STUFF(Name,1,CHARINDEX('.',Name),''))
Output
K1.1 K1 1
K1.2 K1 2
K1.10 K1 10
K3.1 K3 1
K3.5 K3 5
K3.14 K3 14
This works if there is always one char before the first number and the number is not higher than 9:
SELECT name
FROM YourTable
ORDER BY CAST(SUBSTRING(name,2,1) AS INT), --Get the number before dot
CAST(RIGHT(name,LEN(name)-CHARINDEX('.',name)) AS INT) --Get the number after the dot
Perhaps, more verbal, but should do the trick
declare #source as table(num varchar(12));
insert into #source(num) values('K1.1'),('K1.10'),('K1.2'),('K3.1'),('K3.14'),('K3.5');
-- create helper table
with data as
(
select num,
cast(SUBSTRING(replace(num, 'K', ''), 1, CHARINDEX('.', num) - 2) as int) as [first],
cast(SUBSTRING(replace(num, 'K', ''), CHARINDEX('.', num), LEN(num)) as int) as [second]
from #source
)
-- Select and order accordingly
select num
from data
order by [first], [second]
sqlfiddle:
http://sqlfiddle.com/#!6/a9b06/2
The shorter solution is this one :
Select Num
from yourtable
order by cast((Parsename(Num, 1) ) as Int)

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