Sample inputs:
"Hi there how are you"
"What is the #1 pizza place in NYC?"
"Dominoes is number 1"
"Blah blah 123123"
"More blah 12321 123123 123132"
Expected output:
"Hi there how are you"
"What is the #1 pizza place in NYC?"
"Dominoes is number"
"Blah blah"
"More blah"
I'm thinking it's a 2 step process:
Split the entire string into characters, one row per character (including spaces), in reverse order
Loop through, and for each one if it's a space or a number, skip, otherwise add to the start of another array.
And i should end up with the desired result.
I can think of a few quick and dirty ways, but this needs to perform fairly well, as it's a trigger that runs on a busy table, so thought i'd throw it out to the T-SQL pros.
Any suggestions?
This solution should be a bit more efficient because it first checks to see if the string contains a number, then it checks to see if the string ends in a number.
CREATE FUNCTION dbo.trim_ending_numbers(#columnvalue AS VARCHAR(100)) RETURNS VARCHAR(100)
BEGIN
--This will make the query more efficient by first checking to see if it contains any numbers at all
IF #columnvalue NOT LIKE '%[0-9]%'
RETURN #columnvalue
DECLARE #counter INT
SET #counter = LEN(#columnvalue)
IF ISNUMERIC(SUBSTRING(#columnvalue,#counter,1)) = 0
RETURN #columnvalue
WHILE ISNUMERIC(SUBSTRING(#columnvalue,#counter,1)) = 1 OR SUBSTRING(#columnvalue,#counter,1) = ' '
BEGIN
SET #counter = #counter -1
IF #counter < 0
BREAK
END
SET #columnvalue = SUBSTRING(#columnvalue,0,#counter+1)
RETURN #columnvalue
END
If you run
SELECT dbo.trim_ending_numbers('More blah 12321 123123 123132')
It will return
'More blah'
A loop on a busy table will be very unlikely to perform adequately. Use REVERSE and PATINDEX to find the first non digit, begin a SUBSTRING there, then REVERSE the result. This will be plenty slow with no loops.
Your examples imply that you also don't want to match spaces.
DECLARE #t TABLE (s NVARCHAR(500))
INSERT INTO #t (s)
VALUES
('Hi there how are you'),('What is the #1 pizza place in NYC?'),('Dominoes is number 1'),('Blah blah 123123'),('More blah 12321 123123 123132')
select s
, reverse(s) as beginning
, patindex('%[^0-9 ]%',reverse(s)) as progress
, substring(reverse(s),patindex('%[^0-9 ]%',reverse(s)), 1+len(s)-patindex('%[^0-9 ]%',reverse(s))) as [more progress]
, reverse(substring(reverse(s),patindex('%[^0-9 ]%',reverse(s)), 1+len(s)-patindex('%[^0-9 ]%',reverse(s)))) as SOLUTION
from #t
Final answer:
reverse( substring( reverse( #s ), patindex( '%[^0-9 ]%', reverse( #s ) ), 1 + len( #s ) - patindex( '%[^0-9 ]%', reverse( #s ) ) ) )
I believe that the below query is fast and useful
select reverse(substring(reverse(colA),PATINDEX('%[0-9][a-z]%',reverse(colA))+1,
len(colA)-PATINDEX('%[0-9][a-z]%',reverse(colA))))
from TBLA
--DECLARE #String VARCHAR(100) = 'the fat cat sat on the mat'
--DECLARE #String VARCHAR(100) = 'the fat cat 2 sat33 on4 the mat'
--DECLARE #String VARCHAR(100) = 'the fat cat sat on the mat1'
--DECLARE #String VARCHAR(100) = '2121'
DECLARE #String VARCHAR(100) = 'the fat cat 2 2 2 2 sat on the mat2121'
DECLARE #Answer NVARCHAR(MAX),
#Index INTEGER = LEN(#String),
#Character CHAR,
#IncorrectCharacterIndex SMALLINT
-- Start from the end, going to the front.
WHILE #Index > 0 BEGIN
-- Get each character, starting from the end
SET #Character = SUBSTRING(#String, #Index, 1)
-- Regex check.
SET #IncorrectCharacterIndex = PATINDEX('%[A-Za-z-]%', #Character)
-- Is there a match? We're lucky here because it will either match on index 1 or not (index 0)
IF (#IncorrectCharacterIndex != 0)
BEGIN
-- We have a legit character.
SET #Answer = SUBSTRING(#String, 0, #Index + 1)
SET #Index = 0
END
ELSE
SET #Index = #Index - 1 -- No match, lets go back one index slot.
END
PRINT LTRIM(RTRIM(#Answer))
NOTE: I've included a dash in the valid regex match.
Thanks for all the contributions which were very helpful. To go further and extract off JUST the trailing number:
, substring(s, 2 + len(s) - patindex('%[^0-9 ]%',reverse(s)), 99) as numeric_suffix
I needed to sort on the number suffix so had to restrict the pattern to numerics and to get around numbers of different lengths sorting as text (ie I wanted 2 to sort before 19) cast the result:
,cast(substring(s, 2 + len(s) - patindex('%[^0-9]%',reverse(s)),99) as integer) as numeric_suffix
Related
I need to replace a string in a SAS dataset in the following way :
OTTAWA ON should be replaced with OTTAWA ONTARIO
WHATEVER QC should be replaced with WHATEVER QUEBEC
etc.
However, HOUSE ON THE HILL should not become HOUSE ONTARIO THE HILL.
That is, I want to replace all instances of ON with ONTARIO but only if ON exists as the last word in the string
You could use regular expressions to do this. From what you have described, I think the following should work.
myString = prxchange("s/(.*)( ON)$/$1 ONTARIO/",-1,strip(myString));
myString = prxchange("s/(.*)( QC)$/$1 QUEBEC/",-1,strip(myString));
Use a separate control data set to maintain the substitutions (postal code -> province) you want.
Load the control data into a hash
Process the data scanning out the last 'word'
If the word is a key in the hash then replace the word with the province value.
Presuming you are only performing transformations for a 'token' (CA postal code) as the final 'word' an example of the control data, data and transformation is as follows:
data O_Canada(label="Our home and native land");
length postal $2 province $26 ;
input postal& province&; * suffix & means data fields separated by >1 space;
datalines;
ON Ontario
QC Quebec
NS Nova Scotia
NB New Brunswick
MB Manitoba
BC British Columbia
PE Prince Edward Island
SK Saskatchewan
AB Alberta
NL Newfoundland and Labrador
;
data cities(label='Some popular places');
length place $100;
input place $CHAR50.;
datalines;
CALGARY AB
VANCOUVER BC
WINNIPEG MB
MONCTON NB
ST. JONHS NL
HALIFAX NS
TORONTO ON
MONTREAL QC
SAKATOON SK
CHARLOTTETOWN PE
WHITEHORSE YT
YELLOWKNIFE NT
IQALUIT NU
GOLDMINE YUKON
;
data cities;
modify cities;
if _n_ = 1 then do;
length postal $3 province $26; * postal 1 bigger so scanned postal will not always match;
declare hash provinces(dataset:'O_Canada');
provinces.defineKey('postal');
provinces.defineData('province');
provinces.defineDone();
call missing(postal, province);
drop postal province;
end;
postal = scan(place,-1,' ');
if provinces.find() eq 0 then do;
* this inline replacement presumes all postal codes are 2 characters;
* -1 from length will replace starting from found postal;
substr(place,length(place)-1) = province; * inline replacement;
replace;
end;
run;
Result
data GEOGRAPHY;
file datalines truncover;
informat geo $2. graphy $32.;
input geo $ graphy $;
datalines;
ON ONTARIO
QC QUEBEC
;
proc sql;
select whatever_you_want,
case graphy
when '' then myString
else substr(myString, length(myString) - length(geo)) || graphy
end as myString
from HAVE left joion GEOGRAPHY on scan(myString, -1) eq geo;
quit;
scan(myString, -1) returns the last word in myString and trim(myString) removes trailing blanks, so in a data step this does the job:
cutString = substr(myString, length(myString) - 2);
select scan(myString, -1)
when 'ON' myString = cutString || 'ONTARIO';
when 'QC' myString = cutString || 'QUEBEC';
end;
or in SQL
select case scan(myString, -1)
when 'ON' then trim(myString) || 'TARIO'
when 'QC' then substr(myString, length() - 2) || 'QUEBEC'
else myString end as myString
from YOU_KNOW_BETTER_THAN_I_DO;
#Sonny, I think the regular expression is very good. And #astel, there is another easy understading way:
data test;
InText = 'HOUSE ON THE HILL';
output;
InText = 'OTTAWA ON';
output;
run;
data _null_;
set test;
if cats(reverse(InText)) =: 'NO ' then OutText = tranwrd(InText,' ON',' ONTARIO');
put Intext = #30 OutText = ;
run;
The output will be
InText=HOUSE ON THE HILL OutText=
InText=OTTAWA ON OutText=OTTAWA ONTARIO
Reverse the variable so you can easily judge if the new variable is start with NO , that means the raw variable is end with ON. Then do the replace work by using a tranwrd() funtion.
I have an interesting problem and am wondering if oracle has a built-in function to do this or I need to find a fast way to do it in plsql.
Take 2 strings:
s1 = 'abc def hijk'
s2 = 'abc def iosk'
The function needs to return abc def because the strings are exactly the same up to that point.
Another example:
s1 = 'abc def hijk www'
s2 = 'abc def iosk www'
The function needs to return abc def.
The only way I can think of doing this is loop through string1 and compare each character with substr() again the substr of string 2.
Just wondering if Oracle's got something built-in. Performance is pretty important.
After re-reading your question, here would be what you really wanted:
with cte1 as (
select 1 id, 'abc def hijk www' str from dual
union all
select 2 id, 'abc def iosk www' str from dual
), num_gen as (
-- a number generator up to the minimum length of the strings
SELECT level num
FROM dual t
CONNECT BY level <= (select min(length(str)) from cte1)
), cte2 as (
-- build substrings of increasing length
select id, num_gen.num, substr(cte1.str, 1, num_gen.num) sub
from cte1
cross join num_gen
), cte3 as (
-- self join to check if the substrings are equal
select x1.num, x1.sub sub1, x2.sub sub2
from cte2 x1
join cte2 x2 on (x1.num = x2.num and x1.id != x2.id)
), cte4 as (
-- select maximum string length
select max(num) max_num
from cte3
where sub1 = sub2
)
-- finally, get the substring with the max length
select cte3.sub1
from cte3
join cte4 on (cte4.max_num = cte3.num)
where rownum = 1
Essentially, this is what you would do in pl/sql: Build substrings of increasing length and stop at the point at which they are not matching anymore.
I doubt that there is some built-in SQL function, but it can be done in SQL only using regular expressions:
with cte1 as (
select 1 id, 'abc def hijk www' str from dual
union all
select 2 id, 'abc def iosk www' str from dual
), cte2 as (
SELECT distinct id, trim(regexp_substr(str, '[^ ]+', 1, level)) str
FROM cte1 t
CONNECT BY instr(str, ' ', 1, level - 1) > 0
)
select distinct t1.str
from cte2 t1
join cte2 t2 on (t1.str = t2.str and t1.id != t2.id)
I haven't done any performance tests, but my experience tells me this is most likely faster than any pl/sql solution since you are totally avoiding context switches.
You should check the package UTL_MATCH for a similar functionality, but the get exact your request you must write own function.
The binary search for the common substring length provides good performance for long strings.
create or replace function ident_pfx(str1 varchar2, str2 varchar2) return varchar2
as
len_beg PLS_INTEGER;
len_end PLS_INTEGER;
len_mid PLS_INTEGER;
len_result PLS_INTEGER;
begin
if str1 is null or str2 is null then return null; end if;
--
len_result := 0;
len_beg := 0;
len_end := least(length(str1),length(str2));
LOOP
BEGIN
-- use binary search for the common substring length
len_mid := ceil((len_beg + len_end) / 2);
IF (substr(str1,1,len_mid) = substr(str2,1,len_mid))
THEN
len_beg := len_mid; len_result := len_mid;
ELSE
len_end := len_mid;
END IF;
END;
IF (len_end - len_beg) <= 1 THEN
-- check last character
IF (substr(str1,1,len_end) = substr(str2,1,len_end))
THEN
len_result := len_end;
END IF;
EXIT ;
END IF;
END LOOP;
return substr(str1,1,len_result);
end;
/
select ident_pfx('abc def hijk www','abc def iosk www') ident_pfx from dual;
abc def
Another possible solution would be to use the XOR.
If you XOR the two strings together, the result should have a NUL byte whereever the two strings match.
XOR is not a native operator, but i am pretty sure there is support for it in one of the libraries.
If "the performance is pretty important", you should avoid the "looping" on substrings.
Here an alternative using the XOR (as proposed by #EvilTeach).
with string_transform as (
select 'abc def hijk www' str1, 'abc def iosk www' str2 from dual
),
str as (
select
str1, str2,
-- add suffix to handle nulls and identical strings
-- calculate XOR
utl_raw.bit_xor(utl_raw.cast_to_raw(str1||'X'),utl_raw.cast_to_raw(str2||'Y')) str1_xor_str2
from string_transform
), str2 as (
select
str1, str2,
str1_xor_str2,
-- replace all non-identical characters (not 00) with 2D = '-'
utl_raw.translate(str1_xor_str2,
utl_raw.translate(str1_xor_str2,'00','01'),
utl_raw.copies('2D',length(str1_xor_str2))) xor1
from str
), str3 as (
select
str1, str2,
-- replace all identical characters (00) with 2B (= '+') and cast back to string
utl_raw.cast_to_varchar2(utl_raw.translate(xor1,'00','2B')) diff
-- diff = ++++++++---+++++ (+ means identical position; - difference)
from str2
)
select str1, str2,
-- remove the appended suffix character
substr(diff,1,length(diff)-1) diff,
-- calculate the length of the identical prefix
instr(diff,'-')-1 same_prf_length
from str3
;
Basically both strings are first converted to RAW format. XOR sets the identical bytes (characters) to 00. With translate the identical bytes are converted to '+', all other to '-'.
The identical prefix length is the position of the first '-' in the string minus one.
Technically a (different) sufix character is added to both strings to hanlde NULLs and identical strings.
Note that if the string is longer that 2000, some extra processing must be added
due to limitation of UTL_RAW.CAST_TO_VARCHAR2.
If my string is this: 'this is a string', how can I produce all possible combinations by joining each word with its neighboring word?
What this output would look like:
this is a string
thisis a string
thisisa string
thisisastring
thisis astring
this isa string
this isastring
this is astring
What I have tried:
s = 'this is a string'.split()
for i, l in enumerate(s):
''.join(s[0:i])+' '.join(s[i:])
This produces:
'this is a string'
'thisis a string'
'thisisa string'
'thisisastring'
I realize I need to change the s[0:i] part because it's statically anchored at 0 but I don't know how to move to the next word is while still including this in the output.
A simpler (and 3x faster than the accepted answer) way to use itertools product:
s = 'this is a string'
s2 = s.replace('%', '%%').replace(' ', '%s')
for i in itertools.product((' ', ''), repeat=s.count(' ')):
print(s2 % i)
You can also use itertools.product():
import itertools
s = 'this is a string'
words = s.split()
for t in itertools.product(range(len('01')), repeat=len(words)-1):
print(''.join([words[i]+t[i]*' ' for i in range(len(t))])+words[-1])
Well, it took me a little longer than I expected... this is actually tricker than I thought :)
The main idea:
The number of spaces when you split the string is the length or the split array - 1. In our example there are 3 spaces:
'this is a string'
^ ^ ^
We'll take a binary representation of all the options to have/not have either one of the spaces, so in our case it'll be:
000
001
011
100
101
...
and for each option we'll generate the sentence respectively, where 111 represents all 3 spaces: 'this is a string' and 000 represents no-space at all: 'thisisastring'
def binaries(n):
res = []
for x in range(n ** 2 - 1):
tmp = bin(x)
res.append(tmp.replace('0b', '').zfill(n))
return res
def generate(arr, bins):
res = []
for bin in bins:
tmp = arr[0]
i = 1
for digit in list(bin):
if digit == '1':
tmp = tmp + " " + arr[i]
else:
tmp = tmp + arr[i]
i += 1
res.append(tmp)
return res
def combinations(string):
s = string.split(' ')
bins = binaries(len(s) - 1)
res = generate(s, bins)
return res
print combinations('this is a string')
# ['thisisastring', 'thisisa string', 'thisis astring', 'thisis a string', 'this isastring', 'this isa string', 'this is astring', 'this is a string']
UPDATE:
I now see that Amadan thought of the same idea - kudos for being quicker than me to think about! Great minds think alike ;)
The easiest is to do it recursively.
Terminating condition: Schrödinger join of a single element list is that word.
Recurring condition: say that L is the Schrödinger join of all the words but the first. Then the Schrödinger join of the list consists of all elements from L with the first word directly prepended, and all elements from L with the first word prepended with an intervening space.
(Assuming you are missing thisis astring by accident. If it is deliberately, I am sure I have no idea what the question is :P )
Another, non-recursive way you can do it is to enumerate all numbers from 0 to 2^(number of words - 1) - 1, then use the binary representation of each number as a selector whether or not a space needs to be present. So, for example, the abovementioned thisis astring corresponds to 0b010, for "nospace, space, nospace".
I've been working on this part of a saycommand system which is supposed to separate parts of a string and put them in a table which is sent to a function, which is queried at the beginning of the string. This would look like, for example, !save 1 or !teleport 0 1, or !tell 5 "a private message".
I would like this string to turn into a table:
[[1 2 word 2 9 'more words' 1 "and more" "1 2 34"]]
(Every non-quoted part of the string gets its own key, and the quoted parts get grouped into a key)
1 = 1
2 = 2
3 = word
4 = 2
5 = 9
6 = more words
7 = 1
8 = and more
9 = 1 2 34
I've tried doing this with Lua pattern, but I'm stuck trying to find out how to capture both quoted and unquoted pieces of the string. I've tried a lot of things, but nothing helped.
My current pattern attempts look like this:
a, d = '1 2 word 2 9 "more words" 1 "and more" "1 2 34"" ', {}
-- previous attempts
--[[
This one captures quotes
a:gsub('(["\'])(.-)%1', function(a, b) table.insert(d, b) end)
This one captures some values and butchered quotes,
which might have to do with spaces in the string
a:gsub('(["%s])(.-)%1', function(a, b) table.insert(d, b) end)
This one captures every value, but doesn't take care of quotes
a:gsub('(%w+)', function(a) table.insert(d, a) end)
This one tries making %s inside of quotes into underscores to
ignore them there, but it doesn't work
a = a:gsub('([%w"\']+)', '%1_')
a:gsub('(["\'_])(.-)%1', function(a, b) table.insert(d, b) end)
a:gsub('([%w_]+)', function(a) table.insert(d, a) end)
This one was a wild attempt at cracking it, but no success
a:gsub('["\']([^"\']-)["\'%s]', function(a) table.insert(d, a) end)
--]]
-- This one adds spaces, which would later be trimmed off, to test
-- whether it helped with the butchered strings, but it doesn't
a = a:gsub('(%w)(%s)(%w)', '%1%2%2%3')
a:gsub('(["\'%s])(.-)%1', function(a, b) table.insert(d, b) end)
for k, v in pairs(d) do
print(k..' = '..v)
end
This would not be needed for simple commands, but a more complex one like !tell 1 2 3 4 5 "a private message sent to five people" does need it, first to check if it's sent to multiple people and next to find out what the message is.
Further down the line I want to add commands like !give 1 2 3 "component:material_iron:weapontype" "food:calories", which is supposed to add two items to three different people, would benefit greatly from such a system.
If this is impossible in Lua pattern, I'll try doing it with for loops and such, but I really feel like I'm missing something obvious. Am I overthinking this?
You cannot process quoted strings with Lua patterns. You need to parse the string explicitly, as in the code below.
function split(s)
local t={}
local n=0
local b,e=0,0
while true do
b,e=s:find("%s*",e+1)
b=e+1
if b>#s then break end
n=n+1
if s:sub(b,b)=="'" then
b,e=s:find(".-'",b+1)
t[n]=s:sub(b,e-1)
elseif s:sub(b,b)=='"' then
b,e=s:find('.-"',b+1)
t[n]=s:sub(b,e-1)
else
b,e=s:find("%S+",b)
t[n]=s:sub(b,e)
end
end
return t
end
s=[[1 2 word 2 9 'more words' 1 "and more" "1 2 34"]]
print(s)
t=split(s)
for k,v in ipairs(t) do
print(k,v)
end
Lua string patterns and regex for that matter generally aren't well suited when you need to do parsing that requires varying nesting levels or token count balancing like parenthesis ( ). But there is another tool available to Lua that's powerful enough to deal with that requirement: LPeg.
The LPeg syntax is a bit archaic and takes some getting use to so I'll use the lpeg re module instead to make it easier to digest. Keep in mind that anything you can do in one form of the syntax you can also express in the other form as well.
I'll start by defining the grammar for parsing your format description:
local re = require 're'
local cmdgrammar =
[[
saycmd <- '!' cmd extra
cmd <- %a%w+
extra <- (singlequote / doublequote / unquote / .)*
unquote <- %w+
singlequote <- "'" (unquote / %s)* "'"
doublequote <- '"' (unquote / %s)* '"'
]]
Next, compile the grammar and use it to match some of your test examples:
local cmd_parser = re.compile(cmdgrammar)
local saytest =
{
[[!save 1 2 word 2 9 'more words' 1 "and more" "1 2 34"]],
[[!tell 5 "a private message"]],
[[!teleport 0 1]],
[[!say 'another private message' 42 "foo bar" baz]],
}
There are currently no captures in the grammar so re.match returns the last character position in the string it was able to match up to + 1. That means a successful parse will return the full character count of the string + 1 and therefore is a valid instance of your grammar.
for _, test in ipairs(saytest) do
assert(cmd_parser:match(test) == #test + 1)
end
Now comes the interesting part. Once you have the grammar working as desired you can now add captures that automatically extracts the results you want into a lua table with relatively little effort. Here's the final grammar spec + table captures:
local cmdgrammar =
[[
saycmd <- '!' {| {:cmd: cmd :} {:extra: extra :} |}
cmd <- %a%w+
extra <- {| (singlequote / doublequote / { unquote } / .)* |}
unquote <- %w+
singlequote <- "'" { (unquote / %s)* } "'"
doublequote <- '"' { (unquote / %s)* } '"'
]]
Running the tests again and dumping the re.match results:
for i, test in ipairs(saytest) do
print(i .. ':')
dump(cmd_parser:match(test))
end
You should get output similar to:
lua say.lua
1:
{
extra = {
"1",
"2",
"word",
"2",
"9",
"more words",
"1",
"and more",
"1 2 34"
},
cmd = "save"
}
2:
{
extra = {
"5",
"a private message"
},
cmd = "tell"
}
3:
{
extra = {
"0",
"1"
},
cmd = "teleport"
}
4:
{
extra = {
"another private message",
"42",
"foo bar",
"baz"
},
cmd = "say"
}
I want to split string into words like below, the output of all the string should be same:
INPUT:
1. This is a string
2. This is a string
3. This is a string
4. This is a string
OUTPUT:
This is a
Means, that I want first three words from the sentence, irrespective of the spaces.
Try this:
declare #s1 varchar(3000) ;
declare #xml xml,#str varchar(100),#delimiter varchar(10), #out varchar(max);;
select #delimiter =' '
select #s1 = 'This is a string';
select #s1 = 'This is a string ';
select #s1 = 'This is a string ';
select #s1 = 'This is a string';
select #xml = cast(('<X>'+replace(#s1,#delimiter ,'</X><X>')+'</X>') as xml)
select top 3 #out =
COALESCE(#out + ' ', '') + C.value('.', 'varchar(100)')
from #xml.nodes('X') as X(C)
where LEN(C.value('.', 'varchar(10)')) > 0
select #out
Now your case contains two steps:
1. Removing additional spaces and converting them to single space. You can use REPLACE() method to this.
SELECT REPLACE(REPLACE(REPLACE("This is a string",' ','<>'),'><',''),'<>',' ')
Process:
The innermost REPLACE changes all blanks to a less-than greater-than pair.
If there are three spaces between This and is, the innermost REPLACE returns This<><><>is.
The middle REPLACE changes all greater-than less-than pairs to the empty string, which removes them.
The<><><>is becomes The<>is.
The outer REPLACE changes all less-than greater-than pairs to a single blank. The<>is becomes
The is.
Now all the sentences are normalized with one space.
2. Split the words and get the three words.
There are lot of Stackoverflow question which discusses them. I liked the Common Table Expression to split the string : How do I split a string so I can access item x?
Let me know if you require any help in the splitting the words.
Create a Tally Table:
SELECT TOP 11000
IDENTITY( INT,1,1 ) AS Num
INTO dbo.Tally
FROM Master.dbo.SysColumns sc1,
Master.dbo.SysColumns sc2
GO
Create a Table Valued Function:
CREATE FUNCTION dbo.[fnSetSplit]
(
#String VARCHAR(8000),
#Delimiter CHAR(1)
)
RETURNS TABLE
AS
RETURN
( SELECT Num,
SUBSTRING(#String, CASE Num
WHEN 1 THEN 1
ELSE Num + 1
END,
CASE CHARINDEX(#Delimiter, #String,
Num + 1)
WHEN 0
THEN LEN(#String) - Num + 1
ELSE CHARINDEX(#Delimiter,
#String, Num + 1)
- Num
- CASE WHEN Num > 1 THEN 1
ELSE 0
END
END) AS String
FROM dbo.Tally
WHERE Num <= LEN(#String)
AND ( SUBSTRING(#String, Num, 1) = #Delimiter
OR Num = 1 )
)
Query function:
SELECT TOP 3
fss.String
FROM dbo.fnSetSplit('This is a string', ' ') fss
WHERE NOT ( fss.String = '' )
If you need to reconcatenate, look at string concatenation using FOR XML (PATH)
SQL Server 2016 (compatibility level 130) allows to use STRING_SPLIT function:
DECLARE #delimiter varchar(10) = ' '
SELECT STRING_AGG(value, #delimiter)
FROM (SELECT TOP 3 value FROM STRING_SPLIT('This is a string', #delimiter) WHERE LEN(value)>0) inq
SELECT STRING_AGG(value, #delimiter)
FROM (SELECT TOP 3 value FROM STRING_SPLIT('This is a string ', #delimiter) WHERE LEN(value)>0) inq
SELECT STRING_AGG(value, #delimiter)
FROM (SELECT TOP 3 value FROM STRING_SPLIT('This is a string', #delimiter) WHERE LEN(value)>0) inq
SELECT STRING_AGG(value, #delimiter)
FROM (SELECT TOP 3 value FROM STRING_SPLIT('This is a string', #delimiter) WHERE LEN(value)>0) inq
Result:
This is a
This is a
This is a
This is a