Replace every 2nd instance of character in string with TSQL - sql-server

Question

I have a field that contains a string of lat/long co-ordinates that define a geofence (polygon). Each is seperated by a comma.
eg: 'lat,long,lat,long,lat,long'
eg: 148.341158,-21.500773,148.341406,-21.504989,148.375136,-21.513174,148.401674,-21.535247,148.418044,-21.532767,148.408867,-21.511685,148.414075,-21.508461,148.36968,-21.432567,148.349094,-21.438768,148.346862,-21.480187,148.341158,-21.500773,
I'd like to use this with the geography type in MSSQL (http://msdn.microsoft.com/en-us/library/bb933971.aspx)
DECLARE #g geography;
SET #g = geography::STPolyFromText('POLYGON((-122.358 47.653, -122.348 47.649, -122.348 47.658, -122.358 47.658, -122.358 47.653))', 4326);
SELECT #g.ToString();
This seems to require: 'lat long, lat long, lat long' ie: no comma between the pair.
I can't change the source data, as it's used by a vendor program. I need to manipulate the string to remove every 1 out of 2 commas, or failing that, get regex working in TSQL

Answer 1

You could use a numbers table (a very handy tool for many purposes, by the way) to find the positions of all the commas, then, using only the odd position numbers, replace the corresponding commas with spaces in a single SELECT statement. Here's what I am talking about:
WITH CTE AS (
SELECT number, rn = ROW_NUMBER() OVER (ORDER BY number)
FROM master.dbo.spt_values
WHERE type = 'P'
AND number BETWEEN 1 AND LEN(#coords)
AND SUBSTRING(#coords, number, 1) = ','
)
SELECT
#coords = STUFF(#coords, number, 1, ' ')
FROM CTE
WHERE rn % 2 = 1
;
In the above query, the numbers table's "part" is "played" by a subset of system table master.dbo.spt_values. The CTE calculates the positions of all the commas in the #coords string, returning the results as a row set. The main SELECT is used an assignment statement. It takes every number in the CTE set and removes the character at the corresponding position in #coords, replacing it with a space character (all with the help of the STUFF function).
You can use this SQL Fiddle demo to play with the query.

Answer 2

This ended up my Solution:
DECLARE #WorkingCoordList VARCHAR(max) = 'some lat,long,lat,long string'
DECLARE #Processed INT = 0
DECLARE #CommaLoc INT
DECLARE #Count INT = 0
WHILE #Processed = 0
BEGIN
SET #CommaLoc = PATINDEX('%,%', #WorkingCoordList)
IF #Count % 2 = 0
BEGIN
SET #WorkingCoordList = STUFF(#WorkingCoordList, #CommaLoc, 1, ' ') --Convert comma to space
END
ELSE
BEGIN
SET #WorkingCoordList = STUFF(#WorkingCoordList, #CommaLoc, 1, '#') -- Convert comma to hash
END
IF #CommaLoc = LEN(#WorkingCoordList)
BEGIN
SET #WorkingCoordList = LEFT(#WorkingCoordList, LEN(#WorkingCoordList) - 1) -- trim trailing ,
SET #WorkingCoordList = RTRIM(LTRIM(REPLACE(#WorkingCoordList, '#', ', '))) -- Convert all the hashes to commas
SET #Processed = 1
END
SET #Count = #Count + 1
END
END

Answer 3

You could roll your own parser like the following. It uses the commas in the string to find all of the lattitude/ longitude values. It concatenates all the values together using the pattern: lat long, lat long, ...
declare #list varchar(max)
declare #result varchar(max)
declare #word varchar(max)
declare #splitOn varchar(1)
declare #wpos int
declare #cpos int
declare #wordCount int
select #list = '148.341158,-21.500773,148.341406,-21.504989,148.375136,-21.513174,148.401674,-21.535247,148.418044,-21.532767,148.408867,-21.511685,148.414075,-21.508461,148.36968,-21.432567,148.349094,-21.438768,148.346862,-21.480187,148.341158,-21.500773,'
select #splitOn = ','
select #result = ''
select #cpos = 0
select #wpos = 1
select #wordCount = 1
while (#cpos <= len(#list))
begin
select #cpos = charindex(#splitOn, #List, #cpos)
if (#cpos < 1) select #cpos = len(#list) + 1
select #word = substring(#list, #wpos, #cpos - #wpos)
select #result = #result + ' ' + #word
if ((#wordCount % 2) = 0 and (#cpos < len(#list))) select #result = #result + ','
select #cpos = #cpos + 1
select #wpos = #cpos
select #wordCount = #wordCount + 1
end
select #result as result
Which produces the following string:
148.341158 -21.500773, 148.341406 -21.504989, 148.375136 -21.513174, 148.401674 -21.535247, 148.418044 -21.532767, 148.408867 -21.511685, 148.414075 -21.508461, 148.36968 -21.432567, 148.349094 -21.438768, 148.346862 -21.480187, 148.341158 -21.500773

Answer 4

I don't know how your regex works but if you preprocess the string with a regex, it might work with a global search and replace like this:
find: ,([^,]*(?:,|$))
replace: '$1' ie. space plus capture group 1

Answer 5

Thanks for the code :-)
I had a slightly different scenario however created a function using "chue x" code to add a character '#' at every 3rd ';'
/*** My select query ***/
SELECT dbo.fn_AddEveryNthItem(a.MyString, ';','#', 3) AS Expr1
FROM dbo.MyTable as a
Function is below
/*** Function to add character for every nth item ***/
Create function dbo.fn_AddEveryNthItem(#list varchar(1000), #splitOn varchar(1), #addChar varchar(1), #EveryNthItem int)
RETURNS VARCHAR(1000)
AS
BEGIN
declare #word varchar(max)
declare #result varchar(max) = ''
declare #wpos int = 1
declare #cpos int = 0
declare #wordCount int =1
while (#cpos <= len(#list))
begin
select #cpos = charindex(#splitOn, #List, #cpos)
if (#cpos < 1) select #cpos = len(#list) + 1
select #word = substring(#list, #wpos, #cpos - #wpos)
select #result = #result + #splitOn + #word
if ((#wordCount % #EveryNthItem) = 0 and (#cpos < len(#list))) select #result = #result + #addChar
select #cpos = #cpos + 1
select #wpos = #cpos
select #wordCount = #wordCount + 1
end
Return #result
end