Extract a uniform pattern from a file path

Question

Extract a uniform pattern from a file path

pnr8uk

SSCommitted

Points: 1828
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October 10, 2016 at 1:20 am

#313788

Hi All
I have a strange problem here which I can't quite get to the bottom of.
I have a file path in my table which reads like this:
\\cc0dware01\BMF\RMS\REBATERATES-WORKING-2016ToEnd08-20160905180112.csv
I only want to extract 20160905180112 from the string, the name format is strict and will always be the same.
Despite trying many things LEFT, REVERSE, CHARINDEX and getting very close I can't quite pull out the part I want every time.
Anyone got an easy fix for this?
Any help greatly appreciated.

Viewing 7 posts - 1 through 6 (of 6 total)

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Eirikur Eiriksson SSC Guru Points: 182882 More actions · Answer 1

pnr8uk (10/10/2016)
Hi All
I have a strange problem here which I can't quite get to the bottom of.
I have a file path in my table which reads like this:
\\cc0dware01\BMF\RMS\REBATERATES-WORKING-2016ToEnd08-20160905180112.csv
I only want to extract 20160905180112 from the string, the name format is strict and will always be the same.
Despite trying many things LEFT, REVERSE, CHARINDEX and getting very close I can't quite pull out the part I want every time.
Anyone got an easy fix for this?
Any help greatly appreciated.

Quick suggestion

😎

DECLARE @INPUT_STR VARCHAR(1024) = '\\cc0dware01\BMF\RMS\REBATERATES-WORKING-2016ToEnd08-20160905180112.csv';

SELECT

LEFT(RIGHT(@INPUT_STR,CHARINDEX(CHAR(45),REVERSE(@INPUT_STR),1) - 1),14) AS OUT_STR;

Output

OUT_STR

--------------

20160905180112

pnr8uk SSCommitted Points: 1828 More actions · Answer 2

Wow thank you that is amazingly simple.

Many thanks again

Eirikur Eiriksson SSC Guru Points: 182882 More actions · Answer 3

Eirikur Eiriksson

SSC Guru

Points: 182882

October 10, 2016 at 2:46 am

#1905607

You are very welcome.

😎

Luis Cazares SSC Guru Points: 183698 More actions · Answer 4

This might be is simpler.

DECLARE @INPUT_STR VARCHAR(1024) = '\\cc0dware01\BMF\RMS\REBATERATES-WORKING-2016ToEnd08-20160905180112.csv';

SELECT SUBSTRING( @INPUT_STR, LEN( @INPUT_STR) - 17, 14);

I wonder what are the varying parts of the file path. Is it only the path? Maybe WORKING can change to something else? Or maybe the time period? Knowing what can change, we can be sure to prevent any problems.

Luis C.
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Jason A. Long SSC-Insane Points: 23780 More actions · Answer 5

Looks like a time code (fixed at 14 characters)... The following will work even if the file path changes...

DECLARE @input VARCHAR(1000) = '\\cc0dware01\BMF\RMS\REBATERATES-WORKING-2016ToEnd08-20160905180112.csv';

SELECT output = LEFT(RIGHT(@input, 18), 14);

ScottPletcher SSC Guru Points: 100947 More actions · Answer 6

As long as the file name is the only YYYYMMDDHHMMSS string in the file, I think you can just check for that pattern:

SELECT input, SUBSTRING(input, PATINDEX('%[2][01][0-9][0-9][01][0-9][012][0-9][0-5][0-9][0-5][0-9]%', input), 14) AS extract

FROM (

VALUES('\\cc0dware01\BMF\RMS\REBATERATES-WORKING-2016ToEnd08-20160905180112.csv'),

('\\cc0dware01\BMF\RMS\REBATERATES-WORKING-2016ToEnd08-20160905180113.xlsx'),

('c:\20160905180113.')

) AS test_data(input)

Edit: Added SQL code block.

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