Question

I need to get a substring from a path directory made of 2 numbers preceeded and followed by a ‘_’. The string is like:

'P:\pgdfecol\71698384737978\INFENTECONTROL\2011_9_43\2011_9_46_43_29_10.ZIP'

and in this case I want to get the 43 following the 46.

The path is stored following the next rule, after the last backslash:

'\TablesPK_twoCharactersClassification_twoCharactersDocumentType_anything.ZIP'

I want to get the classification. The catch is that the table’s primary key could be more than one field, although I know in each case how many fields are in the pk.

I got with something like this:

select substring(substring(substring('P:\pgdfecol\71698384737978\INFENTECONTROL\2011_9_43\2011_9_46_43_29_10.ZIP' from '([^\\]*(\.ZIP|zip))') from '([^_]*_){4}') from '[0-9]{2}')

But I would like something simpler.

Other cases:

'P:\pgdfecol\71698384737978\INFENTECONTROL\2011_03_46\2011_03_46_46_48_.ZIP'

(need the second 46)

'P:\pgdfecol\71698384737978\INFCONTABLE\2009_05_INBP\2009_05_INBP_22_28_.ZIP'

(need the 22 after INBP)

'P:\pgdfecol\71698384737978\INFOFICIAL\2007_06_MB\2007_06_MB_29_28_.ZIP'

(need the 29)

'P:\pgdfecol\71698384737978\ASOCIADOS\8010625\8010625_02_04_20110111.ZIP'

(02 after 8010625)

In the last case, the pk is only one field, so I’ve changed the sentence as:

select substring(substring(substring('P:\pgdfecol\71698384737978\ASOCIADOS\8010625\8010625_02_04_20110111.ZIP' from '([^\\]*(\.ZIP|zip))') from '([^_]*_){2}') from '[0-9]{2}')

For one Pk, I need the second set of ([^_]*_), for three the fourth, and so on..

select substring(substring(substring('P:\pgdfecol\71698384737978\ACTASCOMITE\ACRE123\ACRE123_17_11_.ZIP' from '([^\\]*(\.ZIP|zip))') from '([^_]*_){2}') from '[0-9]{2}')

(I get 17)

I’m using postgres 9.0.

Answer 1

I am beginning to understand. Consider this test case:

WITH x(txt) AS ( VALUES
     ('P:\pgdfecol\71698384737978\INFENTECONTROL\2011_9_43\2011_9_46_43_29_10.ZIP')  -- 43
    ,('P:\pgdfecol\71698384737978\INFENTECONTROL\2011_03_46\2011_03_46_46_48_.ZIP')  --need the second 46
    ,('P:\pgdfecol\71698384737978\INFCONTABLE\2009_05_INBP\2009_05_INBP_22_28_.ZIP') --need the 22 after INBP
    ,('P:\pgdfecol\71698384737978\INFOFICIAL\2007_06_MB\2007_06_MB_29_28_.ZIP')      --need the 29
    )
SELECT txt, substring(txt, '\\(?:[^_\\]+_){3}(\d\d)_[^\\]*\.(?:ZIP|zip)$')
FROM   x

(?:) .. non-capturing parenthesis
[^_\\].. character class with any character except \ and _
\d .. a digit, same as [0-9] effectively
+ .. 1 or more matches (greedy)
$ .. end of string
[_\d] .. character class with digits and _

The case with only one pk needs a different pattern. Use {1} instead of {3}.

Not sure why you escape the backslashes. In modern versions of PostgreSQL standard_conforming_strings is on by default so you don’t need to escape backslashes in strings – but still in regular expression of course.