Reputation: 33
I have a column of data that looks like this:
58,0:102,56.00
52,0:58,68
58,110
57,440.00
52,0:58,0:106,6105.95
I need to extract the character before the last delimiter (',').
Using the data above, I want to get:
102
58
58
57
106
Upvotes: 1
Views: 3446
Reputation: 29
This worked for me, I just needed the last field. select substring('abc:def:hij:kef', ':[^:]*$');
:kef (1 row)
Upvotes: -1
Reputation: 14932
The following treats the source string as an "array of arrays". It seems each data element can be defined as S(x,y) and the overall string as S1:S2:...Sn. The task then becomes to extract x from Sn.
with as_array as
( select string_to_array(S[n], ',') Sn
from (select string_to_array(col,':') S
, length(regexp_replace(col, '[^:]','','g'))+1 n
from tablename
) t
)
select Sn[array_length(Sn,1)-1] from as_array
The above extends S(x,y) to S(a,b,...,x,y) the task remains to extracting x from Sn. If it is the case that all original sub-strings S are formatted S(x,y) then the last select reduces to select Sn[1]
Upvotes: 0
Reputation: 659367
Might be done with a regular expression in substring()
. If you want:
the longest string of only digits before the last comma:
substring(data, '(\d+)\,[^,]*$')
Or you may want:
the string before the last comma (',') that's delimited at the start either by a colon (':') or the start of the string.
Could be another regexp:
substring(data, '([^:]*)\,[^,]*$')
Or this:
reverse(split_part(split_part(reverse(data), ',', 2), ':', 1))
More verbose but typically much faster than a (expensive) regular expression.
db<>fiddle here
Upvotes: 1
Reputation: 164224
With a CTE that removes everything after the last comma and then splits the rest into an array:
with cte as (
select
regexp_split_to_array(
replace(left(col, length(col) - position(',' in reverse(col))), ':', ','),
','
) arr
from tablename
)
select arr[array_upper(arr, 1)] from cte
See the demo.
Results:
| result |
| ------ |
| 102 |
| 58 |
| 58 |
| 57 |
| 106 |
Upvotes: 0
Reputation: 16417
Can't promise this is the best way to do it, but it is a way to do it:
with splits as (
select string_to_array(bar, ',') as bar_array
from foo
),
second_to_last as (
select
bar_array[cardinality(bar_array)-1] as field
from splits
)
select
field,
case
when field like '%:%' then split_part (field, ':', 2)
else field
end as last_item
from second_to_last
I went a little overkill on the CTEs, but that was to expose the logic a little better.
Upvotes: 0