正则表达式

Question

I FOUND MY MISTAKE!!! This is wrong CASE WHEN regexp_substr(artikel.abez1,'[^/]*$') AS download,

I just deleted 'Case when' and it works! Thank you everyone! :)

I have this data:

1. DE-Internet-LTE

2. AE-Internet-Ethernet-10M/30M

How can I get only the value "10M"? and if there's none, I want it to return to null.

I used this query:

regexp_substr(artikel.abez1,'[^-/]*$') AS upload

On the second row, it gives the result "10M" but on the first row it return to "LTE" instead of "null".

I'm using SQL Tool 1.8 b38.

UPDATE: My full query: `

SELECT DISTINCT artikel.artnr1, L1.lfnr1 AS lfnr, L1.name1 AS lf_name, artikel.abez1, regexp_substr(artikel.abez1,'(. ?){1}(. ?)-', 1, 1, '', 2) AS land, regexp_substr(artikel.abez1,'(. ?-){1}(. ?)-', 1, 1, '', 2) AS Technologie, CASE WHEN regexp_substr(artikel.abez1,'(. ?-){2}(. ?)-', 1, 1, '', 2) IS NULL THEN regexp_substr(artikel.abez1,'[^-] $') ELSE regexp_substr(artikel.abez1,'(. ?-){2}(. ?)-', 1, 1, '', 2) END AS Topologie, regexp_substr(regexp_substr(artikel.abez1, '([^-/]+)/[^-/]+$'), '^[^-/]+') AS upload, CASE WHEN regexp_substr(artikel.abez1,'[^/] $') AS download, bestanfragepos.preis / bestanfrage.bwkurs, 'Anfrage' AS Art, To_Char(bestanfrage.lfdanfrage), bestanfrage.anfragedatum, CASE WHEN InStr(angaufgut.reserve1, '.') > 1 THEN Months_Between( bestanfrage.anfragedatum, To_Date(angaufgut.reserve1)) ELSE To_Number(angaufgut.reserve1) end AS Laufzeit FROM artikel inner join modell ON modell.lfdnr = artikel.lfdmodnr left join bestanfragepos ON artikel.lfdnr = bestanfragepos.lf dartnr left join bestanfrage ON bestanfragepos.lfdanfrage = bestanfrage.lfdanfrage left join lieferant L1 ON L1.liefnr = bestanfrage.lfdliefnr left JOIN angaufgut ON bestanfragepos.lfdangaufgutnr = angaufgut.lfdnr WHERE Lower(modcode) LIKE 'ac%' AND Lower(abez1) NOT LIKE 'cust%' and artikel.mandant = 1 AND bestanfragepos.preis != 0 ORDER BY abez1 /

`

it gives error ora-00920 invalid relational operator.

It only worked on a single data when I used this:

SELECT regexp_substr(regexp_substr(artikelbez, '([^-/]+)/[^-/]+$'), '^[^-/]+') FROM nag_reporting_leitungspreise WHERE art = 'Vertrag' AND REF = 3791

Answer 1

Your code returns 30M. If you really want 10M, then this should do what you want:

select regexp_substr(regexp_substr(t.abez1, '([^-/]+)/[^-/]+$'), '^[^-/]+') AS upload
from t;

Here is a db<>fiddle.

Answer 2

with t(str) as (
select * from table(ku$_vcnt(
'DE-Internet-LTE',
'AE-Internet-Ethernet-10M/30M',
'AE-Internet-Ethernet-20m/40m',
'AE-Internet-Ethernet-300M/500M',
'AE-Internet-Ethernet-4000k/600k',
'AE-Internet-Ethernet-200M'
))
)
select 
  str, 
  regexp_substr(str, '(\d+(k|m))(/\d+(k|m))?',1,1,'i',1) upload,
  regexp_substr(str, '/(\d+(k|m))',1,1,'i',1) download
from t;

Results:

STR                                 UPLOAD  DOWNLOAD
----------------------------------- ------- --------
DE-Internet-LTE
AE-Internet-Ethernet-10M/30M        10M     30M
AE-Internet-Ethernet-20m/40m        20m     40m
AE-Internet-Ethernet-300M/500M      300M    500M
AE-Internet-Ethernet-4000k/600k     4000k   600k
AE-Internet-Ethernet-200M           200M

Answer 3

Just have a look at the below solution:

with t(str) as 
(
    select  column_value 
    from    table(sys.odcivarchar2list(
            'DE-Internet-LTE',
            'AE-Internet-Ethernet-10M/30M',
            'AE-Internet-Ethernet-20m/40m',
            'AE-Internet-Ethernet-300M/500M',
            'AE-Internet-Ethernet-4000k/600k',
            'AE-Internet-Ethernet-200M'
            ))
)
select  str, 
        regexp_substr(str, '-(\d+.*?)(/|$)',1,1,null,1) as upload,
        regexp_substr(str, '/(\d+.*?)$',1,1,null,1) as download,
        regexp_substr(regexp_substr(str, '([^-/]+)/[^-/]+$'), '^[^-/]+') as Gordan_Linoff_upload
from    t;

Output: