I have column with data such as '123456789012'
I want to divide each of each 3 chars from the data with a '/' in between so that the output will be like: "123/456/789/012"
I tried "SELECT TO_CHAR(DATA, '999/999/999/999') FROM TABLE 1" but it does not print out the output as what I wanted. Previously I did "SELECT TO_CHAR(DATA, '$999,999,999,999.99') FROM TABLE 1 and it printed out as "$123,456,789,012.00" so I thought I could do the same for other case as well, but I guess that's not the case.
There is also a case where I also want to put '#' in front of the data so the output will be something like this: #12345678901234. Can I use TO_CHAR for this problem too?
Is these possible? Because when I go through the documentation of oracle about TO_CHAR, it stated a few format that can be use for TO_CHAR function and the format that I want is not listed there.
Thank you in advance. :D
Here is one option with varchar2 datatype:
with test as (
select '123456789012' a from dual
)
select listagg(substr(a,(level-1)*3+1,3),'/') within group (order by rownum) num
from test
connect by level <=length(a)
or
with test as (
select '123456789012.23' a from dual
)
select '$'||listagg(substr((regexp_substr(a,'[0-9]{1,}')),(level-1)*3+1,3),',') within group (order by rownum)||regexp_substr(a,'[.][0-9]{1,}') num
from test
connect by level <=length(a)
output:
1st query
123/456/789/012
2nd query
$123,456,789,012.23
If you wants groups of three then you can use the group separator G, and specify the character to use:
SELECT TO_CHAR(DATA, 'FM999G999G999G999', 'NLS_NUMERIC_CHARACTERS=./') FROM TABLE_1
123/456/789/012
If you want a leading # then you can use the currency indicator L, and again specify the character to use:
SELECT TO_CHAR(DATA, 'FML999999999999', 'NLS_CURRENCY=#') FROM TABLE_1
#123456789012
Or combine both:
SELECT TO_CHAR(DATA, 'FML999G999G999G999', 'NLS_CURRENCY=# NLS_NUMERIC_CHARACTERS=./') FROM TABLE_1
#123/456/789/012
db<>fiddle
The data type is always a string; only the format changes.
Related
So, i have a lot of strings like the ones below in my database:
product1:1stparty:single_aduls:android:
product2:3rdparty:married_adults:ios:
product3:3rdparty:other_adults:android:
I need a regex to get only the text after the product name and before the device category. So, in the first line I'd get 1stparty:single_aduls, in the second 3rdparty:married_adults and in the third 3rdparty:other_adults. I'm stuck and can't find a way to solve that. Could anyone help me please?
As a regular expression, you can use:
select regexp_extract('product1:1stparty:single_aduls:android:', '^[^:]*:(.*):[^:]*:$')
This returns every after the first colon and before the penultimate colon.
We can try using REGEXP_REPLACE here:
SELECT REGEXP_REPLACE(val, r"^.*?:|:[^:]+:$", "") AS output
FROM yourTable;
This approach removes either the leading ...: or trailing :...: from the column, leaving behind the content you want. Here is a demo showing that the regex replacement is working:
Demo
You can also use standard split function and access result array element by index, which is quite clear to read and understand.
with a as (
select split('product1:1stparty:single_aduls:android:', ':') as splitted
)
select splitted[ordinal(2)] || ':' || splitted[ordinal (3)] as subs
from a
Consider below example
with your_table as (
select 'product1:1stparty:single_aduls:android:' txt union all
select 'product2:3rdparty:married_adults:ios:' union all
select 'product3:3rdparty:other_adults:android:'
)
select *,
(
select string_agg(part, ':' order by offset)
from unnest(split(txt, ':')) part with offset
where offset in (1, 2)
) result
from your_table
with output
How to get rest of string after specific char?
I have a string 'a|b|c|2|:x80|3|rr|' and I would like to get result after 3rd occurance of |. So the result should be like 2|:x80|3|rr|
The query
select REGEXP_SUBSTR('a|b|c|2|:x80|3|rr|','[^|]+$',1,4)
from dual
Returned me NULL
Use SUBSTR / INSTR combination
WITH t ( s ) AS (
SELECT 'a|b|c|2|:x80|3|rr|'
FROM dual
) SELECT substr(s,instr(s,'|',1,3) + 1)
FROM t;
Demo
REGEXP_REPLACE() will do the trick. Skip 3 groups of anything followed by a pipe, then replace with the 2nd group, which is the rest of the line (anchored to the end).
SQL> select regexp_replace('a|b|c|2|:x80|3|rr|', '(.*?\|){3}(.*)$', '\2') trimmed
2 from dual;
TRIMMED
------------
2|:x80|3|rr|
SQL>
I suggest a nice by long way by using regexp_substr, regexp_count and listagg together as :
select listagg(str) within group (order by lvl)
as "Result String"
from
(
with t(str) as
(
select 'a|b|c|2|:x80|3|rr|' from dual
)
select level-1 as lvl,
regexp_substr(str,'(.*?)(\||$)',1,level) as str
from dual
cross join t
connect by level <= regexp_count('a|b|c|2|:x80|3|rr|','\|')
)
where lvl >= 3;
Rextester Demo
If you use oracle 11g and above you can specify a subexpression to return like this:
select REGEXP_SUBSTR('a|b|c|2|:x80|3|rr|','([^|]+\|){3}(.+)$',1,1,null,2) from dual
Erkko,
You need to use the combination of SUBSTR and REGEXP_INSTR OR INSTR.
Your query will look like this. (Without Regex)
SELECT SUBSTR('a|b|c|2|:x80|3|rr|',INSTR('a|b|c|2|:x80|3|rr|','|',1,3)+1) from dual;
Your query will look like this. (With Regex as you want to use)
SELECT SUBSTR('a|b|c|2|:x80|3|rr|',REGEXP_INSTR('a|b|c|2|:x80|3|rr|','\|',1,3)+1) from dual;
Explanation:
First, you will need to find the place of the string you want as you mentioned. So in your case | comes at place 6. So that +1 would be your position to start to substring.
Second, from the original string, substring from that position+1 to unlimited.(Where your string ends)
Example:
https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=6fd782db95f575201eded084493232ee
Can anyone help me, I have a problem regarding on how can I get the below result of data. refer to below sample data. So the logic for this is first I want delete the letters before the number and if i get that same thing goes on , I will delete the numbers before the letter so I can get my desired result.
Table:
SALV3000640PIX32BLU
SALV3334470A9CARBONGRY
TP3000620PIXL128BLK
Desired Output:
PIX32BLU
A9CARBONGRY
PIXL128BLK
You need to use a combination of the SUBSTRING and PATINDEX Functions
SELECT
SUBSTRING(SUBSTRING(fielda,PATINDEX('%[^a-z]%',fielda),99),PATINDEX('%[^0-9]%',SUBSTRING(fielda,PATINDEX('%[^a-z]%',fielda),99)),99) AS youroutput
FROM yourtable
Input
yourtable
fielda
SALV3000640PIX32BLU
SALV3334470A9CARBONGRY
TP3000620PIXL128BLK
Output
youroutput
PIX32BLU
A9CARBONGRY
PIXL128BLK
SQL Fiddle:http://sqlfiddle.com/#!6/5722b6/29/0
To do this you can use
PATINDEX('%[0-9]%',FieldName)
which will give you the position of the first number, then trim off any letters before this using SUBSTRING or other string functions. (You need to trim away the first letters before continuing with the next step because unlike CHARINDEX there is no starting point parameter in the PATINDEX function).
Then on the remaining string use
PATINDEX('%[a-z]%',FieldName)
to find the position of the first letter in the remaining string. Now trim off the numbers in front using SUBSTRING etc.
You may find this other solution helpful
SQL to find first non-numeric character in a string
Try this it may helps you
;With cte (Data)
AS
(
SELECT 'SALV3000640PIX32BLU' UNION ALL
SELECT 'SALV3334470A9CARBONGRY' UNION ALL
SELECT 'SALV3334470A9CARBONGRY' UNION ALL
SELECT 'SALV3334470B9CARBONGRY' UNION ALL
SELECT 'SALV3334470D9CARBONGRY' UNION ALL
SELECT 'TP3000620PIXL128BLK'
)
SELECT * , CASE WHEN CHARINDEX('PIX',Data)>0 THEN SUBSTRING(Data,CHARINDEX('PIX',Data),LEN(Data))
WHEN CHARINDEX('A9C',Data)>0 THEN SUBSTRING(Data,CHARINDEX('A9C',Data),LEN(Data))
ELSE NULL END AS DesiredResult FROM cte
Result
Data DesiredResult
-------------------------------------
SALV3000640PIX32BLU PIX32BLU
SALV3334470A9CARBONGRY A9CARBONGRY
SALV3334470A9CARBONGRY A9CARBONGRY
SALV3334470B9CARBONGRY NULL
SALV3334470D9CARBONGRY NULL
TP3000620PIXL128BLK PIXL128BLK
I'm trying to sort the result set of a query where the row is VARCHAR2.
I've tried using just:
ORDER BY
UPPER(SERVER_NAME) ASC
But I get inconstant results, for example:
120157
777555
AKO
a20064
Elilikes
kagan
1200165_DAVID
As you can see, 1200165_DAVID appears last, in addition, I tried using a regular expression like so:
ORDER BY
(CASE WHEN REGEXP_LIKE(UPPER(SERVER_NAME), '^[0-9]+$') THEN 1 ELSE 2 END) ASC,
UPPER(SERVER_NAME) ASC
But I get the same results, I would like to get the following ordring is possible:
120157
1200165_DAVID
777555
a20064
AKO
Elilikes
kagan
Please advise.
Three things.
First: Why do you want 1200165_DAVID to appear AFTER 120157? It should appear before it, if you order alphabetically.
Second: Running your query on your test data, I get the correct result. So I am inclined to believe either your query is different from what you reported, or there is some other error somewhere.
Third: You may have who-knows-what characters in your data. Selecting str and dump(str) side by side (or whatever the name of your expression; I like to use str in my test data) to see what characters are in each string. Look especially at those that seem to be sorted "out of order".
with
inputs ( str ) as (
select '120157' from dual union all
select '777555' from dual union all
select 'AKO' from dual union all
select 'a20064' from dual union all
select 'Elilikes' from dual union all
select 'kagan' from dual union all
select '1200165_DAVID' from dual
)
select str from inputs
order by upper(str);
STR
-------------
1200165_DAVID
120157
777555
a20064
AKO
Elilikes
kagan
7 rows selected.
This is too long for a comment.
Your data would appear to not be all characters that you recognize. In particular, the first character is suspicious.
I would suggest that you run a query like this:
select ASCII(SUBSTR(server_name, 1, 1)) as first_char-ascii,
'|' || SUBSTR(server_name, 1, 1) || '|' as first_char,
COUNT(*), min(server_name), max(server_name)
from t
group by SUBSTR(server_name, 1, 1)
order by count(*) asc;
Then you will see what characters are actually at the beginning of the string. My guess is you will find at least one interesting character. You will then need to modify the data (or the query) to handle that.
I want to select names from a table where the 'name' column contains '%' anywhere in the value. For example, I want to retrieve the name 'Approval for 20 % discount for parts'.
SELECT NAME FROM TABLE WHERE NAME ... ?
You can use like with escape. The default is a backslash in some databases (but not in Oracle), so:
select name
from table
where name like '%\%%' ESCAPE '\'
This is standard, and works in most databases. The Oracle documentation is here.
Of course, you could also use instr():
where instr(name, '%') > 0
One way to do it is using replace with an empty string and checking to see if the difference in length of the original string and modified string is > 0.
select name
from table
where length(name) - length(replace(name,'%','')) > 0
Make life easy on yourselves and just use REGEXP_LIKE( )!
SQL> with tbl(name) as (
select 'ABC' from dual
union
select 'E%FS' from dual
)
select name
from tbl
where regexp_like(name, '%');
NAME
----
E%FS
SQL>
I read the documentation mentioned by Gordon. The relevent sentence is:
An underscore (_) in the pattern matches exactly one character (as opposed to one byte in a multibyte character set) in the value
Here was my test:
select c
from (
select 'a%be' c
from dual) d
where c like '_%'
The value a%be was returned.
While the suggestions of using instr() or length in the other two answers will lead to the correct answer, they will do so slowly. Filtering on function results simply take longer than filtering on fields.