I have string like Order#Confirm####2791 i wanted to fetch 2791 after #### delimiter. I tried below one not getting exact sub-string what i am expecting to return.
SELECT regexp_substr('Order # Confirm ####2791','[^####]+',1,2) regexp_substr
FROM dual;
I would like to return 2791 from above query.
SELECT regexp_substr('Order # Confirm ####2791','####(.*)$',1,1, null, 1) regexp_substr
FROM dual;
If you want to restrict the match to digits:
SELECT regexp_substr('Order # Confirm ####2791','####(\d+)$',1,1, null, 1) regexp_substr
FROM dual;
regexp_replace works too:
SELECT regexp_replace('Order # Confirm ####2791','.*?####(\d+)$', '\1') regexp_replace
FROM dual;
Note with regexp_substr() if a match is not found NULL is returned and with regexp_replace() if a match is not found the original string is returned.
You don't need regular expressions for this.
SELECT substr('Order # Confirm ####2791',
instr('Order # Confirm ####2791', '####') + 4) as your_substr
FROM dual;
Related
In my table, I have data like PAT5DSA-(ALRP), LAR6DAOP-(RAH) etc..But I want to remove the strings like -(xxxx) or -(xxx) which can be any alphabets inside braces. Tried using the below:
select replace(:code,'-(Aa-Zz)',null) from employee;
But this didn't work..Can anyone please help?
We can do a regex replacement using REGEXP_REPLACE:
SELECT REGEXP_REPLACE('PAT5DSA-(ALRP)', '-\(.*?\)', '')
FROM dual;
PAT5DSA
The plain replace() doesn't understand patterns. You could use a regular expression replace, e.g.:
-- CTE for sample data
with cte (code) as (
select 'PAT5DSA-(ALRP)' from dual
union all
select 'LAR6DAOP-(RAH)' from dual
)
select code, regexp_replace(code, '-\(.*?\)$') as result
from cte;
CODE RESULT
-------------- --------------
PAT5DSA-(ALRP) PAT5DSA
LAR6DAOP-(RAH) LAR6DAOP
This will remove anything inside a pair of parentheses which is preceded by a dash, at the end of the original string. If the parentheses to be removed could be anywhere in the string then remove the $.
Use INSTR and SUBSTR:
WITH cteVals AS (SELECT 'PAT5DSA-(ALRP)' AS COL_VAL FROM DUAL UNION ALL
SELECT 'LAR6DAOP-(RAH)' AS COL_VAL FROM DUAL)
SELECT SUBSTR(COL_VAL, 1, INSTR(COL_VAL, '-')-1)
FROM cteVals;
Best of luck.
I want to replace this:
"STORES/KOL#10/8/36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL"
with this:
"STORES/KOL#10#8#36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL"
basically this is conditional based replace I want to replace / with #
like STORES/KOL string should be STORES/KOL
but 10/8/36 string should be 10#8#36
This will replace the 2nd and 3rd / character with a #:
Oracle Setup:
CREATE TABLE test_data ( value ) AS
SELECT '"STORES/KOL#10/8/36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL"'
FROM DUAL;
Query:
SELECT REGEXP_REPLACE(
value,
'^(.*?/.*?)/(.*?)/(.*)$',
'\1#\2#\3'
) AS replacement
FROM test_data
Output:
| REPLACEMENT |
| :---------------------------------------------------------------------------------------------------------------- |
| "STORES/KOL#10#8#36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL" |
db<>fiddle here
Here is one option using REGEXP_REPLACE. We can try targeting the following regex pattern:
#(\d+)/(\d+)/(\d+)#
Then replace using the three capture groups, replacing the path separators with pound signs.
WITH yourTable AS (
SELECT 'STORES/KOL#10/8/36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL' AS input FROM dual
)
SELECT
input,
REGEXP_REPLACE(input, '#(\d+)/(\d+)/(\d+)#', '#\1#\2#\3#') AS output
FROM yourTable;
Demo
Whether or not this regex replacement is specific enough/accurate for the rest of your data depends on that data, which you never showed us.
with s as (select '"STORES/KOL#10/8/36#1718.00#4165570.00#119539388#PT3624496#9902001#04266#6721#PT3624496-11608091-1-55-STORES/KOL"' str from dual)
select
replace(replace(str, '/', '#'), 'STORES#KOL', 'STORES/KOL') result_str_1,
regexp_replace(str, '(\d)/', '\1#') result_str_2
from s;
I am trying to understand a query in my application where it uses substr function.
I have gone through the documentation for substr, which looks simple and clear.
Now below is my query without using substr:
select last_day(to_date(to_char(add_months(TO_DATE('2004/10/25', 'yyyy/mm/dd'),1),'YYYY')||'0201','YYYYMMDD')) from dual;
This gives me result as 2/29/2004. The above query just returns last day of Feb in simple words.
Now I am using substr as below:
select substr(last_day(to_date(to_char(add_months(TO_DATE('2004/10/25', 'yyyy/mm/dd'),1),'YYYY')||'0201','YYYYMMDD')),5,1) from dual;
So here the start value is 5 & length is 1, so I am expecting output as 2 looking at 2/29/2004. but the actual output is E, I am not clear from where this E is coming as result.
You cannot use SUBSTR() on DATE values. SUBSTR() works only on strings!
When you run SUBSTR({DATE_VALUE}, ...) then Oracle actually does following:
SELECT
SUBSTR(
TO_CHAR(
{DATE_VALUE}, (SELECT VALUE FROM nls_session_parameters WHERE parameter = 'NLS_DATE_FORMAT')
), ...
)
FROM dual;
What is the purpose of this query? Do you like to find out whether input year is a leap-year?
Try this instead -
select substr(to_char(last_day(to_date(to_char(add_months(TO_DATE('2004/10/25', 'yyyy/mm/dd'),1),'YYYY')||'0201','YYYYMMDD')),'dd/mm/yyyy'),5,1)
from dual;
I have a string IN-123456; now I need to trim the IN- from that string. I tried as in Oracle
select trim('IN-' from 'IN-123456) from dual;
but I get an error
ORA-30001: trim set should have only one character
30001. 00000 - "trim set should have only one character"
*Cause: Trim set contains more or less than 1 character. This is not
allowed in TRIM function.
How can I solve this?
A simple replace wouldn't do the trick?
select replace('IN-123456', 'IN-', '') from dual;
Thanks for the result...
It can be solved with LTRIM() function
Clearly, TRIM is not the correct function for the job. You need to REPLACE the (sub)string IN- with nothing:
SELECT REPLACE('IN-123456', 'IN-') FROM dual;
Be aware that this will replace all occurrences of IN- anywhere in the string. If that's not appropriate, but the IN- will always be at the start of the string, then you could use SUBSTR instead:
SELECT SUBSTR('IN-123456', 4) FROM dual;
you just forget to complete single quote
select trim('IN-' from 'IN-123456') from dual;
now try this
Trim Function is always remove one char from string
Here is the example -
SELECT TRIM(both 'P' FROM 'PHELLO WORLDP') FROM DUAL
Out put -HELLO WORLD
You may use LEADING /TRAILING insert of Both.
In your case "IN-" holding three char.
I have a sql statement:
select translate('abcdefg', 'abc', '') from dual;
Why the result is nothing?
I think it should be 'defg'.
From the documentation:
You cannot use an empty string for to_string to remove all characters in from_string from the return value. Oracle Database interprets the empty string as null, and if this function has a null argument, then it returns null. To remove all characters in from_string, concatenate another character to the beginning of from_string and specify this character as the to_string. For example, TRANSLATE(expr, 'x0123456789', 'x') removes all digits from expr.
So you can do something like:
select translate('abcdefg', '#abc', '#') from dual;
TRANSLATE('ABCDEFG','#ABC','#')
-------------------------------
defg
... using any character that isn't going to be in your from_string.
select translate('abcdefg', 'abc', '') from dual;
To add to Alex's answer, you could use any character(allowed in SQL) for that matter to concatenate to remove all the characters. So, you could even use a space instead of empty string. An empty string in Oracle is considered as NULL value.
So, you could also do -
SQL> SELECT TRANSLATE('abcdefg', ' abc', ' ') FROM dual;
TRAN
----
defg
SQL>
Which is the same as -
SQL> SELECT TRANSLATE('abcdefg', chr(32)||'abc', chr(32)) FROM dual;
TRAN
----
defg
SQL>
Since the ascii value of space is 32.
It was just a demo, it is better to use any other character than space for better understanding and code readability.