I have data of this pattern:
000000001 ItemA
000000002 ItemB
000000003 ItemC
and I want to get this result:
ItemA
ItemB
ItemC
I tried REGEXP_REPLACE with this regex: #"^[\d-]*\s*" that I found in the accepted answer of this: regex remove digits and - in beginning question like this:
SELECT REGEXP_REPLACE(test_column, '#"^[\d-]*\s*"', '') FROM test_table;
but it doesn't work...
The solution's code snippet is written in C# and utilizes .NET Regular Expressions. However, the Oracle database requires POSIX Extended Regular Expressions (ERE) for compatibility with its REGEXP_REPLACE function. To resolve this issue, you can use an Oracle POSIX-compliant regular expression, like the one below:
^[0-9]*[[:space:]]*
^ - It matches the beginning of the text being searched.
[0-9] - It matches any single character that is a digit (0 to 9).
* - It matches zero or more occurrences of the preceding character.
[[:space:]] - It matches any white-space character, including spaces, tabs, and line breaks.
* - It matches zero or more white-space characters.
You have some small errors, but you can use folowing two queries
CREATE TABLE table1
("data" varchar2(15))
;
INSERT ALL
INTO table1 ("data")
VALUES ('000000001 ItemA')
INTO table1 ("data")
VALUES ('000000002 ItemB')
INTO table1 ("data")
VALUES ('000000003 ItemC')
SELECT * FROM dual
3 rows affected
SELECT REGEXP_REPLACE("data", '^[0-9-]*\s', '') FROM table1;
REGEXP_REPLACE("DATA",'^[0-9-]*\S','')
ItemA
ItemB
ItemC
SELECT REGEXP_REPLACE("data", '^[[:digit:]-]*\s', '') FROM table1;
REGEXP_REPLACE("DATA",'^[[:DIGIT:]-]*\S','')
ItemA
ItemB
ItemC
fiddle
If it doesn't have to be regular expression, try (fast!) substr + instr combination. For small data sets you won't notice any difference, but with A LOT of data, this should behave better.
Sample data:
SQL> with test (col) as
2 (select '000000001 ItemA' from dual union all
3 select '000000002 ItemB' from dual
4 )
Query:
5 select substr(col, instr(col, ' ') + 1) result
6 from test;
RESULT
---------------
ItemA
ItemB
SQL>
Related
I have a table like cust_attbr consists column attbr which has values like:
{"SRCTAXAMT":"11300",เอ็ก10110","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":"0835546003122"}
{"SRCTAXAMT":"11300", กรุงค10110","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":"0835546003122"}
........ ... ...
{"SRCTAXAMT":"11300", กรุงค10110","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":" "}
So, I have to write one select statement which will fetch only VAT_NUMBER value like:
0835546003122
0835546003122
.... ... ..
null
With sample data you posted:
SQL> select * From test;
ID ATTBR
---------- ----------------------------------------------------------------------------------------------------------------
1 "{"SRCTAXAMT":"11300",????10110","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":"0835546003122"}"
2 "{"SRCTAXAMT":"11300", ?????10110","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":"0835546003122"}"
3 "{"SRCTAXAMT":"11300", ?????10110","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":" "}"
this might be one option:
SQL> select id,
2 regexp_substr(regexp_substr(attbr, 'VAT_NUMBER":"(\d+)?'), '\d+$') vat
3 from test;
ID VAT
---------- --------------------
1 0835546003122
2 0835546003122
3
SQL>
Inner regexp_substr returns VAT_NUMBER followed by optional number, while the outer one extracts only the number anchored to the end of the previous substring.
If you're on 18c and the data is actual json (it currently is not because of the double quotes around the curly braces and the ",.กรุงค10110" - It is unclear that this is because of your sample data) you could use json_table function:
WITH t (json_val) AS
(
SELECT '{"SRCTAXAMT":"11300","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":"0835546003122"}' FROM DUAL UNION ALL
SELECT '{"SRCTAXAMT":"11300","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":"0835546003122"}' FROM DUAL UNION ALL
SELECT '{"SRCTAXAMT":"11300","TAXAMT":"11300","LOGID":"190301863","VAT_NUMBER":" "}' FROM DUAL
)
SELECT jt.*
FROM t,
JSON_TABLE(json_val, '$'
COLUMNS (first_name VARCHAR2(50 CHAR) PATH '$."VAT_NUMBER"')) jt;
0835546003122
0835546003122
One option would be converting those column values to JSON syntax an then extract the values of VAT_NUMBER keys provided DB version is 12c Release 1+. Here, we have an issue that there are unrecognized characters, probably an alphabet from far east and those strings are not properly quoted, then we need to remove the part upto TAXAMT key, and then extracting VAT_NUMBER key's value through prefixing an opening curly brace('{') by use of JSON_VALUE() function :
SELECT JSON_VALUE(
'{'||REGEXP_REPLACE(str,'(.*10110",)(.*)','\2'),
'$.VAT_NUMBER'
) AS VAT_NUMBER
FROM tab --> your original data source
Demo
I'm trying to get first string after a character.
Example is like
ABCDEF||GHJ||WERT
I need only
GHJ
I tried to use REGEXP but i couldnt do it.
Can anyone help me with please?
Thank you
Somewhat simpler:
SQL> select regexp_substr('ABCDEF||GHJ||WERT', '\w+', 1, 2) result from dual;
^
RES |
--- give me the 2nd "word"
GHJ
SQL>
which reads as: give me the 2nd word out of that string. Won't work properly if GHJ consists of several words (but that's not what your example suggests).
Something like I interpret with a separator in place, In this case it is || or | example is with oracle database
-- pattern -- > [^] represents non-matching character and + for says one or more character followed by ||
-- 3rd parameter --> starting position
-- 4th parameter --> nth occurrence
WITH tbl(str) AS
(SELECT 'ABCDEF||GHJ||WERT' str FROM dual)
SELECT regexp_substr(str
,'[^||]+'
,1
,2) output
FROM tbl;
I think the most general solution is:
WITH tbl(str) AS (
SELECT 'ABCDEF||GHJ||WERT' str FROM dual UNION ALL
SELECT 'ABC|DEF||GHJ||WERT' str FROM dual UNION ALL
SELECT 'ABClDEF||GHJ||WERT' str FROM dual
)
SELECT regexp_replace(str, '^.*\|\|(.*)\|\|.*', '\1')
FROM tbl;
Note that this works even if the individual elements contain punctuation or a single vertical bar -- which the other solutions do not. Here is a comparison.
Presumably, the double vertical bar is being used for maximum flexibility.
You should use regexp_substr function
select regexp_substr('ABCDEF||GHJ||WERT ', '\|{2}([^|]+)', 1, 1, 'i', 1) str
from dual;
STR
---
GHJ
Simple Question
I have the following type of results in a string field
'Number=123456'
'Number=1234567'
'Number=12345678'
How do I extract the value from the string with regard that the value can change between 5-8 figures
So far I did this but I doubt that fits my requirement
SELECT substring('Size' from 8 for ....
If I can tell it to start from the = sign till the end that would help!
The likely simplest solution is to trim 7 leading characters with right():
right(str, -7)
Demo:
SELECT str, right(str, -7)
FROM (
VALUES ('Number=123456')
, ('Number=1234567')
, ('Number=12345678')
) t(str);
str | right
-----------------+----------
Number=123456 | 123456
Number=1234567 | 1234567
Number=12345678 | 12345678
You could use REPLACE:
SELECT col, REPLACE(col, 'Number=', '')
FROM tab;
DBFiddle Demo
Based on this question:
Split comma separated column data into additional columns
You could probably do the following:
SELECT *, split_part(col, '=', 2)
FROM table;
You may use regexp_matches :
with t(str) as
(
select 'Number=123456' union all
select 'Number=1234567' union all
select 'Number=12345678' union all
select 'Number=12345678x9'
)
select t.str as "String",
regexp_matches(t.str, '=([A-Za-z0-9]+)', 'g') as "Number"
from t;
String Number
-------------- ---------
Number=123456 123456
Number=1234567 1234567
Number=12345678 12345678
Number=12345678x9 12345678x9
--> the last line shows only we look chars after equal sign even if non-digit
Rextester Demo
I need to insert character string after each character in Oracle SQL.
Example:
ABC will A,B,C
DEFG will be D,E,F,G
This question gives only one character in string
Oracle insert character into a string
Edit: As some fellows have mentioned, Oracle does not admit this regex. So my approach would be to do a regex to match all characters, add them a comma after the character and then removing the last comma.
WITH regex AS (SELECT REGEXP_REPLACE('ABC', '(.)', '\1,') as reg FROM dual) SELECT SUBSTR(reg, 1, length(reg)-1) FROM regex;
Note that with the solution of rtrim there could be errors if the string you want to parse has a final ending comma and you don't want to remove it.
Previous solution: (Not working on Oracle)
Check if this does the trick:
SELECT REGEXP_REPLACE('ABC', '(.)(?!$)', '\1,') FROM dual;
It does a regexp_replace of every character, but the last one for the same character followed by a ,
To see how regexp_replace works I recommend you: https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions130.htm
SELECT rtrim(REGEXP_REPLACE('ABC', '(.)', '\1,'),',') "REGEXP_REPLACE" FROM dual;
You could do it using:
REGEXP_REPLACE
RTRIM
For example,
SQL> WITH sample_data AS(
2 SELECT 'ABC' str FROM dual UNION ALL
3 SELECT 'DEFG' str FROM dual UNION ALL
4 SELECT 'XYZ' str FROM dual
5 )
6 -- end of sample_data mimicking a real table
7 SELECT str,
8 rtrim(regexp_replace(str, '(\w?)', '\1,'),',') new_str
9 FROM sample_data;
STR NEW_STR
---- ----------
ABC A,B,C
DEFG D,E,F,G
XYZ X,Y,Z
Since there is no way to negate the end of string in an Oracle regex (that does not support lookarounds), you may use
SELECT REGEXP_REPLACE(
REGEXP_REPLACE('ABC', '([^,])([^,])','\1,\2'),
'([^,])([^,])',
'\1,\2')
AS Result from dual
See the DB Fiddle. The point here is to use REGEXP_REPLACE with ([^,])([^,]) pattern twice to cater for consecutive matches.
The ([^,])([^,]) pattern matches any non-comma char into Group 1 (\1) and then any non-comma char into Group 2 (\2), and inserts a comma in between them.
I have the following problem.
There is a String:
There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235
I need to show only just the last date from this string: 2015.06.07.
I tried with regexp_substr with insrt but it doesn't work.
So this is just test, and if I can solve this after it with this solution I should use it for a CLOB query where there are multiple date, and I need only the last one. I know there is regexp_count, and it is help to solve this, but the database what I use is Oracle 10g so it wont work.
Can somebody help me?
The key to find the solution of this problem is the idea of reversing the words in the string presented in this answer.
Here is the possible solution:
WITH words AS
(
SELECT regexp_substr(str, '[^[:space:]]+', 1, LEVEL) word,
rownum rn
FROM (SELECT 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 2015.06.08 2015.06.17. 2015.07.01. 12345678999 12125235' str
FROM dual) tab
CONNECT BY LEVEL <= LENGTH(str) - LENGTH(REPLACE(str, ' ')) + 1
)
, words_reversed AS
(
SELECT *
FROM words
ORDER BY rn DESC
)
SELECT regexp_substr(word, '\d{4}\.\d{2}\.\d{2}', 1, 1)
FROM words_reversed
WHERE regexp_like(word, '\d{4}\.\d{2}\.\d{2}')
AND rownum = 1;
From the documentation on regexp_substr, I see one problem immediately:
The . (period) matches any character. You need to escape those with a backslash: \. in order to match only a period character.
For reference, I am linking this post which appears to be the approach you are taking with substr and instr.
Relevant documentation from Oracle:
INSTR(string , substring [, position [, occurrence]])
When position is negative, then INSTR counts and searches backward from the end of string. The default value of position is 1, which means that the function begins searching at the beginning of string.
The problem here is that your regular expression only returns a single value, as explained here, so you will be giving the instr function the appropriate match in the case of multiple dates.
Now, because of this limitation, I recommend using the approach that was proposed in this question, namely reverse the entire string (and your regular expression, i.e. \d{2}\.\d{2}\.\d{4}) and then the first match will be the 'last match'. Then, perform another string reversal to get the original date format.
Maybe this isn't the best solution, but it should work.
There are three different PL/SQL functions that will get you there.
The INSTR function will identify where the first "period" in the date string appears.
SUBSTR applied to the entire string using the value from (1) as the start point
TO_DATE for a specific date mask: YYYY.MM.DD will convert the result from (2) into a Oracle date time type.
To make this work in procedural code, the standard blocks apply:
DECLARE
v_position pls_integer;
... other variables
BEGIN
sql code and function calls;
END
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE finddate
(column1 varchar2(11), column2 varchar2(39))
;
INSERT ALL
INTO finddate (column1, column2)
VALUES ('row1', '1234567 242424 2015.06.07. 12125235')
INTO finddate (column1, column2)
VALUES ('string2', '1234567 242424 2015.06.07. 12125235')
SELECT * FROM dual
;
Query 1:
select instr(column2,'.',1) from finddate
where column1 = 'string2'
select substr(column2,(20-4),10) from finddate
select to_date('2015.06.07','YYYY.MM.DD') from finddate
Results:
| TO_DATE('2015.06.07','YYYY.MM.DD') |
|------------------------------------|
| June, 07 2015 00:00:00 |
| June, 07 2015 00:00:00 |
Here's a way using regexp_replace() that should work with 10g, assuming the format of the lines will be the same:
with tbl(col_string) as
(
select 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235'
from dual
)
select regexp_replace(col_string, '^.*(\d{4}\.\d{2}\.\d{2})\. \d*$', '\1')
from tbl;
The regex can be read as:
^ - Match the start of the line
. - followed by any character
* - followed by 0 or more of the previous character (which is any character)
( - Start a remembered group
\d{4}\.\d{2}\.\d{2} - 4 digits followed by a literal period followed by 2 digits, etc
) - End the first remembered group
\. - followed by a literal period
- followed by a space
\d* - followed by any number of digits
$ - followed by the end of the line
regexp_replace then replaces all that with the first remembered group (\1).
Basically describe the whole line as a regular expression, group around what you want to return. You will most likely need to tweak the regex for the end of the line if it could be other characters than digits but this should give you an idea.
For the sake of argument this works too ONLY IF there are 2 occurrences of the date pattern:
with tbl(col_string) as
(
select 'There is something 2015.06.06. in the air 1234567 242424 2015.06.07. 12125235' from dual
)
select regexp_substr(col_string, '\d{4}\.\d{2}\.\d{2}', 1, 2)
from tbl;
returns the second occurrence of the pattern. I expect the above regexp_replace more accurately describes the solution.