1. <0,0><120.96,2000><241.92,4000><362.88,INF>
2. <0,0><143.64,2000><241.92,4000><362.88,INF>
3. <0,0><125.5,2000><241.92,4000><362.88,INF>
4. <0,0><127.5,2000><241.92,4000><362.88,INF>
Above is the data set I have in Oracle 10g. I need output as below
1. 120.96
2. 143.64
3. 125.5
4. 125.5
the output I want is only before "comma" (120.96). I tried using REGEXP_SUBSTR but I could not get any output. It will be really helpful if someone could provide effective way to solve this
Here is one method that first parses out the second element and then gets the first number in it:
select regexp_substr(regexp_substr(x, '<[^>]*>', 1, 2), '[0-9.]+', 1, 1)
Another method just gets the third number in the string:
select regexp_substr(x, '[0-9.]+', 1, 3)
Here is an approach without using Regexp.
Find the index of second occurrence of '<'. Then find the second occurrence of ',' use those values in substring.
with
data as
(
select '<0,0><120.96,2000><241.92,4000><362.88,INF>' x from dual
UNION ALL
select '<0,0><143.64,2000><241.92,4000><362.88,INF>' x from dual
UNION ALL
select '<0,0><125.5,2000><241.92,4000><362.88,INF>' from dual
)
select substr(x, instr(x,'<',1,2)+1, instr(x,',',1,2)- instr(x,'<',1,2)-1)
from data
Approach Using Regexp:
Identify the 2nd occurence of numerical value followed by a comma
Then remove the trailing comma.
with
data as
(
select '<0,0><120.96,2000><241.92,4000><362.88,INF>' x from dual
UNION ALL
select '<0,0><143.64,2000><241.92,4000><362.88,INF>' x from dual
UNION ALL
select '<0,0><125.5,2000><241.92,4000><362.88,INF>' from dual
)
select
trim(TRAILING ',' FROM regexp_substr(x,'[0-9.]+,',1,2))
from data
This example uses regexp_substr to get the string contained within the 2nd occurance of a less than sign and a comma:
SQL> with tbl(id, str) as (
select 1, '<0,0><120.96,2000><241.92,4000><362.88,INF>' from dual union
select 2, '<0,0><143.64,2000><241.92,4000><362.88,INF>' from dual union
select 3, '<0,0><125.5,2000><241.92,4000><362.88,INF>' from dual union
select 4, '<0,0><127.5,2000><241.92,4000><362.88,INF>' from dual
)
select id,
regexp_substr(str, '<(.*?),', 1, 2, null, 1) value
from tbl;
ID VALUE
---------- -------------------------------------------
1 120.96
2 143.64
3 125.5
4 127.5
EDIT: I realized the OP specified 10g and the regexp_substr example I gave used the 6th argument (subgroup) which was added in 11g. Here is an example using regexp_replace instead which should work with 10g:
SQL> with tbl(id, str) as (
select 1, '<0,0><120.96,2000><241.92,4000><362.88,INF>' from dual union
select 2, '<0,0><143.64,2000><241.92,4000><362.88,INF>' from dual union
select 3, '<0,0><125.5,2000><241.92,4000><362.88,INF>' from dual union
select 4, '<0,0><127.5,2000><241.92,4000><362.88,INF>' from dual
)
select id,
regexp_replace(str, '^(.*?)><(.*?),.*$', '\2') value
from tbl;
ID VALUE
---------- ----------
1 120.96
2 143.64
3 125.5
4 127.5
SQL>
Related
I am working with table data that contains strings with decimal and back-slash like below:
info
1/2.2.2
2/1.1.1
3/1.1.11
I need to use a regular expression to replace the data like below:
info
1/2.2
2/1.1
3/1.1
Don't use a (slow) regular expression, use simple (faster) string functions instead:
SELECT info,
CASE
WHEN INSTR(info, '.', 1, 2) > 0
THEN SUBSTR(info, 1, INSTR(info, '.', 1, 2) - 1)
ELSE info
END AS part
FROM table_name;
Which, for the sample data:
CREATE TABLE table_name (info) AS
SELECT '1/2.2.2' FROM DUAL UNION ALL
SELECT '2/1.1.1' FROM DUAL UNION ALL
SELECT '3/1.1.11' FROM DUAL UNION ALL
SELECT '3/1.1' FROM DUAL;
Outputs:
INFO
PART
1/2.2.2
1/2.2
2/1.1.1
2/1.1
3/1.1.11
3/1.1
3/1.1
3/1.1
If you want to update the table then:
UPDATE table_name
SET info = SUBSTR(info, 1, INSTR(info, '.', 1, 2) - 1)
WHERE INSTR(info, '.', 1, 2) > 0
fiddle
For the sake of argument, here's a solution using REGEXP_SUBSTR(). REGEXP_SUBSTR() returns NULL if the pattern is not found. Thanks to MT0 for the CTE so I didn't have to type it up :-)
WITH table_name(ID, info) AS (
SELECT 1, '1/2.2.2' FROM DUAL UNION ALL
SELECT 2, '2/1.1.1' FROM DUAL UNION ALL
SELECT 3, '3/1.1.11' FROM DUAL UNION ALL
SELECT 4, '3/1.1' FROM DUAL UNION ALL
SELECT 5, '4/4' FROM DUAL)
SELECT ID, REGEXP_SUBSTR(info, '\d/\d\.\d') DATA
from table_name;
ID DATA
---------- --------
1 1/2.2
2 2/1.1
3 3/1.1
4 3/1.1
5
5 rows selected.
I need to use regexp_substr, but I can't use it properly
I have column (l.id) with numbers, for example:
1234567891123!123 EXPECTED OUTPUT: 1234567891123
123456789112!123 EXPECTED OUTPUT: 123456789112
12345678911!123 EXPECTED OUTPUT: 12345678911
1234567891123!123 EXPECTED OUTPUT: 1234567891123
I want use regexp_substr before the exclamation mark (!)
SELECT REGEXP_SUBSTR(l.id,'[%!]',1,13) from l.table
is it ok ?
You can try using INSTR() and substr()
DEMO
select substr(l.id,1,INSTR(l.id,'!', 1, 1)-1) from dual
You want to remove the exclamation mark and all following characters it seems. That is simply:
select regexp_replace(id, '!.*', '') from mytable;
Look at it like a delimited string where the bang is the delimiter and you want the first element, even if it is NULL. Make sure to test all possibilities, even the unexpected ones (ALWAYS expect the unexpected)! Here the assumption is if there is no delimiter you'll want what's there.
The regex returns the first element followed by a bang or the end of the line. Note this form of the regex handles a NULL first element.
SQL> with tbl(id, str) as (
select 1, '1234567891123!123' from dual union all
select 2, '123456789112!123' from dual union all
select 3, '12345678911!123' from dual union all
select 4, '1234567891123!123' from dual union all
select 5, '!123' from dual union all
select 6, '123!' from dual union all
select 7, '' from dual union all
select 8, '12345' from dual
)
select id, regexp_substr(str, '(.*?)(!|$)', 1, 1, NULL, 1)
from tbl
order by id;
ID REGEXP_SUBSTR(STR
---------- -----------------
1 1234567891123
2 123456789112
3 12345678911
4 1234567891123
5
6 123
7
8 12345
8 rows selected.
SQL>
If you like to use REGEXP_SUBSTR rather than regexp_replace then you can use
SELECT REGEXP_SUBSTR(l.id,'^\d+')
assuming you have only numbers before !
If I understand correctly, this is the pattern that you want:
SELECT REGEXP_SUBSTR(l.id,'^[^!]+', 1)
FROM (SELECT '1234567891123!123' as id from dual) l
I am working on below query, I am expected to select exact 7 or 10 digit values columns using regular expression, I have used express in regexp_like() function of oracle, but its not working, please help
Query :
select * from
(select '1234567CELL' "a" from dual
union
select '123CaLLAsasd12' "a" from dual
union
select 'as9960488188CELLas12' "a" from dual
union
select '1234567' "a" from dual
union
select '9960488188' "a" from dual
union
select 'asdCELLqw' "a" from dual) b
where b."a" like '%CELL%' and regexp_like(b."a",'^(\d{7}|\d{10})$');
Expected output
"1234567"
"9960488188"
as above two rows, please check
^ and $ match the start and end of a string and the value cannot contain the string CELL and be solely a 7- or 10-digit number. Instead you could use the regular expression (^|\D)(\d{7}|\d{10})($|\D) which will match either the start of the string or a not digit character (^|\D) then either 7- or 10- digits and then either the end of the string or a non digit character ($|\D).
Like this:
WITH data ( a ) AS (
select '1234567CELL' from dual union
select '123CaLLAsasd12' from dual union
select 'as9960488188CELLas12' from dual union
select '1234567' from dual union
select '9960488188' from dual union
select 'asdCELLqw' from dual
)
SELECT a,
REGEXP_SUBSTR( a, '(^|\D)(\d{7}|\d{10})($|\D)', 1, 1, NULL, 2 ) AS val
FROM data
WHERE a LIKE '%CELL%'
AND REGEXP_LIKE( a, '(^|\D)(\d{7}|\d{10})($|\D)');
Output:
A VAL
-------------------- ----------
1234567CELL 1234567
as9960488188CELLas12 9960488188
You may just use
where regexp_like(b."a",'^([[:digit:]]{7}|[[:digit:]]{10})$')
Since the pattern is anchored (^ matches the start of the string and $ matches the end of the string) there can't be CELL inside the entries you fetch, and you can remove where b."a" like '%CELL%' from the query.
I have a sample source string like below, which was in pipe delimited format in that the value obr can be at anywhere. I need to get the second value of the pipe from the first occurrence of obr. So for the below source strings the expected would be,
Source string:
select 'asd|dfg|obr|1|value1|end' text from dual
union all
select 'a|brx|123|obr|2|value2|end' from dual
union all
select 'hfv|obr|3|value3|345|pre|end' from dual
Expected output:
value1
value2
value3
I have tried the below regexp in oracle sql, but it is not working fine properly.
with t as (
select 'asd|dfg|obr|1|value1|end' text from dual
union all
select 'a|brx|123|obr|2|value2|end' from dual
union all
select 'hfv|obr|3|value3|345|pre|end' from dual
)
select text,to_char(regexp_replace(text,'*obr\|([^|]*\|)([^|]*).*$', '\2')) output from t;
It is working fine when the string starts with OBR, but when OBR is in the middle like the above samples it is not working fine.
Any help would be appreciated.
Not sure of how Oracle handles regular expressions, but starting with an asterisk usually implies that you're looking for zero or more null characters.
Have you tried '^.*obr\|([^|]*\|)([^|]*).*$' ?
This handles null elements and is wrapped in a NVL() call which supplies a value if 'obr' is not found or occurs too far toward the end of a record so a value 2 away is not possible:
SQL> with t(id, text) as (
select 1, 'asd|dfg|obr|1|value1|end' from dual
union
select 2, 'a|brx|123|obr|2|value2|end' from dual
union
select 3, 'hfv|obr|3|value3|345|pre|end' from dual
union
select 4, 'hfv|obr||value4|345|pre|end' from dual
union
select 5, 'a|brx|123|obriem|2|value5|end' from dual
union
select 6, 'a|brx|123|obriem|2|value6|obr' from dual
)
select
id,
nvl(regexp_substr(text, '\|obr\|[^|]*\|([^|]*)(\||$)', 1, 1, null, 1), 'value not found') value
from t;
ID VALUE
---------- -----------------------------
1 value1
2 value2
3 value3
4 value4
5 value not found
6 value not found
6 rows selected.
SQL>
The regex basically can be read as "look for a pattern of a pipe, followed by 'obr', followed by a pipe, followed by zero or more characters that are not a pipe, followed by a pipe, followed by zero or more characters that are not a pipe (remembered in a captured group), followed by a pipe or the end of the line". The regexp_substr() call then returns the 1st captured group which is the set of characters between the pipes 2 fields from the 'obr'.
For these strings
RSLR_AIRL19_ID3454_T20030913091226
RSLR_AIRL19_ID3122454_T20030913091226
RSLR_AIRL19_ID34_T20030913091226
How to get the number after ID ?
Or how to get the content between two characters but not include them ?
I use this '/\_ID([^_]+)/' got matches like Array ( [0] => _ID3454 [1] => 3454 )
Is this the right way?
To extract a number after an ID, you could write a similar query.
SQL> with t1 as(
2 select 'RSLR_AIRL19_ID3454_T20030913091226' as col from dual union all
3 select 'RSLR_AIRL19_ID3122454_T20030913091226' from dual union all
4 select 'RSLR_AIRL19_ID34_T20030913091226' from dual
5 )
6 select regexp_substr(col, '^([[:alnum:]]+_){2}ID([[:digit:]]+)_([[:alnum:]]+){1}$', 1, 1, 'i', 2) as ID
7 from t1
8 ;
ID
-------------
3454
3122454
34
Or, if you want to extract digits from a first occurrence of the pattern without verifying if an entire string matches a specific format:
SQL> with t1 as(
2 select 'RSLR_AI_RL19_ID3454_T20030913091226' as col from dual union all
3 select 'RSLR_AIRL19_ID3122454_T20030913091226' from dual union all
4 select 'RSLR_AIRL19_ID34_T20030913091226' from dual
5 )
6 select regexp_substr(col, 'ID([[:digit:]]+)', 1, 1, 'i', 1) as ID
7 from t1
8 ;
ID
--------------
3454
3122454
34
With pcre & perl engines :
ID\K\w+
NOTE
\K "restart" the match.
See http://www.phpfreaks.com/blog/pcre-regex-spotlight-k (php use pcre)