I would like to convert a string containing dates in SQL select from Oracle 11g database.
Original string (CLOB) example:
"1.12.2011 - event 1
2.2.2012 - event 2
13.3.2012 - event 44"
Desired output:
"20111201 - event 1
20120202 - event 2
20120313 - event 44"
Is there a better (faster) way than using 4 separate replacements?
regexp_replace(regexp_replace(regexp_replace(regexp_replace(my_string,
'(\d\d)\.(\d\d)\.(20\d\d)', '\3\2\1'),
'(\d\d)\.(\d)\.(20\d\d)', '\30\2\1'),
'(\d)\.(\d\d)\.(20\d\d)', '\3\20\1'),
'(\d)\.(\d)\.(20\d\d)', '\30\20\1')
Especially if you're using clobs you have to be careful unless you're certain of the data in there.
However, if your clob only looks like that then you need threeregexp_replace in order for this to work; it'll also be much more dynamic. Just explicitly specify digits using [[:digit:]] then specify a minimum and maximum number of times these digits could be there using {1,2}.
Then the following would work:
select regexp_replace(
regexp_replace(
regexp_replace( my_string
, '([[:digit:]]{1,2})\.([[:digit:]]{1,2})\.(20[[:digit:]]{2})'
, '\3-\2-\1')
, '-([[:digit:]]{1}(-|$))'
, '0\1' )
, ('-')
, '')
from dual
This means:
match ( group 1 ) 1 or 2 digits
match a full stop.
match ( group 2 ) 1 or 2 digits
match a full stop
match ( group 3 ) 20 + 2 digits.
Then take out only groups 1, 2 and 3, i.e. ignoring the full stops and return then in the order 3, 2, 1 padded with a hyphen
Then replace any [digit] that is followed by either a hyphen or the end of the string, i.e. the number of digits is only 1 with -0[digit].
Lastly replace all the hyphens.
Separately from that I agree with tbone. It would make a lot more sense to store this data in a separate table (event_id number, event_date date). Any string transformations are easy with no chance of getting it wrong, unlike in this situation, and the data is easy to query and compare.
there are no better options (both correct and readable) with better performance - or if there are, no one cares..
i prefer a 2-level regexp_replace for date part:
select regexp_replace(
regexp_replace( my_string,
'([[:digit:]]{1,2})\.([[:digit:]]{1,2})\.(20[[:digit:]]{2})',
'\3-0\2-0\1' ),
'(20[[:digit:]]{2})-0?([[:digit:]]{2})-0?([[:digit:]]{2})',
'\3\2\1' )
from dual;
Demo
Maybe try doing:
select to_char(to_date('13.3.2011', 'DD.MM.YYYY'),'YYYYMMDD') from dual;
Related
I have a query similar to this:
SELECT YEAR_CODE FROM YEAR_CODES
and it returns several records: typically 1 but sometimes 2 or 3. The returned records look like this: 2018FOO, 2019BAR
I need to get the matching previous year of the returned codes. For instance:
2018FOO becomes 2017FOO
2019BAR becomes 2018BAR
Looking for something similar to:
REGEX_REPLACE(SELECT YEAR_CODE FROM YEAR_CODES, 4th character, 4th character minus 1)
You don't need regexp_replace(), using substr() string operator with concat() function (or concatenation operators ||) is enough :
with year_codes(year_code) as
(
select '2018FOO' from dual union all
select '2019BAR' from dual
)
select concat(substr(year_code,1,4) - 1,substr(year_code,-3)) as year_code
from year_codes;
YEAR_CODE
---------
2017FOO
2018BAR
to_number() conversion is redundant, since Oracle implicitly considers a string as a number which is completely composed of digits for an arithmetic operation.
You can do use string operations:
with c as (
<your query here>
)
select
from year_code yc
where to_number(substr(yc.code, 1, 4)) = to_number(substr(c.code)) - 1 and
substr(yc.code, 5) = substr(c.code, 5)
I have to select only the IDs which have only even digits (an ID looks like: p19 ,p20 etc). That is, p20 is good (both 2 and 0 are even digits); p18 is not.
I thought to use substr to get each number from the IDs and then see if it's even .
select from profs
where to_number(substr(id_prof,2,2))%2=0 and to_number(substr(id_prof,3,2))%2=0;
IF you need all rows consist of 'p' in beginning and even digits on tail It should look like:
select *
from profs
where regexp_like (id_prof, '^p[24680]+$');
with
profs ( prof_id ) as (
select 'p18' from dual union all
select 'p24' from dual union all
select 'p53' from dual
)
-- End of test data; what is above this line is NOT part of the solution.
-- The solution (SQL query) begins here.
select *
from profs
where length(prof_id) = length(translate(prof_id, '013579', '0'));
PROF_ID
-------
p24
This solution should work faster than anything using regular expressions. All it does is to replace 0 with itself and DELETE all odd digits from the input string. (The '0' is included due to a strange but documented behavior of translate() - the third argument can't be empty). If the length of the input string doesn't change after the translation, that means the input string didn't have any odd digits.
where mod(to_number(regexp_replace(id_prof, '[^[:digit:]]', '')),2) = 0
In a simplified form, I'm attempting to retrieve either the first occurrence of the '.*?=(.*?);.*' regex, or the second, or the third -- that is, either x or y or z (that is, I want to be able to hardcode in this query that I want the first, second or third values) in this following example:
select regexp_replace(
'margin=x;margin=y;margin=z;',
'.*?=(.*?);.*',
'\1',
1 -- occurrences. I thought that picking 1, 2 or 3 would solve my problem?
) from dual;
-- This returns "xyz", which is terrible. I was expecting it to return "x", in this case.
Looking at the Oracle documentation, I thought this would be relatively straightforward, as the last parameter (occurrences), apparently allows me to select which groups to take into consideration. But it doesn't! Why?
Thanks
i´m goingoff to another completly different solution. Would combining a hierarchial substring select with a regexp_replace be an option for your needs?
This way you could create an option to either select one or multiple values, depending on your needs. You wouldn´t need to write a concatinating regex value and you could adjust the select a bit more to your needs
select regexp_replace(subselect.val, '.*=(.*?);', '\1') -- remove "margin="
from (select regexp_substr(
'margin=x;margin=y;margin=z;',
'.*?=(.*?);',
1,
level) val,
level lvl
from dual
connect by regexp_substr('margin=x;margin=y;margin=z;',
'.*?=(.*?);',
1,
level) is not null) subselect -- This select represents each margin=T as a single row
where lvl = 1; -- cou could define multiple values to select aswell.
You need a regex that will match 1 to n occurrences of the whole group. E.g.
([^=]*=([^;]*);){2}.*
(replaced with \2 backreference) will get the 2nd attribute value. Your regex can also be used (though it is quite synonymous to the above pattern): (.*?=(.*?);){2}.*.
See the regex demo
If you define the index variable as IDX, you can use something like
select regexp_replace(
'margin=x;margin=y;margin=z;',
CONCAT('([^=]*=([^;]*);){', IDX, '}.*'),
'\2'
) from dual;
NOTE: If you want to get an empty string as a result of trying to obtain a non-existing value, add |.* at the end of the regex:
(.*?=(.*?);){4}.*|.*
See this regex demo (with your input string, the result will be empty string).
Perhaps all you need is this.... The fourth parameter is NOT the occurrence but the POSITION from which the search starts. The FIFTH parameter is the occurrence.
https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions130.htm
Also, are you sure you want REPLACE and not SUBSTR?
EDITED: To clarify (it seems at least one person was confused). I show a possible solution to what you need (perhaps) at the end, but first let's look at REGEXP_REPLACE. I rewrote your query to use different occurrences; I put the index in a CTE, but you can instead make idx into a bind variable, or any other mechanism you need to use. As you will see, the output makes no sense.
with t1 ( idx ) as (select 1 from dual union all select 2 from dual
union all select 3 from dual)
select idx,
regexp_replace('margin=x;margin=y;margin=z;', '.*?=(.*?);.*', '\1', 1, idx) as val
from t1;
Output:
IDX VAL
---------- -----------------------
1 xmargin=y;margin=z;
2 margin=x;ymargin=z;
3 margin=x;margin=y;z
3 rows selected.
I guess this is not what you needed - but it demonstrates what was wrong in your query. The fourth argument to REGEXP_REPLACE, 1 in all cases in the above query, is the position from which the search begins. The fifth argument, idx, is the occurrence. This query replaces the first, second, third occurrence with the subexpression - probably not what you wanted.
If you need to extract x, or y, or z, depending on the occurrence number, you must use REGEXP_SUBSTR, not REGEXP_REPLACE. Note also that I changed the match pattern - the .*? at the beginning and the .* at the end are unnecessary. If you want to find x, y or z in something like margin=x; but not in length=x; then you must make that explicit, the match pattern should be 'margin=(.*?);'.
with t1 ( idx ) as (select 1 from dual union all select 2 from dual
union all select 3 from dual)
select idx,
regexp_replace('margin=x;margin=y;margin=z;', '=(.*?);', '\1', 1, idx) as val
from t1;
Output:
IDX VAL
---------- -------
1 x
2 y
3 z
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.
I am trying to locate some problematic records in a very large Oracle table. The column should contain all numeric data even though it is a varchar2 column. I need to find the records which don't contain numeric data (The to_number(col_name) function throws an error when I try to call it on this column).
I was thinking you could use a regexp_like condition and use the regular expression to find any non-numerics. I hope this might help?!
SELECT * FROM table_with_column_to_search WHERE REGEXP_LIKE(varchar_col_with_non_numerics, '[^0-9]+');
To get an indicator:
DECODE( TRANSLATE(your_number,' 0123456789',' ')
e.g.
SQL> select DECODE( TRANSLATE('12345zzz_not_numberee',' 0123456789',' '), NULL, 'number','contains char')
2 from dual
3 /
"contains char"
and
SQL> select DECODE( TRANSLATE('12345',' 0123456789',' '), NULL, 'number','contains char')
2 from dual
3 /
"number"
and
SQL> select DECODE( TRANSLATE('123405',' 0123456789',' '), NULL, 'number','contains char')
2 from dual
3 /
"number"
Oracle 11g has regular expressions so you could use this to get the actual number:
SQL> SELECT colA
2 FROM t1
3 WHERE REGEXP_LIKE(colA, '[[:digit:]]');
COL1
----------
47845
48543
12
...
If there is a non-numeric value like '23g' it will just be ignored.
In contrast to SGB's answer, I prefer doing the regexp defining the actual format of my data and negating that. This allows me to define values like $DDD,DDD,DDD.DD
In the OPs simple scenario, it would look like
SELECT *
FROM table_with_column_to_search
WHERE NOT REGEXP_LIKE(varchar_col_with_non_numerics, '^[0-9]+$');
which finds all non-positive integers. If you wau accept negatiuve integers also, it's an easy change, just add an optional leading minus.
SELECT *
FROM table_with_column_to_search
WHERE NOT REGEXP_LIKE(varchar_col_with_non_numerics, '^-?[0-9]+$');
accepting floating points...
SELECT *
FROM table_with_column_to_search
WHERE NOT REGEXP_LIKE(varchar_col_with_non_numerics, '^-?[0-9]+(\.[0-9]+)?$');
Same goes further with any format. Basically, you will generally already have the formats to validate input data, so when you will desire to find data that does not match that format ... it's simpler to negate that format than come up with another one; which in case of SGB's approach would be a bit tricky to do if you want more than just positive integers.
Use this
SELECT *
FROM TableToSearch
WHERE NOT REGEXP_LIKE(ColumnToSearch, '^-?[0-9]+(\.[0-9]+)?$');
After doing some testing, i came up with this solution, let me know in case it helps.
Add this below 2 conditions in your query and it will find the records which don't contain numeric data
and REGEXP_LIKE(<column_name>, '\D') -- this selects non numeric data
and not REGEXP_LIKE(column_name,'^[-]{1}\d{1}') -- this filters out negative(-) values
Starting with Oracle 12.2 the function to_number has an option ON CONVERSION ERROR clause, that can catch the exception and provide default value.
This can be used for the test of number values. Simple set NULL when the conversion fails and filer all not NULL values.
Example
with num as (
select '123' vc_col from dual union all
select '1,23' from dual union all
select 'RV12P2000' from dual union all
select null from dual)
select
vc_col
from num
where /* filter numbers */
vc_col is not null and
to_number(vc_col DEFAULT NULL ON CONVERSION ERROR) is not null
;
VC_COL
---------
123
1,23
From http://www.dba-oracle.com/t_isnumeric.htm
LENGTH(TRIM(TRANSLATE(, ' +-.0123456789', ' '))) is null
If there is anything left in the string after the TRIM it must be non-numeric characters.
I've found this useful:
select translate('your string','_0123456789','_') from dual
If the result is NULL, it's numeric (ignoring floating point numbers.)
However, I'm a bit baffled why the underscore is needed. Without it the following also returns null:
select translate('s123','0123456789', '') from dual
There is also one of my favorite tricks - not perfect if the string contains stuff like "*" or "#":
SELECT 'is a number' FROM dual WHERE UPPER('123') = LOWER('123')
After doing some testing, building upon the suggestions in the previous answers, there seem to be two usable solutions.
Method 1 is fastest, but less powerful in terms of matching more complex patterns.
Method 2 is more flexible, but slower.
Method 1 - fastest
I've tested this method on a table with 1 million rows.
It seems to be 3.8 times faster than the regex solutions.
The 0-replacement solves the issue that 0 is mapped to a space, and does not seem to slow down the query.
SELECT *
FROM <table>
WHERE TRANSLATE(replace(<char_column>,'0',''),'0123456789',' ') IS NOT NULL;
Method 2 - slower, but more flexible
I've compared the speed of putting the negation inside or outside the regex statement. Both are equally slower than the translate-solution. As a result, #ciuly's approach seems most sensible when using regex.
SELECT *
FROM <table>
WHERE NOT REGEXP_LIKE(<char_column>, '^[0-9]+$');
You can use this one check:
create or replace function to_n(c varchar2) return number is
begin return to_number(c);
exception when others then return -123456;
end;
select id, n from t where to_n(n) = -123456;
I tray order by with problematic column and i find rows with column.
SELECT
D.UNIT_CODE,
D.CUATM,
D.CAPITOL,
D.RIND,
D.COL1 AS COL1
FROM
VW_DATA_ALL_GC D
WHERE
(D.PERIOADA IN (:pPERIOADA)) AND
(D.FORM = 62)
AND D.COL1 IS NOT NULL
-- AND REGEXP_LIKE (D.COL1, '\[\[:alpha:\]\]')
-- AND REGEXP_LIKE(D.COL1, '\[\[:digit:\]\]')
--AND REGEXP_LIKE(TO_CHAR(D.COL1), '\[^0-9\]+')
GROUP BY
D.UNIT_CODE,
D.CUATM,
D.CAPITOL,
D.RIND ,
D.COL1
ORDER BY
D.COL1