I have column in a table which contains special characters with an attached string. The same column contains numbers also. I am only intresed in extracting numbers out of that column e.g
Name-3445 => 3445; Out-90 => 90; 786 => 786
How would i do this in SQL or PL/SQL?
SELECT regexp_replace(some_column, '[^0-9]*', '') as clean_value
FROM your_table
PL/SQL has a REGEX_REPLACE function, which you could use to replace anything that isn't a digit with an empty string. Details on REGEX_REPLACE can be found here: http://psoug.org/reference/regexp.html
Without knowing the integrity of your data, something like this might do what you're asking:
select CAST(SUBSTRING(_COLUMNNAME_,CHARINDEX('-', _COLUMNNAME_),1000), Integer) as ColumnName
from tblTable where _COLUMNNAME_ like '%-%'
union all select CAST(_COLUMNNAME, Integer) as ColumnName
from tblTable where _COLUMNNAME_ not like '%-%'
You can use the regexp_substr function:
http://docs.oracle.com/cd/B14117_01/server.101/b10759/functions116.htm
REGEXP_REPLACE(<Your_String>,'[^[:alnum:]'' '']', NULL)
Example:
SELECT REGEXP_REPLACE('##$$$123&&!!__!','[^[:alnum:]'' '']', NULL) FROM dual;
Output:
123
Related
I am using PostgreSQL. Data in the table looks like below there is ''. I would like to remove '' with the empty value.
Query:
select 'Patriots Colony Family Monthly'
Actual Result Screenshot:
Expected result:
Abcd
You can use regex_replace to remove non alphanumeric chars
select regexp_replace(yourColumn, '[^[:alnum:]]', ' ', 'g')
from yourTable;
You can also do like this
select regexp_replace(yourColumn, '[^a-zA-Z0-9]+',' ','g')
from yourTable;
I would like to keep rows only with Numeric Character i.e. 0-9. My source data can have any type of character e.g. 2,%,( .
Input (postcode)
3453gds sdg3
454232
sdg(*d^
452
Expected Output (postcode)
454232
452
I have tried using WHERE REGEXP_LIKE(postcode, '^[[:digit:]]+$');
however in my version of Oracle I get an error saying
function regexp_like(character varying, "unknown") does not exist
You want regexp_like() and your version should work:
select t.*
from t
where regexp_like(t.postcode, '^[0-9]+$');
However, your error looks more like a Postgres error, so perhaps this will work:
t.postcode ~ '^[0-9]+$'
For Oracle 10 or higher you can use regexp functions. In earlier versions translate function will help you :
SELECT postcode
FROM table_name
WHERE length(translate(postcode,'0123456789','1')) is null
AND postcode IS NOT NULL;
OR
SELECT translate(postcode, '0123456789' || translate(postcode,'x123456789','x'),'0123456789') nums
FROM table_name ;
the above answer also works for me
SELECT translate('1234bsdfs3#23##PU', '0123456789' || translate('1234bsdfs3#23##PU','x123456789','x'),'0123456789') nums
FROM dual ;
Nums:
1234323
For an alternative to the Gordon Linoff answer, we can try using REGEXP_REPLACE:
SELECT *
FROM yourTable
WHERE REGEXP_REPLACE(postcode, '[0-9]+', '') IS NULL;
The idea here is to strip away all digit characters, and then assert that nothing were left behind. For a mixed digit-letter value, the regex replacement would result in a non-empty string.
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.
I need to create an Oracle DB function that takes a string as parameter. The string contains letters and numbers. I need to extract all the numbers from this string. For example, if I have a string like RO1234, I need to be able to use a function, say extract_number('RO1234'), and the result would be 1234.
To be even more precise, this is the kind of SQL query which this function would be used in.
SELECT DISTINCT column_name, extract_number(column_name)
FROM table_name
WHERE extract_number(column_name) = 1234;
QUESTION: How do I add a function like that to my Oracle database, in order to be able to use it like in the example above, using any of Oracle SQL Developer or SQLTools client applications?
You'd use REGEXP_REPLACE in order to remove all non-digit characters from a string:
select regexp_replace(column_name, '[^0-9]', '')
from mytable;
or
select regexp_replace(column_name, '[^[:digit:]]', '')
from mytable;
Of course you can write a function extract_number. It seems a bit like overkill though, to write a funtion that consists of only one function call itself.
create function extract_number(in_number varchar2) return varchar2 is
begin
return regexp_replace(in_number, '[^[:digit:]]', '');
end;
You can use regular expressions for extracting the number from string. Lets check it. Suppose this is the string mixing text and numbers 'stack12345overflow569'. This one should work:
select regexp_replace('stack12345overflow569', '[[:alpha:]]|_') as numbers from dual;
which will return "12345569".
also you can use this one:
select regexp_replace('stack12345overflow569', '[^0-9]', '') as numbers,
regexp_replace('Stack12345OverFlow569', '[^a-z and ^A-Z]', '') as characters
from dual
which will return "12345569" for numbers and "StackOverFlow" for characters.
This works for me, I only need first numbers in string:
TO_NUMBER(regexp_substr(h.HIST_OBSE, '\.*[[:digit:]]+\.*[[:digit:]]*'))
the field had the following string: "(43 Paginas) REGLAS DE PARTICIPACION".
result field: 43
If you are looking for 1st Number with decimal as string has correct decimal places, you may try regexp_substr function like this:
regexp_substr('stack12.345overflow', '\.*[[:digit:]]+\.*[[:digit:]]*')
To extract charecters from a string
SELECT REGEXP_REPLACE(column_name,'[^[:alpha:]]') alpha FROM DUAL
In order to extract month and a year from a string 'A0807' I did the following in PL/SQL:
DECLARE
lv_promo_code VARCHAR2(10) := 'A0807X';
lv_promo_num VARCHAR2(5);
lv_promo_month NUMBER(4);
lv_promo_year NUMBER(4);
BEGIN
lv_promo_num := REGEXP_SUBSTR(lv_promo_code, '(\d)(\d)(\d)(\d)');
lv_promo_month := EXTRACT(month from to_date(lv_promo_num, 'MMYY'));
DBMS_OUTPUT.PUT_LINE(lv_promo_month);
lv_promo_year := EXTRACT(year from to_date(lv_promo_num, 'MMYY'));
DBMS_OUTPUT.PUT_LINE(lv_promo_year);
END;
Maybe this sounds a little bit crazy, but I need to come up with a query to retrieve only letters out of an alphanumeric field.
For example:
TABLE
1234ADD
3901AC
1812OPA
82711AUU
RESULTS EXPECTED
ADD
AC
OPA
AUU
Thank you!
Looks like you only want to remove numbers. You can use REGEXP_REPLACE for that in 10g or 11g:
SELECT REGEXP_REPLACE( your_column, '[0-9]*', '' ) FROM your_table;
SELECT REGEXP_REPLACE('1234ADD 3901AC 1812OPA 82711AUU', '[0-9]', '')
FROM dual
Try
SELECT TRANSLATE('1234ADD 3901AC 1812OPA 82711AUU', 'A1234567890', 'A') FROM dual;
and in general see: http://www.psoug.org/reference/translate_replace.html