I found a solution to search a string from a blob but how we can ignore case sensitive?
if I have string foo in capital case so how it will match?
source: StackOverflow
select *
from table1
where dbms_lob.instr (t1, -- the blob
utl_raw.cast_to_raw ('foo'), -- the search string cast to raw
1, -- where to start. i.e. offset
1 -- Which occurrence i.e. 1=first
) > 0 -- location of occurrence. Here I don't care. Just find any
;
Related
I am looking a way to find the values in a column that has non alfa numeric values...
I tried
select 'kjh$' not RLIKE '([0-9][a-z]|[A-Z])*')
but does not work
Thanks for your help
You can use REGEXP '^[A-Za-z0-9]+$' or RLIKE '^[A-Za-z0-9]+$'.
Sample SQL -
select * from my table where mycol not RLIKE '^[A-Za-z0-9]+$'
^ - determines start of the string
$ - end of the string
+ - match the preceding character one or more times
[A-Za-z0-9] - to check alphanumeric or not
I ran a simple select statement to check if a string has alphanumeric or not using regex and here is the output.
select 'Aa90$$bc' ,'Aa90$$bc' rlike '^[A-Za-z0-9]+$'
From the String ES-123456-PSA Spain-101, I need to extract only ES-123456-101 Delimiter position is fixed.
Tried REGEXP_SUBSTR('ES-123456-PSA Spain-101','[^-]+',2,3 ) which gives PSA Spain.
Is there a way to ignore those specific characters and returns rest of them.
If you want ES-123456-101 then use this:
SELECT REGEXP_REPLACE('ES-123456-PSA Spain-101', '[^-]+-', '', 1, 3 )
FROM dual;
If you want ES-12345-101 then could you explain the logic for 12345 not 123456? Typo or omit the last character?
you can also use subtr and instr
with t as
(
select 'ES-123456-PSA Spain-101' as text from dual
)
select substr(text,1,instr(text,'-',1,2)) -- ES-123456-
||substr(text,instr(text,'-',1,3)+1) -- 101
from t
In plsql is there a way to split a string into an associative array?
Sample string: 'test1:First string, test2: Second string, test3: Third string'
INTO
TYPE as_array IS TABLE OF VARCHAR2(50) INDEX BY VARCHAR2(50);
a_array as_array;
dbms_output.put_line(a_array('test1')); // Output 'First string'
dbms_output.put_line(a_array('test2')); // Output 'Second string'
dbms_output.put_line(a_array('test3')); // Output 'Third string'
The format of the string does not matter for my purposes. It could be 'test1-First string; test2-Second string; test3-Third string'. I could do this with a very large function manually splitting by commas first and then splitting each of those but I'm wondering if there is something built in to the language.
Like I said, I am not looking to do it through a large function (especially using substr and making it look messy). I am looking for something that does my task simpler.
There is no built in function for such a requirement.
But you can easily build a query like below to parse these strings:
SELECT y.*
FROM (
select trim(regexp_substr(str,'[^,]+', 1, level)) as str1
from (
SELECT 'test1:First string, test2: Second string, test3: Third string' as Str
FROM dual
)
connect by regexp_substr(str, '[^,]+', 1, level) is not null
) x
CROSS APPLY(
select trim(regexp_substr(str1,'[^:]+', 1, 1)) as key,
trim(regexp_substr(str1,'[^:]+', 1, 2)) as value
from dual
) y
KEY VALUE
------ --------------
test1 First string
test2 Second string
test3 Third string
Then you may use this query in your function and pass it's result to the array.
I leave this exercise for you, I believe you can manage it (tip: use Oracle's bulk collect feature)
This method handles NULL list elements if you need to still show that element 2 is NULL for example. Note the second element is NULL:
-- Original data with multiple delimiters and a NULL element for testing.
with orig_data(str) as (
select 'test1:First string,, test3: Third string' from dual
),
--Split on first delimiter (comma)
Parsed_data(rec) as (
select regexp_substr(str, '(.*?)(,|$)', 1, LEVEL, NULL, 1)
from orig_data
where str is not null
CONNECT BY LEVEL <= REGEXP_COUNT(str, ',') + 1
)
-- For testing-shows records based on 1st level delimiter
--select rec from parsed_data;
-- Split the record into columns
select trim(regexp_replace(rec, '^(.*):.*', '\1')) key,
trim(regexp_replace(rec, '^.*:(.*)', '\1')) value
from Parsed_data;
Watch out for the regex form of [^,]+ for parsing delimited strings, it fails on NULL elements. More Information
i have a string 'MCDONALD_YYYYMMDD.TXT' i need to use regular expressions and append the '**' after the letter 'D' in the string given . (i.e In the string at postion 9 i need to append '*' based on a column value 'star_len'
if the star_len = 2 the o/p = ''MCDONALD??_YYYYMMDD.TXT'
if the star_len = 1 the o/p = ''MCDONALD?_YYYYMMDD.TXT'
with
inputs ( filename, position, symbol, len ) as (
select 'MCDONALD_20170812.TXT', 9, '*', 2 from dual
)
-- End of simulated inputs (for testing purposes only, not part of the solution).
-- SQL query begins BELOW THIS LINE.
select substr(filename, 1, position - 1) || rpad(symbol, len, symbol)
|| substr(filename, position) as new_str
from inputs
;
NEW_STR
-----------------------
MCDONALD**_20170812.TXT
select regexp_replace('MCDONALD_YYYYMMDD.TXT','MCDONALD','MCDONALD' ||
decode(star_len,1,'*',2,'**'))
from dual
This is how you could do it. I don't think you need it as a regular expression though if it is always going to be "MCDONALD".
EDIT: If you need to be providing the position in the string as well, I think a regular old substring should work.
select substr('MCDONALD_YYYYMMDD.TXT',1,position-1) ||
decode(star_len,1,'*',2,'**') || substr('MCDONALD_YYYYMMDD.TXT',position)
from dual
Where position and star_len are both columns in some table you provide(instead of dual).
EDIT2: Just to be more clear, here is another example using a with clause so that it runs without adding a table in.
with testing as
(select 'MCDONALD_YYYYMMDD.TXT' filename,
9 positionnum,
2 star_len
from dual)
select substr(filename,1,positionnum-1) ||
decode(star_len,1,'*',2,'**') ||
substr(filename,positionnum)
from testing
For the fun of it, here's a regex_replace solution. I went with a star since that what your variable was called even though your example used a question mark. The regex captures the filename string in 2 parts, the first being from the start up to 1 character before the position value, the second the rest of the string. The replace puts the captured parts back together with the stars in between.
with tbl(filename, position, star_len ) as (
select 'MCDONALD_20170812.TXT', 9, 2 from dual
)
select regexp_replace(filename,
'^(.{'||(position-1)||'})(.*)$', '\1'||rpad('*', star_len, '*')||'\2') as fixed
from tbl;
I'm doing a query that returns a VARCHAR2 and some other fields. I'm ordering my results by this VARCHAR2 and having some problems related to the linguistic sort, as I discovered on Oracle documentation. For example:
SELECT id, my_varchar2 from my_table ORDER BY MY_VARCHAR2;
Will return:
ID MY_VARCHAR2
------ -----------
3648 A
3649 B
6504 C
7317 D
3647 0
I need it to return the string "0" as the first element on this sequence, as it would be comparing ASCII values. The string can have more than one character so I can't use the ascii function as it ignores any characters except for the first one.
What's the best way to do this?
For that case, you should be able to just order by the BINARY value of your characters;
SELECT id, my_varchar2
FROM my_table
ORDER BY NLSSORT(MY_VARCHAR2, 'NLS_SORT = BINARY')
SQLFiddle here.