I wrote a query which splits a string and show me as a value I want using SUBSTR
SELECT SUBSTR ('imagelocation/r1.jpg', 15) AS image_location FROM dual
I am getting the output as r1.jpg but I only want the value to come as r1. Please help
If you want something more universal, regexp_replace might be of use.
SELECT
regexp_replace('imagelocation/r1.jpg','^[^/]*/([^.]+).*$','\1') AS image_location
FROM dual;
select SUBSTR (
'imagelocation/r1.jpg',
INSTR('imagelocation/r1.jpg', '/')+1,
LENGTH('imagelocation/r1.jpg') - INSTR('imagelocation/r1.jpg', '.') - 1
) AS image_location
FROM dual
SUBSTR Function in Oracle
SQL Fiddle
Try this:
SELECT SUBSTR('imagelocation/r1.jpg',15,2) AS image_location FROM dual
Related
I have a string with double quotes inside.
EG:
<cosmtio :ff "intermit"ksks>
I need the data between the ""
I have tried the regexp_substr but still couldn't get the value between double-quotes.
We could try using REGEXP_REPLACE here:
SELECT
string,
REGEXP_REPLACE(string, '.*"([^"]+)".*', '\1') AS quoted_term
FROM yourTable;
Data:
WITH yourTable AS (
SELECT '<cosmtio :ff "intermit"ksks>' AS string FROM dual
)
Demo
Another option, using REGEXP_SUBSTR:
SELECT
string,
TRIM(BOTH '"' FROM REGEXP_SUBSTR(string, '".*"'))
FROM yourTable;
But this approach requires nesting two function calls, which means it might not outperform the REGEXP_REPLACE version.
You need to use REGEXP_SUBSTR:
SELECT REGEXP_SUBSTR('<cosmtio :ff "intermit"ksks>', '"([^"]+)"', 1, 1, NULL, 1) AS Result FROM DUAL
See the online demo.
The regex is simple: "([^"]+)" matches ", then captures any 1+ chars other than " into Group 1 and then matches ". The last argument is 1 telling Oracle REGEXP_SUBSTR to return the Group 1 values. The first (position) and the second (occurrence) arguments are default, 1. NULL means no specific options need to be passed to the regex engine.
You can try the following:
SELECT REGEXP_REPLACE('<cosmtio :ff "intermit"ksks>', '^[^"]*("([^"]*)")?.*', '\2') FROM dual
It is possible with regexp_substr as following:
Select
regexp_substr('<cosmtio :ff "intermit"ksks>', '[^"]+', 1, 2)
from dual;
Cheers!!
How to get rest of string after specific char?
I have a string 'a|b|c|2|:x80|3|rr|' and I would like to get result after 3rd occurance of |. So the result should be like 2|:x80|3|rr|
The query
select REGEXP_SUBSTR('a|b|c|2|:x80|3|rr|','[^|]+$',1,4)
from dual
Returned me NULL
Use SUBSTR / INSTR combination
WITH t ( s ) AS (
SELECT 'a|b|c|2|:x80|3|rr|'
FROM dual
) SELECT substr(s,instr(s,'|',1,3) + 1)
FROM t;
Demo
REGEXP_REPLACE() will do the trick. Skip 3 groups of anything followed by a pipe, then replace with the 2nd group, which is the rest of the line (anchored to the end).
SQL> select regexp_replace('a|b|c|2|:x80|3|rr|', '(.*?\|){3}(.*)$', '\2') trimmed
2 from dual;
TRIMMED
------------
2|:x80|3|rr|
SQL>
I suggest a nice by long way by using regexp_substr, regexp_count and listagg together as :
select listagg(str) within group (order by lvl)
as "Result String"
from
(
with t(str) as
(
select 'a|b|c|2|:x80|3|rr|' from dual
)
select level-1 as lvl,
regexp_substr(str,'(.*?)(\||$)',1,level) as str
from dual
cross join t
connect by level <= regexp_count('a|b|c|2|:x80|3|rr|','\|')
)
where lvl >= 3;
Rextester Demo
If you use oracle 11g and above you can specify a subexpression to return like this:
select REGEXP_SUBSTR('a|b|c|2|:x80|3|rr|','([^|]+\|){3}(.+)$',1,1,null,2) from dual
Erkko,
You need to use the combination of SUBSTR and REGEXP_INSTR OR INSTR.
Your query will look like this. (Without Regex)
SELECT SUBSTR('a|b|c|2|:x80|3|rr|',INSTR('a|b|c|2|:x80|3|rr|','|',1,3)+1) from dual;
Your query will look like this. (With Regex as you want to use)
SELECT SUBSTR('a|b|c|2|:x80|3|rr|',REGEXP_INSTR('a|b|c|2|:x80|3|rr|','\|',1,3)+1) from dual;
Explanation:
First, you will need to find the place of the string you want as you mentioned. So in your case | comes at place 6. So that +1 would be your position to start to substring.
Second, from the original string, substring from that position+1 to unlimited.(Where your string ends)
Example:
https://dbfiddle.uk/?rdbms=oracle_11.2&fiddle=6fd782db95f575201eded084493232ee
I have String like a123bcd-e2343fg-hij-dfgh
and I want OUTPUT e2343fg-hij-dfgh using Regular_expression in oracle.
select regexp_substr('abcd-efg-hij','-[^-]+'1) from dual;
You may apply regexp_substr with [^-]+[-^] pattern and then ltrim as :
select ltrim('a123bcd-e2343fg-hij-dfgh',
regexp_substr('a123bcd-e2343fg-hij-dfgh','[^-]+[-^]')) as output_string
from dual;
or better to call with bind variable :
select ltrim('&str', regexp_substr('&str','[^-]+[-^]')) as output_string
from dual;
where &str may be replaced with a123bcd-e2343fg-hij-dfgh after prompted.
Rextester Demo
Why regular expression, when a trivial SUBSTR + INSTR does the job nicely & quickly? True, it will look smarter, but I can't see any other benefit.
SQL> with test (col) as
2 (select 'a123bcd-e2343fg-hij-dfgh' from dual)
3 select substr(col, instr(col, '-') + 1) result
4 from test;
RESULT
----------------
e2343fg-hij-dfgh
SQL>
select substr('abcd-efg-hij',
regexp_instr('abcd-efg-hij','-[^-]+')+1,length('abcd-efg-hij'))
from dual;
try this
For the sake of argument, regexp_replace works too. This regex matches anything up to and including the first dash, and remembers the rest which it returns.
with tbl(str) as (
select 'a123bcd-e2343fg-hij-dfgh' from dual
)
select regexp_replace(str, '^.*?-(.*)', '\1')
from tbl;
Keep in mind if regexp_substr() does not find a match, it returns NULL but if regexp_replace() does not find a match it return the original string.
I've got a string that ends with a certain number of '=' characters at the end. It's basically a base 64 string.
How can I get this count of '=' characters at the end? A built-in SQL function or regex would be preferred.
I know about the instr function, but it doesn't seem like it could be applied here. I'm not sure if a regex would apply here either.
Use length and replace
select length(some_column) - length(replace(some_column, '=', ''))
from your_table
SELECT REGEXP_COUNT('hello world==', '=') cnt FROM dual
/
SELECT count(distinct(Instr('hello world==','=', LEVEL))) cnt
FROM dual
CONNECT BY LEVEL <= Length('hello world==')
ORDER BY 1
/
Can you please help to get this code for SQL?
I have column name INFO_01 which contain info like:
D10-52247-479-245 HALL SO
and I would like to extract only
D10-52247-479
I want the part of the text before the third "-" dash.
You'll need to get the position of the third dash (using instr) and then use substr to get the necessary part of the string.
with temp as (
select 'D10-52247-479-245 HALL SO' test_string from dual)
select test_string,
instr(test_string,1,3) third_dash,
substr(test_string,1,instr(test_string,1,3)-1) result
from temp
);
Here is a simple statement that should work:
SELECT SUBSTR(column, 1, INSTR(column,'-',1,3) ) FROM table;
Using a combination of SUBSTR and INSTR will return what you want:
SELECT SUBSTR('D10-52247-479-245', 0, INSTR('D10-52247-479-245', '-', -1, 1)-1) AS output
FROM DUAL
Result:
output
-------------
D10-52247-479
Use:
SELECT SUBSTR(t.column, 0, INSTR(t.column, '-', -1, 1)-1) AS output
FROM YOUR_TABLE t
Reference:
SUBSTR
INSTR
Addendum
If using Oracle10g+, you can use regex via REGEXP_SUBSTR.
I'm assuming MySQL, let me know if I'm wrong here. But using SUBSTRING_INDEX you could do the following:
SELECT SUBSTRING_INDEX(column, '-', 3)
EDIT
Appears to be oracle. Looks like we may have to resort to REGEXP_SUBSTR
SELECT REGEXP_SUBSTR(column, '^((?.*\-){2}[^\-]*)')
Can't test, so not sure what kind of result that will have...