Lets say I have following string: 'product=1627;color=45;size=7' in some field of the table.
I want to query for the color and get 45.
With this query:
SELECT REGEXP_SUBSTR('product=1627;color=45;size=7', 'color\=([^;]+);?') "colorID"
FROM DUAL;
I get :
colorID
---------
color=45;
1 row selected
.
Is it possible to get part of the matched string - 45 for this example?
One way to do it is with REGEXP_REPLACE. You need to define the whole string as a regex pattern and then use just the element you want as the replace string. In this example the ColorID is the third pattern in the entire string
SELECT REGEXP_REPLACE('product=1627;color=45;size=7'
, '(.*)(color\=)([^;]+);?(.*)'
, '\3') "colorID"
FROM DUAL;
It is possible there may be less clunky regex solutions, but this one definitely works. Here's a SQL Fiddle.
Try something like this:
SELECT REGEXP_SUBSTR(REGEXP_SUBSTR('product=1627;color=45;size=7', 'color\=([^;]+);?'), '[[:digit:]]+') "colorID"
FROM DUAL;
From Oracle 11g onwards we can specify capture groups in REGEXP_SUBSTR.
SELECT REGEXP_SUBSTR('product=1627;color=45;size=7', 'color=(\d+);', 1, 1, 'i', 1) "colorID"
FROM DUAL;
Related
I have a column as varchar2 datatype, the data in it is in format:
100323.3819823.222
100.323123.443422
1001010100.233888
LOL12333.DDD33.44
I need to remove the whole part after the first occurrence of '.'
In the end it should look like this:
100323
100
1001010100
LOL12333
I cant seem to find the exact substring expression due to the fact that there is not any fix length of the first part.
One way is to use REGEXP_SUBSTR:
SELECT REGEXP_SUBSTR(column_name,'^[^.]*') FROM table
The other way is to combine SUBSTR with INSTR, which is a bit faster, but will result in NULL if the data doesn't contain a dot, so you'll have to add a switch if needed:
SELECT SUBSTR(column_name, 1, INSTR(column_name,'.') - 1) FROM table
For oracle you can try this:
select substr (i,1,Instr(i,'.',i)-1) from Table name.
I have a column with assignment numbers like - 11827,27266,91717,09818-2,726252-3,8716151-0,827272,18181
Now i am selecting the records like
select assignment_number from table;
But now i want that the column detail is retreived in such a way that numbers are only retrieved without -2 -3 etc like
726252-3---> 726252 8716151-0-->8716151
I know i can use regex for this but i do not know how to use it
This will select everthing before the character -:
^([^-]+)
From 726252-3 will match 726252
You would use regexp() substr:
select regexp_substr(assignmentnumber, '[0-9]+')
This will return the first string of numbers encountered in the string.
I have a table in Oracle where one of the column contains UserIds which are in the form of \. For eg "fin\george", "sales\andy" etc. How can I use REGEXP_SUBSTR function to get only the from the UserIds. ie I want to fetch only "george", "andy" etc. I have achieved the desired reult using SUBSTR function but I want to use REGEXP_SUBSTR in this case.
I tried doing this:
SELECT REGEXP_SUBSTR('fin\george','\[^\]+,') "UserName" FROM DUAL;
but it did'nt help. Can anyone please point out my mistake ?
I believe you want to use a regexp_replace with a backreference. I'm assuming that all the characters before and after the \ are alphabetic. If you allow numbers, you'd want to use the [[:alnum:]] rather than [[:alpha::].
1* SELECT REGEXP_replace('fin\george',
'([[:alpha:]]+\\)([[:alpha:]]+)$',
'\2') "UserName"
FROM DUAL
SQL> /
UserNa
------
george
SQL> SELECT REGEXP_SUBSTR('fin\george', '[^\]+', 1, 2) AS userId from dual;
USERID
------
george
See this Oracle Base article
select regexp_replace( 'fin\george', '.*\\', null ) from dual;
returns george.
The regex will match any character followed by the \ (which is escaped), as many times as possible (greedy).
So it will match everything up to the final \.
Then the matching string is replaced with null.
null is the default so
select regexp_replace('fin\george', '.*\\' ) from dual;
does the same thing
Same expression can extract filename from the end of pathname e.g.
select regexp_replace ('fin\fin2\fin3\fin4\george', '.*\\' ) from dual;
will also return george.
You have to use escaping: \\ instead of \
The easiest way (IMHO) to do this is the following:
SELECT REGEXP_SUBSTR('fin\george', '[^\\]+$') AS "UserName" FROM DUAL;
The issues with your original query were (a) that the \ character was not escaped and (b) there was an extraneous comma in the regular expression. I've used the end-of-string anchor $ here, assuming that there are not more than two elements delimited by \. If there are more than two, and you need only the second one, you can use the following:
SELECT REGEXP_SUBSTR('fin\george\ringo', '[^\\]+', 1, 2) AS "UserName"
FROM DUAL;
This tells Oracle to start looking at the first character of the string and return the second match.
I got this query and want to extract the value between the brackets.
select de_desc, regexp_substr(de_desc, '\[(.+)\]', 1)
from DATABASE
where col_name like '[%]';
It however gives me the value with the brackets such as "[TEST]". I just want "TEST". How do I modify the query to get it?
The third parameter of the REGEXP_SUBSTR function indicates the position in the target string (de_desc in your example) where you want to start searching. Assuming a match is found in the given portion of the string, it doesn't affect what is returned.
In Oracle 11g, there is a sixth parameter to the function, that I think is what you are trying to use, which indicates the capture group that you want returned. An example of proper use would be:
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]', 1,1,NULL,1) from dual;
Where the last parameter 1 indicate the number of the capture group you want returned. Here is a link to the documentation that describes the parameter.
10g does not appear to have this option, but in your case you can achieve the same result with:
select substr( match, 2, length(match)-2 ) from (
SELECT regexp_substr('abc[def]ghi', '\[(.+)\]') match FROM dual
);
since you know that a match will have exactly one excess character at the beginning and end. (Alternatively, you could use RTRIM and LTRIM to remove brackets from both ends of the result.)
You need to do a replace and use a regex pattern that matches the whole string.
select regexp_replace(de_desc, '.*\[(.+)\].*', '\1') from DATABASE;
Say I have a table column that has results like:
ABC_blahblahblah
DEFGH_moreblahblahblah
IJKLMNOP_moremoremoremore
I would like to be able to write a query that selects this column from said table, but only returns the substring up to the Underscore (_) character. For example:
ABC
DEFGH
IJKLMNOP
The SUBSTRING function doesn't seem to be up to the task because it is position-based and the position of the underscore varies.
I thought about the TRIM function (the RTRIM function specifically):
SELECT RTRIM('listofchars' FROM somecolumn)
FROM sometable
But I'm not sure how I'd get this to work since it only seems to remove a certain list/set of characters and I'm really only after the characters leading up to the Underscore character.
Using a combination of SUBSTR, INSTR, and NVL (for strings without an underscore) will return what you want:
SELECT NVL(SUBSTR('ABC_blah', 0, INSTR('ABC_blah', '_')-1), 'ABC_blah') AS output
FROM DUAL
Result:
output
------
ABC
Use:
SELECT NVL(SUBSTR(t.column, 0, INSTR(t.column, '_')-1), t.column) AS output
FROM YOUR_TABLE t
Reference:
SUBSTR
INSTR
Addendum
If using Oracle10g+, you can use regex via REGEXP_SUBSTR.
This can be done using REGEXP_SUBSTR easily.
Please use
REGEXP_SUBSTR('STRING_EXAMPLE','[^_]+',1,1)
where STRING_EXAMPLE is your string.
Try:
SELECT
REGEXP_SUBSTR('STRING_EXAMPLE','[^_]+',1,1)
from dual
It will solve your problem.
You need to get the position of the first underscore (using INSTR) and then get the part of the string from 1st charecter to (pos-1) using substr.
1 select 'ABC_blahblahblah' test_string,
2 instr('ABC_blahblahblah','_',1,1) position_underscore,
3 substr('ABC_blahblahblah',1,instr('ABC_blahblahblah','_',1,1)-1) result
4* from dual
SQL> /
TEST_STRING POSITION_UNDERSCORE RES
---------------- ------------------ ---
ABC_blahblahblah 4 ABC
Instr documentation
Susbtr Documentation
SELECT REGEXP_SUBSTR('STRING_EXAMPLE','[^_]+',1,1) from dual
is the right answer, as posted by user1717270
If you use INSTR, it will give you the position for a string that assumes it contains "_" in it. What if it doesn't? Well the answer will be 0. Therefore, when you want to print the string, it will print a NULL.
Example: If you want to remove the domain from a "host.domain". In some cases you will only have the short name, i.e. "host". Most likely you would like to print "host". Well, with INSTR it will give you a NULL because it did not find any ".", i.e. it will print from 0 to 0. With REGEXP_SUBSTR you will get the right answer in all cases:
SELECT REGEXP_SUBSTR('HOST.DOMAIN','[^.]+',1,1) from dual;
HOST
and
SELECT REGEXP_SUBSTR('HOST','[^.]+',1,1) from dual;
HOST
Another possibility would be the use of REGEXP_SUBSTR.
In case if String position is not fixed then by below Select statement we can get the expected output.
Table Structure
ID VARCHAR2(100 BYTE)
CLIENT VARCHAR2(4000 BYTE)
Data-
ID CLIENT
1001 {"clientId":"con-bjp","clientName":"ABC","providerId":"SBS"}
1002
--
{"IdType":"AccountNo","Id":"XXXXXXXX3521","ToPricingId":"XXXXXXXX3521","clientId":"Test-Cust","clientName":"MFX"}
Requirement - Search ClientId string in CLIENT column and return the corresponding value. Like From "clientId":"con-bjp" --> con-bjp(Expected output)
select CLIENT,substr(substr(CLIENT,instr(CLIENT,'"clientId":"')+length('"clientId":"')),1,instr(substr(CLIENT,instr(CLIENT,'"clientId":"')+length('"clientId":"')),'"',1 )-1) cut_str from TEST_SC;
--
CLIENT cut_str
----------------------------------------------------------- ----------
{"clientId":"con-bjp","clientName":"ABC","providerId":"SBS"} con-bjp
{"IdType":"AccountNo","Id":"XXXXXXXX3521","ToPricingId":"XXXXXXXX3521","clientId":"Test-Cust","clientName":"MFX"} Test-Cust
Remember this if all your Strings in the column do not have an underscore
(...or else if null value will be the output):
SELECT COALESCE
(SUBSTR("STRING_COLUMN" , 0, INSTR("STRING_COLUMN", '_')-1),
"STRING_COLUMN")
AS OUTPUT FROM DUAL
To find any sub-string from large string:
string_value:=('This is String,Please search string 'Ple');
Then to find the string 'Ple' from String_value we can do as:
select substr(string_value,instr(string_value,'Ple'),length('Ple')) from dual;
You will find result: Ple