Regular Expression, problem when the literals have content "I_"
Hi,
I guess this seems to be an Oracle bug/defect. If I am doing it wrongly, please let me know.
I need to remove the "_VERSION" which comes at the last of a string column. For e.g, If the string is WINDOWS_XP_VERSION, then the output should be WINDOWS_XP.
But if you run the below query,
SELECT
REGEXP_SUBSTR('AA_TEST_3_XI_VERSION', '\w+[^_VERSION]') t1,
REGEXP_SUBSTR('AA_TEST_3_Xi_VERSION', '\w+[^_VERSION]') t2,
REGEXP_SUBSTR('AA_TEST_3_XIA_VERSION', '\w+[^_VERSION]') t3
FROM DUAL;
T1 T2 T3
AA_TEST_3_X AA_TEST_3_Xi AA_TEST_3_XIA
1 rows selected
It seems the issue exists only when I use "I_" and works fine with "i_". I checked with few other characters and it all worked fine.
Any ideas? I am using 11g.
Thanks,
Sharmin
Folks,
Thanks for all your helps.
MichaelS, REGEXO_SUBSTR (str, '(\w+)_VERSION$',1,1,null,1) works, but if any string which does not have "_VERSION", it would return NULL. Also, I had used'$' in my expression and missed to put it in the forum. I am sorry not to make it clear that my column can/cannot end with "_VERSION".
hm, Thank you for making me understand how the regexp works. I thought if you do a '^' and supply a word, it would be considered together.
What needs to be done for making negation for a specific word?
WhiteHat,
REGEXP_REPLACE is a better option. But in this case, it helps me the same way with the regular REPLACE function as my column will not contain any other '_VERSION' substring.
I am going to use a regular replace function. select REPLACE('AA_TEST_3_XE_VERSION_VERSION','_VERSION','') substring from dual;
Thanks once again for all your helps.
Thanks,
Sharmin
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