I am trying to find the pattern using oracle sql developer. Include numbers that are "x0000"," 0000",".0000"
AND REGEXP_INSTR(string=>FTP.TEXT, pattern=>'[x]\d\d\d\d\s', match_parameter=>'i')
Use
where REGEXP_LIKE(str,'[x.[:space:]]\d{4}','i')
The [x.[:space:]]\d{4} expression matches x, . or any whitespace, then any four digit characters.
You can use REGEXP_LIKE as follows:
SQL> --sample data
SQL> with your_data(id,str) as
2 (select 1, 'x0000' from dual union all
3 select 2, '.0000' from dual union all
4 select 3, ' 0000'from dual union all
5 select 4, '0000' from dual) -- this should not be in the result
6 -- your query starts from here
7 select * from your_data
8 where REGEXP_like(str,'[x| |\.][0-9]{4}','i');
ID STR
---------- -----
1 x0000
2 .0000
3 0000
SQL>
Related
Could you please tell me the expression regexp_like to check if a string has the format which begins with one or two letters max followed by numbers. For example, like 'A25', 'AB567'.
Thanks
How about
SQL> with test (col) as
2 (select 'A25' from dual union all
3 select 'AB567' from dual union all
4 select 'A12A532A' from dual union all
5 select 'ABC123' from dual union all
6 select '12XYZ34' from dual
7 )
8 select col,
9 case when regexp_like(col, '^[[:alpha:]]{1,2}[[:digit:]]+$') then 'OK'
10 else 'Wrong'
11 end result
12 from test;
COL RESULT
---------- ----------
A25 OK
AB567 OK
A12A532A Wrong
ABC123 Wrong
12XYZ34 Wrong
SQL>
The regular expression would be ^[A-Z]{1,2}[0-9]+$.
Working demo on db<>fiddle here
This match a string that contain only numbers and letters, that start with a letter, when there is never more than 2 letters in a row.
^(([A-Za-z]){1,2}([0-9])+)*\2{0,2}$
I want to get the time pattern along with AM or PM from the given string Aaaaa_gggg_ne_A030_66788_Abcd_Oct_24_0329PM.csv
I tried the following:
Select regexp_substr(filename,'\d{4}',1,3)
From
(Select 'Aaaaa_gggg_ne_A030_66788_Abcd_Oct_24_0329PM.csv' filename from dual);
which only gives me the last number, e.g. 0329, but I need 0329PM.
Using this form of REGEXP_SUBSTR() will get what you need in one call. It returns the first group, which is the set of characters after the last underscore and before the literal period of 1 or more numbers followed by an A or P then an M.
with tbl(filename) as (
Select 'Aaaaa_gggg_ne_A030_66788_Abcd_Oct_24_0329PM.csv'
from dual
)
select regexp_substr(filename, '_(\d+[AP]M)\.', 1, 1, NULL, 1)
From tbl;
Actually, to tighten up the match you could make it case-insensitive and add the extension:
select regexp_substr(filename, '_(\d+[AP]M)\.csv', 1, 1, 'i', 1)
From tbl;
Note if a match is not found NULL will be returned.
Nested substr is one option (if data always looks like this; you didn't say it doesn't):
SQL> with test (col) as
2 (select 'Aaaaa_gggg_ne_A030_66788_Abcd_Oct_24_0329PM.csv' from dual)
3 select substr(substr(col, -10), 1, 6) result from test
4 /
RESULT
------
0329PM
SQL>
the inner substr returns the last 10 characters (0329PM.csv)
the outer substr returns the first 6 characters out of it (0329PM)
Or, using regular expressions:
SQL> with test (col) as
2 (select 'Aaaaa_gggg_ne_A030_66788_Abcd_Oct_24_0329PM.csv' from dual)
3 select regexp_substr(translate(col, '_.', ' '), '\S+',
4 1,
5 regexp_count(translate(col, '_.', ' '), '\S+') - 1
6 ) result
7 from test;
RESULT
------
0329PM
SQL>
line #3: translate replaces underlines and dots with a space
line #4: start from the beginning
line #5: return substring which is one before the last one
There is a column in my table having values of a pattern like 'A=xxx^B=xxx^C=xxx^D=xxx^' i need to update all the columns having this pattern to a pattern like 'C=xxx^D=xxx^', where x is a number.
Would something like this help? REGEXP_LIKE returns rows which satisfy the condition, while an ordinary SUBSTR returns the desired result.
SQL> with test (col) as
2 (select 'A=123^B=123^C=123^D=123^' from dual union
3 select 'A=123^B=456^C=789^D=987^' from dual union
4 select 'A=333^C=333^D=333^' from dual union
5 select 'C=987^D=987^' from dual union
6 select 'B=876^' from dual union
7 select 'A=123^B=123^C=123^D=123^E=123^' from dual
8 )
9 select col,
10 substr(col, instr(col, 'C')) result
11 from test
12 where regexp_like(col, '^A=\d+{3}\^B=\d+{3}\^C=\d+{3}\^D=\d+{3}\^$');
COL RESULT
------------------------------ ------------------------------
A=123^B=123^C=123^D=123^ C=123^D=123^
A=123^B=456^C=789^D=987^ C=789^D=987^
SQL>
I managed to come up with a solution since i'm looking for a pattern which starts from 'A=' i used REGEXP_LIKE to find that particular pattern. Then i used SUBSTR to extract the value from the string which should start from the 2nd '^' character.
Update MYTABLE t set t.key = SUBSTR(t.key,INSTR(t.key,'^',1,2)+1) WHERE REGEXP_LIKE(t.key_ref,'^A=') and t.dno = 'xxxxx';
I am trying to get all the name which has special char and numbers in it. I want to know if there is a better way of writing this query. Thanks
select Name from TABLE1
WHERE NAME LIKE '%-%'
OR NAME LIKE '%$%'
OR NAME LIKE '%4%' etc
Try this:
SELECT Name FROM Test WHERE REGEXP_LIKE(Name,'(-|$|4)');
The | (OR or Alternation) operator allows the expression to return true if the value contains an '-' OR an '$' OR an '4' etc.
Use square brackets to define a character class. Also a full regex to anchor the pattern to the beginning of the string, any number of any characters before a character class consisting of a dash or a 4 or a dollar sign then any number of any characters anchored to the end of the string:
SQL> with tbl(name) as (
2 select 'efs' from dual
3 union
4 select 'abc-' from dual
5 union
6 select 'a4bd' from dual
7 union
8 select 'gh$dll' from dual
9 union
10 select 'xy5zzy' from dual
11 )
12 select name from tbl
13 where regexp_like(name, '^.*[-|$|4].*$');
NAME
------
a4bd
abc-
gh$dll
SQL>
SELECT REGEXP_SUBSTR('one,two,three',',[^,]+') AS reg_result FROM DUAL;
REG_RESULT
,two
SELECT REGEXP_SUBSTR('eight,nineteen,five',',[^,]+') AS reg_result FROM DUAL;
REG_RESULT
,nineteen
I have to remove the " , " from the result. Also I want the last string
as the output. i.e. three from 'one,two,three' & five from 'eight,nineteen,five'.
How can I do it??
If want to just get the last word without checking if your string meets particular pattern:
SQL> with t1 as(
2 select 'one,two,three' as str from dual
3 )
4 select regexp_substr(str, '([[:alpha:]]+)$') last_word
5 from t1
6 ;
LAST_WORD
---------
three
Response to the comment
how to get string two from the first one & nineteen from the second??
Fourth parameter of the regexp_substr function is the occurrence of a pattern. So to get the second word in the string we can use regexp_substr(str, '[^,]+', 1, 2)
SQL> with t1 as(
2 select 'one,two,three' as str from dual
3 )
4 select regexp_substr(str, '[^,]+', 1, 2) as Second_Word
5 from t1;
Second_Word
---------
two
If there is a need to extract every word from your strings:
-- sample of data from your question
SQL> with t1 as(
2 select 'one,two,three' as str from dual union all
3 select 'eight,nineteen,five' from dual
4 ), -- occurrences of the pattern
5 occurrence as(
6 select level as ps
7 from ( select max(regexp_count(str, '[^,]+')) mx
8 from t1
9 ) s
10 connect by level <= s.mx
11 ) -- the query
12 select str
13 , regexp_substr(str, '[^,]+', 1, o.ps) word
14 , o.ps as word_num
15 from t1 t
16 cross join occurrence o
17 order by str
18 ;
STR WORD WORD_NUM
------------------- ----------- ----------
eight,nineteen,five eight 1
eight,nineteen,five nineteen 2
eight,nineteen,five five 3
one,two,three three 3
one,two,three one 1
one,two,three two 2
6 rows selected
SELECT REGEXP_SUBSTR(REGEXP_SUBSTR('one,two,three',',[^,]+$'),'[^,]+') AS reg_result FROM DUAL;
I'm not certain if Oracle has lookbehinds, but you can try this as well:
SELECT REGEXP_SUBSTR('one,two,three','(?<=,)[^,]+$') AS reg_result FROM DUAL;