I have this table of rows
RowA
______
ABC123
DEF432
WER677
JKL342
how can I add a '_' in between the record using oracle to this?
Assuming to add on the last 4 character.
RowA
______
ABC_123
DEF_432
WER_677
JKL_342
You would try something like:
Update Table_name set table_column = substr(table_column, 1, 3) || '_' || substr(table_column, 4);
The SUBSTR functions allows you to extract a substring from a string.
The syntax for the SUBSTR function is:
SUBSTR( string, start_position, [ length ] )
string is the source string.
start_position is the position for extraction. The first position in the string is always 1.
length is optional. It is the number of characters to extract. If this parameter is omitted, the SUBSTR function will return the entire string.
Another approach, using regexp_replace() regular expression function:
select regexp_replace(RowA, '^([[:alpha:]]{3})', '\1_') as res
from your_table
Result:
RES
----------
ABC_123
DEF_432
WER_677
JKL_342
SQLFiddle Demo
Related
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 would like to extract following string in Oracle. How can I do that?
Original String: 011113584378(+) CARD, STAFF
Expected String: STAFF CARD
I presume you have the luxury of writing a PL/SQL function? Then just use "SUBSTR", and/or "INSTR", and || concatenation operator to parse your input.
Here is an example:
https://www.techonthenet.com/oracle/questions/parse.php
...The field may contain the following value:
F:\Siebfile\YD\S_SR_ATT_1-60SS_1-AM3L.SAF
In this case, I need to return the value of '1-60SS', as this is the value that resides between the 3rd and 4th underscores.
SOLUTION:
create or replace function parse_value (pValue varchar2)
return varchar2
is
v_pos3 number;
v_pos4 number;
begin
/* Return 3rd occurrence of '_' */
v_pos3 := INSTR (pValue, '_', 1, 3) + 1;
/* Return 4rd occurrence of '_' */
v_pos4 := INSTR (pValue, '_', 1, 4);
return SUBSTR (pValue, v_pos3, v_pos4 - v_pos3);
end parse_value;
Ok, I'll bite. This example uses REGEXP_REPLACE to describe the string, saving the parts you need in order to rearrange them before returning them. It would be better if you showed some real-world examples of the data you are dealing with as I can only guarantee this example will work with the one line you provided.
The regular expression matches any characters starting at the beginning of the string and ending with a close paren-space. The next set of any characters up to but not including the comma-space is "remembered" by enclosing them in parens. This is called a captured group. The next captured group is the set of characters after that comma-space separator until the end of the line (the dollar sign). The captured groups are referred to by their order from left to right. The 3rd argument is the string to return, which is the 2nd and 1st captured groups, in that order, separated by a space.
SQL> with tbl(str) as (
select '+011113584378(+) CARD, STAFF' from dual
)
select regexp_replace(str, '^.*\) (.*), (.*)$', '\2 \1') formatted
from tbl;
FORMATTED
----------
STAFF CARD
SQL>
I have a problem to get a result as example below using oracle. I have a lot of different data in a field A, and need to do a few step (as below) to become a result in field B.
example:
LM2963NAMBLK-P/NOPB/SA and the result: LM2963NAM
remove all characters after '/'
remove character '-P'
remove character 'BLK'
The below should work:
select A,
regexp_replace(
regexp_replace(A, '-P|/.*', ''),
'^(.*)BLK$', '\1'
) B
--remainder of query
The nested regexp_replace will replace -P or / followed by 0 or more occurrences of any characters with the empty string, leaving you with your desired output plus a possible BLK at the end. The | represents "or." I'm not logged in to oracle at the moment, but the - and / shouldn't need to be escaped as / is not a special character in Oracle regex and - is only a special character inside of [ ]. The outside regexp_replace will replace BLK at the end with the empty string if it is there, or just return your string if it is not there.
Well, this is one way...
select substr( substr( substr( 'LM2963NAMBLK-P/NOPB',1, slash_pos-1 ), 1, dash_p_pos-1 ), 1, blk_pos-1 )
from
(
select 'LM2963NAMBLK-P/NOPB'
, instr( 'LM2963NAMBLK-P/NOPB', '/' ) slash_pos
, instr( 'LM2963NAMBLK-P/NOPB', '-P' ) dash_p_pos
, instr( 'LM2963NAMBLK-P/NOPB', 'BLK' ) blk_pos
from dual
);
SUBSTR(SU
---------
LM2963NAM
I would like to get:
82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
from the following expression
LASTNAME_FIRSTNAME_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
Does someone know how I can get this using regexp_substr ?
EDIT
Basically I have a field which has 7 sets each separated by _ . The string I gave is just one example. I wanted to retrieve everything after the second _ . There is no fixed character length so I can not use a substr function. Hence I was using regexp_substr. I was able to get away by using a simplified version
Select FILE_NAME, ( (REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,3)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,4)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,5)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+_',1,6)) ||
(REGEXP_SUBSTR(FILE_NAME,'[^_]+',1,7)) ) as RegExp
from tbl
Here is some more data from the FILE_NAME field
LAST_FIRST_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
SMITH_JOHN_82961_0130BPQX9QZN9G4P5RDTPA9HR4R_MR_1_1of1
LASTNAME_FIRSTNAME_99999_01V0MU4XUQK0Y24Y9RYTFA7W1CM_MR_3_1of1
To get everything after the second underscore, you do not need regular expressions, but can use something like the following:
select substr(FILE_NAME, instr(FILE_NAME, '_', 1, 2) +1 ) from tbl
The instr returns the position of the second occurrence of '_', starting by the first character; the substr simply gets everything starting from the position given by instr + 1
From your Requirement, you can just go ahead and use the simple SUBSTRfunction. Its faster, and it addresses the simple need to remove the String LASTNAME_FIRSTNAME.
select substr('LASTNAME_FIRSTNAME_82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1', 20) data_string
from dual;
Output:
data_string
-----------------
82961_01B04WZXQQSUGJ4YMRRT2A7TRHK_MR_2_1of1
Unless you have another underlying logic you need to address?
Kindly clarify so i can edit the answer accordingly.
i want to trim value of the given string till specified string in oracle pl/sql.
some thing like below.
OyeBuddy$$flex-Flex_Image_Rotator-1443680885520.
In the above string i want to trim till $$ so that i will get "flex-Flex_Image_Rotator-1443680885520".
You can use different ways; here are two methods, with and without regexp:
with test(string) as ( select 'OyeBuddy$$flex-Flex_Image_Rotator-1443680885520.' from dual)
select regexp_replace(string, '(.*)(\$\$)(.*)', '\3')
from test
union all
select substr(string, instr(string, '$$') + length('$$'))
from test
You want to do a SUBSTR where the starting position is going to be the position of '$$' + 2 . +2 is because the string '$$' is of length 2, and we don't want to include that string in the result.
Something like -
SELECT SUBSTR (
'ABCDEF$$some_big_text',
INSTR ('ABCDEF$$some_big_text', '$$') + 2)
FROM DUAL;