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Can I use some function call on REGEXP_REPLACE back-reference value?
For example I want to call chr() or any other function on back-reference value, but this
SELECT REGEXP_REPLACE('a 98 c 100', '(\d+)', ASCII('\1')) FROM dual;
just returns ASCII value of '\':
'a 92 c 92'
I want that the last parameter (replacement string) to be evaluated first and then to replace string. So result would be:
'a b c d'
Just for fun really, you could do the tokenization, conversion of numbers to characters, and aggregation using XPath:
select *
from xmltable(
'string-join(
for $t in tokenize($s, " ")
return if ($t castable as xs:integer) then codepoints-to-string(xs:integer($t)) else $t,
" ")'
passing 'a 98 c 100' as "s"
);
Result Sequence
--------------------------------------------------------------------------------
a b c d
The initial string value is passed in as $s; tokenize() splits that up using a space as the delimiter; each $t that generates is evaluated to see if it's an integer, and if it is then it's converted to the equivalent character via codepoints-to-string, otherwise it's left alone; then all the tokens are recombined with string-join().
If the original has runs of multiple spaces those will collapse to a single space (as they will with Littlefoot's regex).
I'm not that smart to do it using one regular expression, but - step-by-step, something like this might help. It splits the source string into rows, checks whether part of it is a number and - if so - selects CHR of it. Finally, everything is aggregated back to a single string.
SQL> with test (col) as
2 (select 'a 98 c 100' from dual),
3 inter as
4 (select level lvl,
5 regexp_substr(col, '[^ ]+', 1, level) c_val
6 from test
7 connect by level <= regexp_count(col, ' ') + 1
8 ),
9 inter_2 as
10 (select lvl,
11 case when regexp_like(c_val, '^\d+$') then chr(c_val)
12 else c_val
13 end c_val_2
14 from inter
15 )
16 select listagg(c_val_2, ' ') within group (order by lvl) result
17 from inter_2;
RESULT
--------------------
a b c d
SQL>
It can be shortened for one step (I intentionally left it as is so that you could execute one query at a time and check the result, to make things clearer):
SQL> with test (col) as
2 (select 'a 98 c 100' from dual),
3 inter as
4 (select level lvl,
5 case when regexp_like(regexp_substr(col, '[^ ]+', 1, level), '^\d+$')
6 then chr(regexp_substr(col, '[^ ]+', 1, level))
7 else regexp_substr(col, '[^ ]+', 1, level)
8 end c_val
9 from test
10 connect by level <= regexp_count(col, ' ') + 1
11 )
12 select listagg(c_val, ' ') within group (order by lvl) result
13 from inter;
RESULT
--------------------
a b c d
SQL>
[EDIT: what if input looks differently?]
That is somewhat simpler. Using REGEXP_SUBSTR, extract digits: ..., 1, 1 returns the first one, ... 1, 2 the second one. Pure REPLACE then replaces numbers with their CHR values.
SQL> with test (col) as
2 (select 'a98c100e' from dual)
3 select
4 replace(replace(col, regexp_substr(col, '\d+', 1, 1), chr(regexp_substr(col, '\d+', 1, 1))),
5 regexp_substr(col, '\d+', 1, 2), chr(regexp_substr(col, '\d+', 1, 2))) result
6 from test;
RESULT
--------------------
abcde
SQL>
I have a column "col1" value like : 'a,b,x,y,z' (ordered string)
Another column "col2" is like : 'a,x' or 'b,y,z' (ordered string)
All the string values in "col2" are generated by a subquery. So it is not constant.
But, the value in "col1" is constant. That is col1='a,b,x,y,z'
create table test (col1 varchar2(20), col2 varchar2(10));
insert into test values ('a,b,x,y,z','a,x');
insert into test values ('a,b,x,y,z','b,y,z');
Need help with the replacing in one sql.
Need help to replace the elements on "col1" with "col2".
For example,
when col2='a,x', the result should be : 'b,y,z'
when col2='b,y,z', the result should be : 'a,x'
Here is a fun way to do this:
select col1, col2,
ltrim(regexp_replace(translate(replace(col1,','),','||replace(col2,','),',')
,'(.)',',\1'),',') as col3
from test;
That is: (reading the function calls as they are executed, from inside out)
Remove the commas from both strings
Use TRANSLATE() to remove the characters of the second string from the first string
Use REGEXP_REPLACE to add commas before each character of the remaining string
Trim the leading comma
Here's one option; I included the ID column to make it simpler. I hope that a column similar to it exists in your real case.
The idea is:
split each column (col1, col2) to rows
CTE one represents col1's rows
CTE two represents col2's rows
using the MINUS set operator, subtract those two sets of rows
using LISTAGG, aggregate the result
SQL> select * From test;
ID COL1 COL2
---------- -------------------- ----------
1 a,b,x,y,z a,x
2 a,b,x,y,z b,y,z
SQL> with
2 one as
3 (select id, regexp_substr(col1, '[^,]+', 1, column_value) col
4 from test,
5 table(cast(multiset(select level from dual
6 connect by level <= regexp_count(col1, ',') + 1
7 ) as sys.odcinumberlist))
8 ),
9 two as
10 (select id, regexp_substr(col2, '[^,]+', 1, column_value) col
11 from test,
12 table(cast(multiset(select level from dual
13 connect by level <= regexp_count(col2, ',') + 1
14 ) as sys.odcinumberlist))
15 ),
16 t_minus as
17 (select id, col from one
18 minus
19 select id, col from two
20 )
21 select id, listagg(col, ',') within group (order by col) result
22 From t_minus
23 group by id;
ID RESULT
---------- --------------------
1 b,y,z
2 a,x
SQL>
I have a Table - Product In Oracle, wherein p_spc_cat_id is stored as comma separated values.
p_id p_name p_desc p_spc_cat_id
1 AA AAAA 26,119,27,15,18
2 BB BBBB 0,0,27,56,57,4
3 BB CCCC 26,0,0,15,3,8
4 CC DDDD 26,0,27,7,14,10
5 CC EEEE 26,119,0,48,75
Now I want to search p_name which have p_spc_cat_id in '26,119,7' And this search value are not fixed it will some time '7,27,8'. The search text combination change every time
my query is:
select p_id,p_name from product where p_spc_cat_id in('26,119,7');
when i execute this query that time i can't find any result
I am little late in answering however i hope that i understood the question correctly.
Read further if: you have a table storing records like
1. 10,20,30,40
2. 50,40,20,70
3. 80,60,30,40
And a search string like '10,60', in which cases it should return rows 1 & 3.
Please try below, it worked for my small table & data.
create table Temp_Table_Name (some_id number(6), Ab varchar2(100))
insert into Temp_Table_Name values (1,'112,120')
insert into Temp_Table_Name values (2,'7,8,100,26')
Firstly lets breakdown the logic:
The table contains comma separated data in one of the columns[Column AB].
We have a comma separated string which we need to search individually in that string column. ['26,119,7,18'-X_STRING]
ID column is primary key in the table.
1.) Lets multiple each record in the table x times where x is the count of comma separated values in the search string [X_STRING]. We can use below query to create the cartesian join sub-query table.
Select Rownum Sequencer,'26,119,7,18' X_STRING
from dual
CONNECT BY ROWNUM <= (LENGTH( '26,119,7,18') - LENGTH(REPLACE( '26,119,7,18',',',''))) + 1
Small note: Calculating count of comma separated values =
Length of string - length of string without ',' + 1 [add one for last value]
2.) Create a function PARSING_STRING such that PARSING_STRING(string,position). So If i pass:
PARSING_STRING('26,119,7,18',3) it should return 7.
CREATE OR REPLACE Function PARSING_STRING
(String_Inside IN Varchar2, Position_No IN Number)
Return Varchar2 Is
OurEnd Number; Beginn Number;
Begin
If Position_No < 1 Then
Return Null;
End If;
OurEnd := Instr(String_Inside, ',', 1, Position_No);
If OurEnd = 0 Then
OurEnd := Length(String_Inside) + 1;
End If;
If Position_No = 1 Then
Beginn := 1;
Else
Beginn := Instr(String_Inside, ',', 1, Position_No-1) + 1;
End If;
Return Substr(String_Inside, Beginn, OurEnd-Beginn);
End;
/
3.) Main query, with the join to multiply records.:
select t1.*,PARSING_STRING(X_STRING,Sequencer)
from Temp_Table_Name t1,
(Select Rownum Sequencer,'26,119,7,18' X_STRING from dual
CONNECT BY ROWNUM <= (Select (LENGTH( '26,119,7,18') - LENGTH(REPLACE(
'26,119,7,18',',',''))) + 1 from dual)) t2
Please note that with each multiplied record we are getting 1 particular position value from the comma separated string.
4.) Finalizing the where condition:
Where
/* For when the value is in the middle of the strint [,value,] */
AB like '%,'||PARSING_STRING(X_STRING,Sequencer)||',%'
OR
/* For when the value is in the start of the string [value,]
parsing the first position comma separated value to match*/
PARSING_STRING(AB,1) = PARSING_STRING(X_STRING,Sequencer)
OR
/* For when the value is in the end of the string [,value]
parsing the last position comma separated value to match*/
PARSING_STRING(AB,(LENGTH(AB) - LENGTH(REPLACE(AB,',',''))) + 1) =
PARSING_STRING(X_STRING,Sequencer)
5.) Using distinct in the query to get unique ID's
[Final Query:Combination of all logic stated above: 1 Query to find them all]
select distinct Some_ID
from Temp_Table_Name t1,
(Select Rownum Sequencer,'26,119,7,18' X_STRING from dual
CONNECT BY ROWNUM <= (Select (LENGTH( '26,119,7,18') - LENGTH(REPLACE( '26,119,7,18',',',''))) + 1 from dual)) t2
Where
AB like '%,'||PARSING_STRING(X_STRING,Sequencer)||',%'
OR
PARSING_STRING(AB,1) = PARSING_STRING(X_STRING,Sequencer)
OR
PARSING_STRING(AB,(LENGTH(AB) - LENGTH(REPLACE(AB,',',''))) + 1) = PARSING_STRING(X_STRING,Sequencer)
You can use like to find it:
select p_id,p_name from product where p_spc_cat_id like '%26,119%'
or p_spc_cat_id like '%119,26%' or p_spc_cat_id like '%119,%,26%' or p_spc_cat_id like '%26,%,119%';
Use the Oracle function instr() to achieve what you want. In your case that would be:
SELECT p_name
FROM product
WHERE instr(p_spc_cat_id, '26,119') <> 0;
Oracle Doc for INSTR
If the string which you are searching will always have 3 values (i.e. 2 commas present) then you can use below approach.
where p_spc_cat_id like regexp_substr('your_search_string, '[^,]+', 1, 1)
or p_spc_cat_id like regexp_substr('your_search_string', '[^,]+', 1, 2)
or p_spc_cat_id like regexp_substr('your_search_string', '[^,]+', 1, 3)
If you cant predict how many values will be there in your search string
(rather how many commas) in that case you may need to generate dynamic query.
Unfortunately sql fiddle is not working currently so could not test this code.
SELECT p_id,p_name
FROM product
WHERE p_spc_cat_id
LIKE '%'||'&i_str'||'%'`
where i_str is 26,119,7 or 7,27,8
This solution uses CTE's. "product" builds the main table. "product_split" turns products into rows so each element in p_spc_cat_id is in it's own row. Lastly, product_split is searched for each value in the string '26,119,7' which is turned into rows by the connect by.
with product(p_id, p_name, p_desc, p_spc_cat_id) as (
select 1, 'AA', 'AAAA', '26,119,27,15,18' from dual union all
select 2, 'BB', 'BBBB', '0,0,27,56,57,4' from dual union all
select 3, 'BB', 'CCCC', '26,0,0,15,3,8' from dual union all
select 4, 'CC', 'DDDD', '26,0,27,7,14,10' from dual union all
select 5, 'CC', 'EEEE', '26,119,0,48,75' from dual
),
product_split(p_id, p_name, p_spc_cat_id) as (
select p_id, p_name,
regexp_substr(p_spc_cat_id, '(.*?)(,|$)', 1, level, NULL, 1)
from product
connect by level <= regexp_count(p_spc_cat_id, ',')+1
and prior p_id = p_id
and prior sys_guid() is not null
)
-- select * from product_split;
select distinct p_id, p_name
from product_split
where p_spc_cat_id in(
select regexp_substr('26,119,7', '(.*?)(,|$)', 1, level, NULL, 1) from dual
connect by level <= regexp_count('26,119,7', ',') + 1
)
order by p_id;
P_ID P_
---------- --
1 AA
3 BB
4 CC
5 CC
SQL>
I have problem with my REGEXP expression which I want to loop and every iteration deletes text after slash. My expression looks like this now
REGEXP_SUBSTR('L1161148/1/10', '.*(/)')
I'm getting L1161148/1/ instead of L1161148/1
You said you wanted to loop.
CAVEAT: Both of these solutions assume there are no NULL list elements (all slashes have a value in between them).
SQL> with tbl(data) as (
select 'L1161148/1/10' from dual
)
select level, nvl(substr(data, 1, instr(data, '/', 1, level)-1), data) formatted
from tbl
connect by level <= regexp_count(data, '/') + 1 -- Loop # of delimiters +1 times
order by level desc;
LEVEL FORMATTED
---------- -------------
3 L1161148/1/10
2 L1161148/1
1 L1161148
SQL>
EDIT: To handle multiple rows:
SQL> with tbl(rownbr, col1) as (
select 1, 'L1161148/1/10/2/34/5/6' from dual
union
select 2, 'ALKDFJV1161148/123/456/789/1/2/3' from dual
)
SELECT rownbr, column_value substring_nbr,
nvl(substr(col1, 1, instr(col1, '/', 1, column_value)-1), col1) formatted
FROM tbl,
TABLE(
CAST(
MULTISET(SELECT LEVEL
FROM dual
CONNECT BY LEVEL <= REGEXP_COUNT(col1, '/')+1
) AS sys.OdciNumberList
)
)
order by rownbr, substring_nbr desc
;
ROWNBR SUBSTRING_NBR FORMATTED
---------- ------------- --------------------------------
1 7 L1161148/1/10/2/34/5/6
1 6 L1161148/1/10/2/34/5
1 5 L1161148/1/10/2/34
1 4 L1161148/1/10/2
1 3 L1161148/1/10
1 2 L1161148/1
1 1 L1161148
2 7 ALKDFJV1161148/123/456/789/1/2/3
2 6 ALKDFJV1161148/123/456/789/1/2
2 5 ALKDFJV1161148/123/456/789/1
2 4 ALKDFJV1161148/123/456/789
2 3 ALKDFJV1161148/123/456
2 2 ALKDFJV1161148/123
2 1 ALKDFJV1161148
14 rows selected.
SQL>
You can try removing the string after the last slash:
select regexp_replace('L1161148/1/10', '/([^/]*)$', '') from dual
You are trying to go as far as the last / and then "look back" and retain what was before it. With regular expressions you can do that with a subexpression, like this:
select regexp_substr('L1161148/1/10', '(.*)/.*', 1, 1, null, 1) from dual;
Here, as usual, the first argument "1" means where to start the search, the second "1" means which matching substring to choose, "null" means no special matching modifiers (like case-insensitive matching and such - not needed here), and the last "1" means return the first subexpression - the first thing in parentheses in the "match pattern."
However, regular expressions should only be used when you can't do it with the standard substr and instr (and translate) functions. Here the job is quite easy:
instr(text_string, '/', -1)
will give you the position of the LAST / in text_string (the -1 means find the last occurrence, instead of the first: count from the end of the string). So the whole thing can be written as:
select substr('L1161148/1/10', 1, instr('L1161148/1/10', '/', -1) - 1) from dual;
Edit: In the spirit of Gary_W's solution, here is a generalization to several strings and stripping successive layers from each input string; still not using regular expressions (resulting in slightly faster performance) and using a recursive CTE, available since Oracle version 11; I believe Gary's solution works only from Oracle 12c on.
Query: (I changed Gary's second input string a bit, to make sure the query works properly)
with tbl(item_id, input_str) as (
select 1, 'L1161148/1/10/2/34/5/6' from dual union all
select 2, 'ALKD/FJV11/61148/123/456/789/1/2/3' from dual
),
r (item_id, proc_string, stage) as (
select item_id, input_str, 0 from tbl
union all
select item_id, substr(proc_string, 1, instr(proc_string, '/', -1) - 1), stage + 1
from r
where instr(proc_string, '/') > 0
)
select * from r
order by item_id, stage;
Output:
ITEM_ID PROC_STRING STAGE
---------- ---------------------------------------- ----------
1 L1161148/1/10/2/34/5/6 0
1 L1161148/1/10/2/34/5 1
1 L1161148/1/10/2/34 2
1 L1161148/1/10/2 3
1 L1161148/1/10 4
1 L1161148/1 5
1 L1161148 6
2 ALKD/FJV11/61148/123/456/789/1/2/3 0
2 ALKD/FJV11/61148/123/456/789/1/2 1
2 ALKD/FJV11/61148/123/456/789/1 2
2 ALKD/FJV11/61148/123/456/789 3
2 ALKD/FJV11/61148/123/456 4
2 ALKD/FJV11/61148/123 5
2 ALKD/FJV11/61148 6
2 ALKD/FJV11 7
2 ALKD 8
For these strings
RSLR_AIRL19_ID3454_T20030913091226
RSLR_AIRL19_ID3122454_T20030913091226
RSLR_AIRL19_ID34_T20030913091226
How to get the number after ID ?
Or how to get the content between two characters but not include them ?
I use this '/\_ID([^_]+)/' got matches like Array ( [0] => _ID3454 [1] => 3454 )
Is this the right way?
To extract a number after an ID, you could write a similar query.
SQL> with t1 as(
2 select 'RSLR_AIRL19_ID3454_T20030913091226' as col from dual union all
3 select 'RSLR_AIRL19_ID3122454_T20030913091226' from dual union all
4 select 'RSLR_AIRL19_ID34_T20030913091226' from dual
5 )
6 select regexp_substr(col, '^([[:alnum:]]+_){2}ID([[:digit:]]+)_([[:alnum:]]+){1}$', 1, 1, 'i', 2) as ID
7 from t1
8 ;
ID
-------------
3454
3122454
34
Or, if you want to extract digits from a first occurrence of the pattern without verifying if an entire string matches a specific format:
SQL> with t1 as(
2 select 'RSLR_AI_RL19_ID3454_T20030913091226' as col from dual union all
3 select 'RSLR_AIRL19_ID3122454_T20030913091226' from dual union all
4 select 'RSLR_AIRL19_ID34_T20030913091226' from dual
5 )
6 select regexp_substr(col, 'ID([[:digit:]]+)', 1, 1, 'i', 1) as ID
7 from t1
8 ;
ID
--------------
3454
3122454
34
With pcre & perl engines :
ID\K\w+
NOTE
\K "restart" the match.
See http://www.phpfreaks.com/blog/pcre-regex-spotlight-k (php use pcre)