Related
I have a string in Oracle DB in the format 'a|b|c' , '|' being the separator between characters. Want to write an SQL query to transform it into a string in the format 'a,b&c'. First occurrence of '|' to ',' 2nd occurrence to '&'.
If suppose the string comes in the format 'a|b' then output should be 'a&b'.
I'm using multiple regex_replace queries to achieve this right now.
select REGEXP_REPLACE ('a|b|c', '[|]', ',', 1, 1)
from dual
Is there any other solution using one single query?
Nested replaces (see line #6):
SQL> with test (col) as
2 (select 'a|b|c' from dual union all
3 select 'a|b' from dual
4 )
5 select col,
6 regexp_replace(regexp_replace(col, '\|', ',', 1, 1), '\|', '&', 2, 1) result
7 from test;
COL RESULT
----- --------------------
a|b|c a,b&c
a|b a,b
SQL>
If you want to replace the last | in the list with & and all the preceding |s with , then you can use:
SELECT value,
REPLACE(
REGEXP_REPLACE( value, '\|([^|]*)$', '&\1' ),
'|',
','
) AS replaced
FROM table_name;
Which, for the sample data:
CREATE TABLE table_name (value) AS
SELECT 'A|B|C|D' FROM DUAL UNION ALL
SELECT 'A|B|C' FROM DUAL UNION ALL
SELECT 'A|B' FROM DUAL UNION ALL
SELECT 'A' FROM DUAL;
Outputs:
VALUE
REPLACED
A|B|C|D
A,B,C&D
A|B|C
A,B&C
A|B
A&B
A
A
fiddle
I have column in table looking like this:
PATTERN
{([option1]+[option2])*([option3]+[option4])}
{([option1]+[option2])*([option3]+[option4])*([option6]+[option7])}
{[option1]+[option6]}
{([option1]+[option2])*([option8]+[option9])}
{([option1]+[option2])*[option4]}
{[option10]}
Every option has a number of value.
There is a table - let's call it option_set and records look like
OPTION VALUE
option1 3653265
option2 26452
option3 73552
option3 100
option4 1235
option5 42565
option6 2330
option7 544
option9 2150
I want to replace option name to number in 1st table, if exists of course, if not exists then =0.
I have done this in PLSQL (get the pattern, go through every option, and if exists - regexp_replace),
but I am wondering if this could be done in SQL??
My goal is to replace values for all patterns for current OPTION_SET and get only records, where all equations would be greater than 0. Of course - I couldn't run this equation in SQL, so I think of something like
for rec in
(
SELECT...
)
loop
execute immediate '...';
if above_equation > 0 then ..
end loop;
Any ideas would be appreciated
You can do a loop-like query in SQL with the recursive CTE, replacing new token on each iteration, so this will let you to replace all the tokens.
The only way I know to execute a dynamic query inside SQL statement in Oracle is DBMS_XMLGEN package, so you can evaluate the expression and filter by the result value without PL/SQL. But all this is viable for low cardinality tables with patterns and options.
Here's the code:
with a as (
select 1 as id, '{([option1]+[option2])*([option3]+[option4])}' as pattern from dual union all
select 2 as id, '{([option1]+[option2])*([option3]+[option4])*([option6]+[option7])}' as pattern from dual union all
select 3 as id, '{[option1]+[option6]}' as pattern from dual union all
select 4 as id, '{([option1]+[option2])*([option8]+[option9])}' as pattern from dual union all
select 5 as id, '{([option1]+[option2])*[option4]}' as pattern from dual union all
select 6 as id, '{[option10]}]' as pattern from dual
)
, opt as (
select 'option1' as opt, 3653265 as val from dual union all
select 'option2' as opt, 26452 as val from dual union all
select 'option3' as opt, 73552 as val from dual union all
select 'option3' as opt, 100 as val from dual union all
select 'option4' as opt, 1235 as val from dual union all
select 'option5' as opt, 42565 as val from dual union all
select 'option6' as opt, 2330 as val from dual union all
select 'option7' as opt, 544 as val from dual union all
select 'option9' as opt, 2150 as val from dual
)
, opt_ordered as (
/*Order options to iterate over*/
select opt.*, row_number() over(order by 1) as rn
from opt
)
, rec (id, pattern, repl_pattern, lvl) as (
select
id,
pattern,
pattern as repl_pattern,
0 as lvl
from a
union all
select
r.id,
r.pattern,
/*Replace each part at new step*/
replace(r.repl_pattern, '[' || o.opt || ']', o.val),
r.lvl + 1
from rec r
join opt_ordered o
on r.lvl + 1 = o.rn
)
, out_prepared as (
select
rec.*,
case
when instr(repl_pattern, '[') = 0
/*When there's no more not parsed expressions, then we can try to evaluate them*/
then dbms_xmlgen.getxmltype(
'select ' || replace(replace(repl_pattern, '{', ''), '}', '')
|| ' as v from dual'
)
/*Otherwise SQL statement will fail*/
end as parsed_expr
from rec
/*Retrieve the last step*/
where lvl = (select max(rn) from opt_ordered)
)
select
id,
pattern,
repl_pattern,
extractvalue(parsed_expr, '/ROWSET/ROW/V') as calculated_value
from out_prepared o
where extractvalue(parsed_expr, '/ROWSET/ROW/V') > 0
ID | PATTERN | REPL_PATTERN | CALCULATED_VALUE
-: | :------------------------------------------------------------------ | :---------------------------------------- | :---------------
1 | {([option1]+[option2])*([option3]+[option4])} | {(3653265+26452)*(73552+1235)} | 275194995279
2 | {([option1]+[option2])*([option3]+[option4])*([option6]+[option7])} | {(3653265+26452)*(73552+1235)*(2330+544)} | 790910416431846
3 | {[option1]+[option6]} | {3653265+2330} | 3655595
5 | {([option1]+[option2])*[option4]} | {(3653265+26452)*1235} | 4544450495
db<>fiddle here
Here is one way to do this. There's a lot to unpack, so hang on tight.
I include the test data in the with clause. Of course, you won't need that; simply remove the two "tables" and use your actual table and column names in the query.
From Oracle 12.1 on, we can define PL/SQL functions directly in the with clause, right at the top; if we do so, the query must be terminated with a slash (/) instead of the usual semicolon (;). If your version is earlier than 12.1, you can define the function separately. The function I use takes an "arithmetic expression" (a string representing a compound arithmetic operation) and returns its value as a number. It uses native dynamic SQL (the "execute immediate" statement), which will cause the query to be relatively slow, as a different cursor is parsed for each row. If speed becomes an issue, this can be changed, to use a bind variable (so that the cursor is parsed only once).
The recursive query in the with clause replaces each placeholder with the corresponding value for the "options" table. I use 0 either if a "placeholder" doesn't have a corresponding option in the table, or if it does but the corresponding value is null. (Note that your sample data shows option3 twice; that makes no sense, and I removed one occurrence from my sample data.)
Instead of replacing one placeholder at a time, I took the opposite approach; assuming the patterns may be long, but the number of "options" is small, this should be more efficient. Namely: at each step, I replace ALL occurrences of '[optionN]' (for a given N) in a single pass. Outside the recursive query, I replace all the placeholders for "non-existent" options with 0.
Note that recursive with clause requires Oracle 11.2. If your version is even earlier than that (although it shouldn't be), there are other ways; you would likely need to do that in PL/SQL also.
So, here it is - a single SELECT query for the whole thing:
with
function expr_eval(pattern varchar2) return number as
x number;
begin
execute immediate 'select ' || pattern || ' from dual' into x;
return x;
end;
p (id, pattern) as (
select 1, '{([option1]+[option2])*([option3]+[option4])}' from dual union all
select 2, '{([option1]+[option2])*([option3]+[option4])*([option6]+[option7])}' from dual union all
select 3, '{[option1]+[option6]}' from dual union all
select 4, '{([option1]+[option2])*([option8]+[option9])}' from dual union all
select 5, '{([option1]+[option2])*[option4]}' from dual union all
select 6, '{[option10]}' from dual union all
select 7, '{[option2]/([option3]+[option8])-(300-[option2])/(0.1 *[option3])}' from dual
)
, o (opt, val) as (
select 'option1', 3653265 from dual union all
select 'option2', 26452 from dual union all
select 'option3', 100 from dual union all
select 'option4', 1235 from dual union all
select 'option5', 42565 from dual union all
select 'option6', 2330 from dual union all
select 'option7', 544 from dual union all
select 'option9', 2150 from dual
)
, n (opt, val, rn, ct) as (
select opt, val, rownum, count(*) over ()
from o
)
, r (id, pattern, rn, ct) as (
select id, substr(pattern, 2, length(pattern) - 2), 1, null
from p
union all
select r.id, replace(r.pattern, '[' || n.opt || ']', nvl(to_char(n.val), 0)),
r.rn + 1, n.ct
from r join n on r.rn = n.rn
)
, ae (id, pattern) as (
select id, regexp_replace(pattern, '\[[^]]*]', '0')
from r
where rn = ct + 1
)
select id, expr_eval(pattern) as result
from ae
order by id
/
Output:
ID RESULT
---- ---------------
1 4912422195
2 14118301388430
3 3655595
4 7911391550
5 4544450495
6 0
7 2879.72
I am new to oracle and to this forum. I have searched and found answers on how to do this with a column of just numbers but this has txt at the beginning then a sequenced number.
I have a table that has a varchar2 column named myid which has characters with a number at the end which is in order the number at the end is always 6 digits with leading zeros.
Hello_002190
Hello_002188
Bye_000187
Bye_000185
Bye_000184
Get_008133
Get_008131
Gone_001112
Gone_001110
Gone_001109
I need an Oracle SQL script that will show me all the missing rows.
The result for the above should be:
Hello_002189
Bye_000186
Get_008132
Gone_001111
Thanks in advance for the help
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE table_name ( value ) AS
SELECT 'Hello_002190' FROM DUAL UNION ALL
SELECT 'Hello_002188' FROM DUAL UNION ALL
SELECT 'Bye_000187' FROM DUAL UNION ALL
SELECT 'Bye_000185' FROM DUAL UNION ALL
SELECT 'Bye_000184' FROM DUAL UNION ALL
SELECT 'Get_008133' FROM DUAL UNION ALL
SELECT 'Get_008131' FROM DUAL UNION ALL
SELECT 'Gone_001112' FROM DUAL UNION ALL
SELECT 'Gone_001110' FROM DUAL UNION ALL
SELECT 'Gone_001109' FROM DUAL;
Query 1:
WITH data ( prefix, suffix ) AS (
SELECT SUBSTR( value, 1, INSTR( value, '_' ) ),
TO_NUMBER( SUBSTR( value, INSTR( value, '_' ) + 1 ) )
FROM table_name
),
bounds ( prefix, min_suffix, max_suffix ) AS (
SELECT prefix, MIN( suffix ), MAX( suffix )
FROM data
GROUP BY prefix
)
SELECT prefix || TO_CHAR( column_value, 'FM000000' ) AS missing_value
FROM bounds b
CROSS JOIN
TABLE(
CAST(
MULTISET(
SELECT b.min_suffix + LEVEL - 1
FROM DUAL
CONNECT BY b.min_suffix + LEVEL - 1 <= b.max_suffix
) AS SYS.ODCINUMBERLIST
)
)
MINUS
SELECT value FROM table_name
Results:
| MISSING_VALUE |
|---------------|
| Bye_000186 |
| Get_008132 |
| Gone_001111 |
| Hello_002189 |
I need your help with the regexp_replace function. I have a table which has a column for concatenated string values which contain duplicates. How do I eliminate them?
Example:
Ian,Beatty,Larry,Neesha,Beatty,Neesha,Ian,Neesha
I need the output to be
Ian,Beatty,Larry,Neesha
The duplicates are random and not in any particular order.
Update--
Here's how my table looks
ID Name1 Name2 Name3
1 a b c
1 c d a
2 d e a
2 c d b
I need one row per ID having distinct name1,name2,name3 in one row as a comma separated string.
ID Name
1 a,c,b,d,c
2 d,c,e,a,b
I have tried using listagg with distinct but I'm not able to remove the duplicates.
The easiest option I would go with -
SELECT ID, LISTAGG(NAME_LIST, ',')
FROM (SELECT ID, NAME1 NAME_LIST FROM DATA UNION
SELECT ID, NAME2 FROM DATA UNION
SELECT ID, NAME3 FROM DATA
)
GROUP BY ID;
Demo.
So, try this out...
([^,]+),(?=.*[A-Za-z],[] ]*\1)
I don't think you can do it just with regexp_replace if the repeated values are not next to each other. One approach is to split the values up, eliminate the duplicates, and then put them back together.
The common method to tokenize a delimited string is with regexp_substr and a connect by clause. Using a bind variable with your string to make the code a bit clearer:
var value varchar2(100);
exec :value := 'Ian,Beatty,Larry,Neesha,Beatty,Neesha,Ian,Neesha';
select regexp_substr(:value, '[^,]+', 1, level) as value
from dual
connect by regexp_substr(:value, '[^,]+', 1, level) is not null;
VALUE
------------------------------
Ian
Beatty
Larry
Neesha
Beatty
Neesha
Ian
Neesha
You can use that as a subquery (or CTE), get the distinct values from it, then reassemble it with listagg:
select listagg(value, ',') within group (order by value) as value
from (
select distinct value from (
select regexp_substr(:value, '[^,]+', 1, level) as value
from dual
connect by regexp_substr(:value, '[^,]+', 1, level) is not null
)
);
VALUE
------------------------------
Beatty,Ian,Larry,Neesha
It's a bit more complicated if you're looking at multiple rows in a table as that confused the connect-by syntax, but you can use a non-determinisitic reference to avoid loops:
with t42 (id, value) as (
select 1, 'Ian,Beatty,Larry,Neesha,Beatty,Neesha,Ian,Neesha' from dual
union all select 2, 'Mary,Joe,Mary,Frank,Joe' from dual
)
select id, listagg(value, ',') within group (order by value) as value
from (
select distinct id, value from (
select id, regexp_substr(value, '[^,]+', 1, level) as value
from t42
connect by regexp_substr(value, '[^,]+', 1, level) is not null
and id = prior id
and prior dbms_random.value is not null
)
)
group by id;
ID VALUE
---------- ------------------------------
1 Beatty,Ian,Larry,Neesha
2 Frank,Joe,Mary
Of course this wouldn't be necessary if you were storing relational data properly; having a delimited string in a column is not a good idea.
There is a way to find duplicates in this case, but it is a problem to remove them if there are more than one duplicated name within a string per id. Here is code that can deal with one duplicate per id.
Sample data:
WITH
tbl AS
(
Select 1 "ID", 'a' "NAME_1", 'b' "NAME_2", 'c' "NAME_3" From Dual Union All
Select 1 "ID", 'c' "NAME_1", 'd' "NAME_2", 'a' "NAME_3" From Dual Union All
Select 2 "ID", 'd' "NAME_1", 'e' "NAME_2", 'a' "NAME_3" From Dual Union All
Select 2 "ID", 'c' "NAME_1", 'd' "NAME_2", 'b' "NAME_3" From Dual
),
lists AS
(
Select 1 "ID", 'a,c,b,d,c' "NAME" From Dual Union All
Select 2 "ID", 'd,c,e,a,b' "NAME" From Dual
),
Creating CTE that compares your LISTAGG sttring with original data finding duplicate values:
grid AS
(
Select DISTINCT l.ID, l.NAME,
CASE WHEN ( Length(l.NAME || ',') - Length(Replace(l.NAME || ',', t.NAME_1 || ',', '')) ) / Length(t.NAME_1 || ',') > 1 THEN NAME_1 END "NAME_1",
CASE WHEN ( Length(l.NAME || ',') - Length(Replace(l.NAME || ',', t.NAME_2 || ',', '')) ) / Length(t.NAME_2 || ',') > 1 THEN NAME_2 END "NAME_2",
CASE WHEN ( Length(l.NAME || ',') - Length(Replace(l.NAME || ',', t.NAME_3 || ',', '')) ) / Length(t.NAME_3 || ',') > 1 THEN NAME_3 END "NAME_3"
From
lists l
Inner Join
tbl t ON(t.ID = l.ID)
)
ID NAME NAME_1 NAME_2 NAME_3
---------- --------- ------ ------ ------
2 d,c,e,a,b
1 a,c,b,d,c c
1 a,c,b,d,c c
Main SQL, using Union, builds new string (removing second appearance) where the duplicate was found and then puts that new string after comparison with the old one.
SELECT DISTINCT l.ID, Nvl(g.NAME, l.NAME) NAME
FROM
lists l
LEFT JOIN
(
SELECT ID, CASE WHEN NAME_1 Is Not Null
THEN REPLACE(NAME, NAME, COALESCE( REPLACE( SubStr(NAME, 1, InStr(NAME, NAME_1, 1, 2) - 1) || SubStr(NAME, InStr(NAME, NAME_1, 1, 2) + Length(NAME_1)), ',,', ','), NULL ) )
END "NAME"
FROM grid
WHERE COALESCE(NAME_1, NAME_2, NAME_3) IS NOT NULL
UNION ALL
SELECT ID, CASE WHEN NAME_2 Is Not Null
THEN REPLACE(NAME, NAME, COALESCE( REPLACE( SubStr(NAME, 1, InStr(NAME, NAME_2, 1, 2) - 1) || SubStr(NAME, InStr(NAME, NAME_2, 1, 2) + Length(NAME_2)), ',,', ','), NULL ) )
END "NAME"
FROM grid
WHERE COALESCE(NAME_1, NAME_2, NAME_3) IS NOT NULL
UNION ALL
SELECT ID, CASE WHEN NAME_3 Is Not Null
THEN REPLACE(NAME, NAME, COALESCE( REPLACE( SubStr(NAME, 1, InStr(NAME, NAME_3, 1, 2) - 1) || SubStr(NAME, InStr(NAME, NAME_3, 1, 2) + Length(NAME_3)), ',,', ','), NULL ) )
END "NAME"
FROM grid
WHERE COALESCE(NAME_1, NAME_2, NAME_3) IS NOT NULL
) g ON(g.ID = l.ID And Length(g.NAME) < Length(l.NAME))
R e s u l t :
ID NAME
---------- -------------
2 d,c,e,a,b
1 a,c,b,d
For multiple occurences within a string or for multiplicated different names there should be done some recursions or multiplied nestings to get it done...
I have data like this in a table
NAME PRICE
A 2
B 3
C 5
D 9
E 5
I want to display all the values in one row; for instance:
A,2|B,3|C,5|D,9|E,5|
How would I go about making a query that will give me a string like this in Oracle? I don't need it to be programmed into something; I just want a way to get that line to appear in the results so I can copy it over and paste it in a word document.
My Oracle version is 10.2.0.5.
-- Oracle 10g --
SELECT deptno, WM_CONCAT(ename) AS employees
FROM scott.emp
GROUP BY deptno;
Output:
10 CLARK,MILLER,KING
20 SMITH,FORD,ADAMS,SCOTT,JONES
30 ALLEN,JAMES,TURNER,BLAKE,MARTIN,WARD
I know this is a little late but try this:
SELECT LISTAGG(CONCAT(CONCAT(NAME,','),PRICE),'|') WITHIN GROUP (ORDER BY NAME) AS CONCATDATA
FROM your_table
Usually when I need something like that quickly and I want to stay on SQL without using PL/SQL, I use something similar to the hack below:
select sys_connect_by_path(col, ', ') as concat
from
(
select 'E' as col, 1 as seq from dual
union
select 'F', 2 from dual
union
select 'G', 3 from dual
)
where seq = 3
start with seq = 1
connect by prior seq+1 = seq
It's a hierarchical query which uses the "sys_connect_by_path" special function, which is designed to get the "path" from a parent to a child.
What we are doing is simulating that the record with seq=1 is the parent of the record with seq=2 and so fourth, and then getting the full path of the last child (in this case, record with seq = 3), which will effectively be a concatenation of all the "col" columns
Adapted to your case:
select sys_connect_by_path(to_clob(col), '|') as concat
from
(
select name || ',' || price as col, rownum as seq, max(rownum) over (partition by 1) as max_seq
from
(
/* Simulating your table */
select 'A' as name, 2 as price from dual
union
select 'B' as name, 3 as price from dual
union
select 'C' as name, 5 as price from dual
union
select 'D' as name, 9 as price from dual
union
select 'E' as name, 5 as price from dual
)
)
where seq = max_seq
start with seq = 1
connect by prior seq+1 = seq
Result is: |A,2|B,3|C,5|D,9|E,5
As you're in Oracle 10g you can't use the excellent listagg(). However, there are numerous other string aggregation techniques.
There's no particular need for all the complicated stuff. Assuming the following table
create table a ( NAME varchar2(1), PRICE number);
insert all
into a values ('A', 2)
into a values ('B', 3)
into a values ('C', 5)
into a values ('D', 9)
into a values ('E', 5)
select * from dual
The unsupported function wm_concat should be sufficient:
select replace(replace(wm_concat (name || '#' || price), ',', '|'), '#', ',')
from a;
REPLACE(REPLACE(WM_CONCAT(NAME||'#'||PRICE),',','|'),'#',',')
--------------------------------------------------------------------------------
A,2|B,3|C,5|D,9|E,5
But, you could also alter Tom Kyte's stragg, also in the above link, to do it without the replace functions.
Here is another approach, using model clause:
-- sample of data from your question
with t1(NAME1, PRICE) as(
select 'A', 2 from dual union all
select 'B', 3 from dual union all
select 'C', 5 from dual union all
select 'D', 9 from dual union all
select 'E', 5 from dual
) -- the query
select Res
from (select name1
, price
, rn
, res
from t1
model
dimension by (row_number() over(order by name1) rn)
measures (name1, price, cast(null as varchar2(101)) as res)
(res[rn] order by rn desc = name1[cv()] || ',' || price[cv()] || '|' || res[cv() + 1])
)
where rn = 1
Result:
RES
----------------------
A,2|B,3|C,5|D,9|E,5|
SQLFiddle Example
Something like the following, which is grossly inefficient and untested.
create function foo returning varchar2 as
(
declare bar varchar2(8000) --arbitrary number
CURSOR cur IS
SELECT name,price
from my_table
LOOP
FETCH cur INTO r;
EXIT WHEN cur%NOTFOUND;
bar:= r.name|| ',' ||r.price || '|'
END LOOP;
dbms_output.put_line(bar);
return bar
)
Managed to get till here using xmlagg: using oracle 11G from sql fiddle.
Data Table:
COL1 COL2 COL3
1 0 0
1 1 1
2 0 0
3 0 0
3 1 0
SELECT
RTRIM(REPLACE(REPLACE(
XMLAgg(XMLElement("x", col1,',', col2, col3)
ORDER BY col1), '<x>'), '</x>', '|')) AS COLS
FROM ab
;
Results:
COLS
1,00| 3,00| 2,00| 1,11| 3,10|
* SQLFIDDLE DEMO
Reference to read on XMLAGG