I have a string value in a column in a table like
001|3880000005376|Personal ID| ||15-MAY-2006
and I want to replace the fourth value by another string value 'ABCDEF' , can it be possible by a single update or by PL/SQL program?
Here's one option:
SQL> with test (id, col) as
2 (select 1, '001|3880000005376|Personal ID| ||15-MAY-2006' from dual union all
3 select 2, '002|3880000005376|Personal ID|XXX||15-MAY-2007' from dual
4 )
5 select
6 id,
7 regexp_replace(col, '[^|]+', 'NEW STRING', 1, 4) result
8 from test;
ID RESULT
---------- ------------------------------------------------------------
1 001|3880000005376|Personal ID|NEW STRING||15-MAY-2006
2 002|3880000005376|Personal ID|NEW STRING||15-MAY-2007
SQL>
It replaces 4th occurrence of the '[^|]+' pattern with a NEW STRING value.
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, '\|([^|]*)$', '&\1' ),
'|',
','
) AS replaced
FROM table_name;
Which, for the sample data:
CREATE TABLE table_name (value) AS
SELECT 'A|B|C|D' FROM DUAL UNION ALL
SELECT 'A|B|C' FROM DUAL UNION ALL
SELECT 'A|B' FROM DUAL UNION ALL
SELECT 'A' FROM DUAL;
Outputs:
VALUE
REPLACED
A|B|C|D
A,B,C&D
A|B|C
A,B&C
A|B
A&B
A
A
fiddle
For the above query, I am using regexp_count but in SQL command line I am getting regexp_count :invaild identifier;
select * from table_name WHERE (REGEXP_COUNT(column_name, 'A')) >2;
Is this query works?
I'm not sure what you really want; is it to return names whose 2nd letter is "a", or names that have two or more letters "a" within.
Anyway, pick the one you find appropriate.
SQL> create table test as
2 (select 'saritha' col from dual union all
3 select 'mamatha' from dual union all
4 select 'vaisnavi' from dual union all
5 select 'sai' from dual union all
6 select 'vijaya' from dual union all
7 select 'kumar' from dual
8 );
Table created.
2 or more letters "a":
SQL> select col
2 from test
3 where regexp_count(col, 'a') >= 2;
COL
--------
saritha
mamatha
vaisnavi
vijaya
2nd letter is "a":
SQL> select col
2 from test
3 where substr(col, 2, 1) = 'a';
COL
--------
saritha
mamatha
vaisnavi
sai
SQL>
On Oracle 10g, which doesn't support REGEXP_COUNT function, one option is to replace all letters a with an empty string (basically, you'd remove all letters a) and fetch rows whose difference of full column length and "replaced" column length is >= 2. Something like this:
SQL> select col
2 from test
3 where length(col) - length(replace(col, 'a', '')) >= 2;
COL
--------
saritha
mamatha
vaisnavi
vijaya
SQL>
How can I search for a specific number in an array using REGEXP?
I have an array and need to verify if it has a specific number.
Ex: [5,2,1,4,6,19] and I am looking for number 1, but just the number 1 and not any number that contain the digit 1.
I had to do this:
case when REGEXP_INSTR(JSON_QUERY(MY_JSON_COLUMN,'$.path') , '[[]{1}[1][,]')<>0
or REGEXP_INSTR(JSON_QUERY(MY_JSON_COLUMN,'$.path') , '[,]{1}[1][,]{1}')<>0
or REGEXP_INSTR(JSON_QUERY(MY_JSON_COLUMN,'$.path') , '[,]{1}[1][]]')<>0
or REGEXP_INSTR(JSON_QUERY(MY_JSON_COLUMN,'$.path') , '[[]{1}[1][]]') <>0
then 'DIGIT_ONE' else 'NO_DIGIT_ONE'
end
Is there anything simpler?
You can use
(^|\D)1(\D|$)
This will seach for 1 not enclosed with other digits.
See this regex demo.
Details
(^|\D) - start of string or non-digit
1 - a 1 char
(\D|$) - non-digit or end of string.
Do NOT use regular expressions, use a proper JSON parser and then filter for the number you want:
SELECT my_json_column,
CASE
WHEN JSON_EXISTS( my_json_column, '$?(#.path[*] == 1)' )
THEN 'DIGIT ONE'
ELSE 'NO DIGIT ONE'
END AS has_one
FROM table_name;
or (if you are using Oracle 12.1 and cannot use path filter expressions with JSON_EXISTS, which is only available from Oracle 12.2):
SELECT my_json_column,
CASE
WHEN EXISTS(
SELECT 'X'
FROM JSON_TABLE(
t.my_json_column,
'$.path[*]'
COLUMNS (
value NUMBER PATH '$'
)
)
WHERE value = 1
)
THEN 'DIGIT ONE'
ELSE 'NO DIGIT ONE'
END
FROM table_name t;
Which, for the sample data:
CREATE TABLE table_name (
my_json_column CHECK ( my_json_column IS JSON )
) AS
SELECT '{"path":[5,2,1,4,6,19],"not_this_path":[1,2,3,4,5]}' FROM DUAL UNION ALL
SELECT '{"path":[5,2,4,6,19],"not_this_path":[1,2,3,4,5]}' FROM DUAL UNION ALL
SELECT '{"path":[11],"not_this_path":[1]}' FROM DUAL UNION ALL
SELECT '{"path":[2],"not_this_path":[1]}' FROM DUAL UNION ALL
SELECT '{"path":[1,11]}' FROM DUAL;
Both output:
MY_JSON_COLUMN | HAS_ONE
:-------------------------------------------------- | :-----------
{"path":[5,2,1,4,6,19],"not_this_path":[1,2,3,4,5]} | DIGIT ONE
{"path":[5,2,4,6,19],"not_this_path":[1,2,3,4,5]} | NO DIGIT ONE
{"path":[11],"not_this_path":[1]} | NO DIGIT ONE
{"path":[2],"not_this_path":[1]} | NO DIGIT ONE
{"path":[1,11]} | DIGIT ONE
db<>fiddle here
Alternatively, with a little bit more typing (a little bit? Am I kidding?!), splitting the string into rows and comparing values to the search string:
SQL> with test (col) as
2 (select '[5,2,1,4,6,19]' from dual)
3 select t.col,
4 case when '&par_search_string' in
5 (select regexp_substr(substr(col, 2, length(col) - 1), '[^,]+', 1, level) val
6 from test
7 connect by level <= regexp_count(col, ',') + 1
8 )
9 then 'Search string exists'
10 else 'Search string does not exist'
11 end result
12 from test t;
Enter value for par_search_string: 1
COL RESULT
-------------- ----------------------------
[5,2,1,4,6,19] Search string exists
SQL> /
Enter value for par_search_string: 24
COL RESULT
-------------- ----------------------------
[5,2,1,4,6,19] Search string does not exist
SQL>
I have two columns with a hashtag delimited value, i.e. Email#Web#Telephone#SMS#MMS & 0#0#0#1#0 Note that each delimited value of the second column matches up with its corresponding delimited value in the first column, i.e. Email = 0, Web = 0, Telephone = 0, SMS = 1 etc.
Based on a parameter, I want to return the matching value of the second column. i.e. incoming param = Web#Telephone#SMS, thus the value that I want to return is 0#0#1.
This need to be done in PL SQL, and I have no clue where to start, which explains the lack of sample code.
Any help please?
There are a couple of very useful utility functions in an Oracle package called APEX_UTIL. (This package concerns Oracle Application Express aka APEX, but can be used anywhere). They are:
apex_util.string_to_table
apex_util.table_to_string
Using string_to_table you can convert the delimited string into a table of values:
declare
v_table apex_application_global.vc_arr2; -- This is the table type apex_util uses
begin
v_table := apex_util.table_to_string ('Email#Web#Telephone#SMS#MMS', '#');
end;
You now have an array with 5 elements ('Email', 'Web', 'Telephone', 'SMS', 'MMS');
You can do the same with the values string to get a table with elements ('0', '0', '0', '1', 0'). And you can do the same with the parameter to get a table with elements ('Web', 'Telephone', 'SMS').
You can then use PL/SQL logic to build a new array with elements for the values you need to return, i.e. ('0', '0', '1'). I have left this part to you!
Finally you can turn that back into a delimited string:
return apex_util.table_to_string (v_return_table, '#');
Firstly, you should normalize the table and have the attributes in different columns rather than delimited strings in a single column.
Anyway, you could do it in many ways using the techniques to Split comma delimited strings in a table
For example, using REGEXP_SUBSTR and CONNECT BY clause:
SQL> WITH DATA(attr, val) AS(
2 SELECT 'Email#Web#Telephone#SMS#MMS', '0#0#0#1#0' FROM dual
3 )
4 SELECT lines.COLUMN_VALUE,
5 trim(regexp_substr(t.attr, '[^#]+', 1, lines.COLUMN_VALUE)) attr,
6 trim(regexp_substr(t.val, '[^#]+', 1, lines.COLUMN_VALUE)) val
7 FROM data t,
8 TABLE (CAST (MULTISET
9 (SELECT LEVEL FROM dual CONNECT BY LEVEL <= regexp_count(t.attr, '#')+1
10 ) AS sys.odciNumberList ) ) lines
11 /
COLUMN_VALUE ATTR VAL
------------ --------------------------- ---------
1 Email 0
2 Web 0
3 Telephone 0
4 SMS 1
5 MMS 0
SQL>
Now, you can get the respective values for each attribute.
You could put the entire logic in a FUNCTION and return the corresponding values of each attribute and call the function in SELECT statement.
For example,
SQL> CREATE OR REPLACE
2 FUNCTION get_val_from_attr(
3 attr_name VARCHAR2)
4 RETURN NUMBER
5 IS
6 var_val NUMBER;
7 BEGIN
8 WITH DATA(attr, val) AS
9 ( SELECT 'Email#Web#Telephone#SMS#MMS', '0#0#0#1#0' FROM dual
10 ),
11 t2 AS
12 (SELECT lines.COLUMN_VALUE,
13 trim(regexp_substr(t.attr, '[^#]+', 1, lines.COLUMN_VALUE)) attr,
14 trim(regexp_substr(t.val, '[^#]+', 1, lines.COLUMN_VALUE)) val
15 FROM data t,
16 TABLE (CAST (MULTISET
17 (SELECT LEVEL FROM dual CONNECT BY LEVEL <= regexp_count(t.attr, '#')+1
18 ) AS sys.odciNumberList ) ) lines
19 )
20 SELECT val INTO var_val FROM t2 WHERE attr = attr_name;
21 RETURN var_val;
22 END;
23 /
Function created.
Let's call the function:
SQL> SELECT get_val_from_attr('Email') FROM dual;
GET_VAL_FROM_ATTR('EMAIL')
--------------------------
0
SQL> SELECT get_val_from_attr('SMS') FROM dual;
GET_VAL_FROM_ATTR('SMS')
------------------------
1
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)