Inserting the Comma Separated Strings into Table

Question

Hi Seniors,
i had two string and i want to insert the records in the Table COMMENT . In this way.
would u please give some programe to insert the records.
The Data and the Table
( 901,902,903,904 )
( 'hai','nice','good & mail is [email protected] ','excellent and the phone [phone]' )
comm_id loc_id company_name comments
1      10 901      Hai
2      10 902      nice
3 10      903      good & mail is [email protected]
4      10 904      excellent and the phone [phone]
Thanks
Seenu

Answer 1

Hi, Seenu,
In Oracle 10 (and up) you can easily split a comma-delimited list using REGEXP_SUBSTR.
INSTR and SUBSTR can do the same thing in any version, but it's more complicated.
See the general instructions below:
/*     How to Split a Delimited String
This shows how to take a single row with a delimited string, such as
     Animal     amoeba,bat,cedusa,dodo
and transform it into multiple rows:
     Animal     1     amoeba
     Animal     2     bat
     Animal     3     cedusa
     Animal     4     dodo
PROMPT     ==========  -1. sep_char parameter  ==========
VARIABLE     sep_char     VARCHAR2 (10)
EXECUTE     :sep_char := ',';
SELECT     :sep_char     AS sep_char
FROM     dual;
PROMPT     ==========  0. string_test table  ==========
DROP TABLE     string_test;
CREATE TABLE     string_test
(     grp_name     VARCHAR2 (10)
,     list_txt     VARCHAR2 (50)
INSERT INTO string_test (grp_name, list_txt) VALUES ('Animal',     'amoeba,bat,cedusa,dodo');
INSERT INTO string_test (grp_name, list_txt) VALUES ('Date',     '15-Oct-1582,16-Oct-2008');
INSERT INTO string_test (grp_name, list_txt) VALUES ('Nothing',     NULL);
INSERT INTO string_test (grp_name, list_txt) VALUES ('Place',     'New York');
INSERT INTO string_test (grp_name, list_txt) VALUES ('Skip',     'Hop,,Jump');
SELECT     *
FROM     string_test
ORDER BY     grp_name;
PROMPT     ==========  Q1.  Oracle 11 Query  ==========
WITH     cntr     AS          -- Requires Oracle 9
(     -- Begin sub-query cntr, to generate n (1, 2, 3, ...)
     SELECT     LEVEL     AS n     -- Requires Oracle 9
     FROM     dual
     CONNECT BY     LEVEL     <= 1 +     (
                         SELECT     MAX ( REGEXP_COUNT (list_txt, :sep_char) )     -- Requires Oracle 11
                         FROM     string_test
)     -- End sub-query cntr, to generate n (1, 2, 3, ...)
SELECT     grp_name
,     n
,     REGEXP_SUBSTR     ( list_txt     -- Requires Oracle 10
               , '[^' || :sep_char || ']'     -- Anything except sep_char ...
                    || '+'               -- ... one or more times
               , 1
               , n
               )     AS item_txt
FROM     string_test
JOIN     cntr                                   -- Requires Oracle 9
ON     n     <= 1 + REGEXP_COUNT (list_txt, :sep_char)     -- Requires Oracle 11
ORDER BY     grp_name
,          n;
/*     Notes:
     REGEXP_SUBSTR (s, '[^,]+', 1, n)
returns the n-th item in a comma-delimited list s.
If there are fewer than n items, it returns NULL.
One or more consecutive characters other than comma make an item, so
'Hop,,Jump' has two items, the second one being 'Jump'.
The sub-query cntr produces a list of integers 1, 2, 3, ..., w
where w is the worst-case (the largest number of items in any list).
This actually counts separators, not items, (e.g., it counts both
commas in 'Hop,,Jump', even though), so the w it produces may be
larger than is really necessary.  No real harm is done.
PROMPT     ==========  Q2. Possible Problems Fixed  ==========
WITH     cntr     AS
(     -- Begin sub-query cntr, to generate n (1, 2, 3, ...)
     SELECT     LEVEL     AS n
     FROM     dual
     CONNECT BY     LEVEL     <= 1 +     (
                         SELECT     MAX ( REGEXP_COUNT (list_txt, :sep_char) )
                         FROM     string_test
)     -- End sub-query cntr, to generate n (1, 2, 3, ...)
SELECT     grp_name
,     n
,     REGEXP_SUBSTR     ( list_txt
               , '[^' || :sep_char || ']'     -- Anything except sep_char ...
                    || '+'               -- ... one or more times
               , 1
               , n
               )     AS item_txt
FROM     string_test
JOIN     cntr          ON n     <= 1 + NVL     ( REGEXP_COUNT (list_txt, :sep_char)     -- Problem (1)
                              , 0
WHERE     REGEXP_SUBSTR     ( list_txt     -- Problem (2)
               , '[^' || :sep_char || ']'     -- Anything except sep_char ...
                    || '+'               -- ... one or more times
               , 1
               , n
               )     IS NOT NULL
OR     list_txt     IS NULL          -- Problems (1) and (2) together
ORDER BY     grp_name
,          n;
     (Possible) Problems and Fixes
(1) If list_txt IS NULL, then REGEXP_COUNT (list_txt, :sep_char)
     returns NULL, the join condition fails, and the output
     contains nothing corresponding to the row from string_test.
     If you want a NULL item to appear in the results, use
     NVL to make sure the expression returns 0 instead of NULL.
(2) If list_txt contains multiple consecutive sep_chars (or if it
     begins or ends with sep_char, then the original query
     will return NULL items.  To suppress these, add a WHERE
     clause to test that the item_txt to be displayed IS NOT NULL.
PROMPT     ==========  Q3. Oracle 8.1 Query  ===========
SELECT     grp_name
,     n
,     SUBSTR     ( list_txt
          , begin_pos
          , end_pos - begin_pos
          )     AS item_txt
FROM     (     -- Begin sub-query to compute begin_pos and end_pos
     SELECT     grp_name
     ,     n
     ,     list_txt
     ,     INSTR     ( :sep_char || list_txt
               , :sep_char
               , 1
               , n
               )     AS begin_pos
     ,     INSTR     ( list_txt || :sep_char
               , :sep_char
               , 1
               , n
               )     AS end_pos
     FROM     string_test
     ,     (     -- Begin sub-query cntr, to generate n (1, 2, 3, ...)
          SELECT     ROWNUM     AS n
          FROM     all_objects
          WHERE     ROWNUM     <= 1 +     (
                         SELECT     MAX     ( LENGTH (list_txt)
                                   - LENGTH (REPLACE (list_txt, :sep_char))
                         FROM     string_test
          )     -- End sub-query cntr, to generate n (1, 2, 3, ...)
          cntr
     WHERE     n     <= 1 +     ( LENGTH (list_txt)
                    - LENGTH (REPLACE (list_txt, :sep_char))
     )     -- End sub-query to compute begin_pos and end_pos
ORDER BY     grp_name
,          n;
/*     Version-Dependent Features and Work-Arounds
The code above, Q3, runs in Oracle 8.1.
The following changes were made to Q1:
(11) REGEXP_COUNT was introduced in Oracle 11.
     In earlier versions, to find the number of sep_chars in list_txt,
     see how much the LENGTH changes when sep_chars are removed.
(10) REGEXP_SUBSTR was introduced in Oracle 10.
     In earlier versions, use INSTR to find where the sep_chars are,
     and use SUBSTR to get the sub-strings between them.
     (Using this technique, 'Hop,,Jump' still contains three items,
     but now item 2 IS NULL and item 3 is 'Jump'.)
(9.a) The WITH-clause was introduced in Oracle 9
     In earlier versions, use in-line views.
(9.b) "CONNECT BY LEVEL < constant" doesn't work in Oracle 8.
     Use ROWNUM from any sufficiently large table or view instead.
(9.c) ANSII join notation (JOIN table_name ON ...) was introduced in Oracle 9
     In earlier versions, join condition go in a WHERE-clause.
*/