How to skip comma from csv using double quotes

Question

I am getting a string in the below format after reading data from a csv file

v_lastline = '29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX';

I just want to convert it into an array while will contain 6 values, the comma before the , Inc needs to be escaped. Can any one please suggest whats the best way to do it in PL/SQL?

Answer 1

This is similar to this question, but you have empty elements in your list; and a simple translation of one of the patterns I tried there skips those:

var v_lastline varchar2(50);
exec :v_lastline := '29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX';

select level as lvl,
  regexp_substr(:v_lastline, '("[^"]*"|[^,]+)', 1, level) as element
from dual
connect by level <= regexp_count(:v_lastline, '("[^"]*"|[^,]+)');

       LVL ELEMENT                                
---------- ----------------------------------------
         1 29218368                                
         2 8062115                                 
         3 " Benedict Canyon Equities, Inc"        
         4 CLS                                     
         5 FAX                                     

If you can identify a special character that will never appear in the data then you can work around that by putting that into the empty elements by changing every comma to comma+character, and then removing it after the split:

select level as lvl,
  replace(regexp_substr(replace(:v_lastline, ',', ',§'),
    '(§"[^"]*"|[^,]+)', 1, level), '§', null) as element
from dual
connect by regexp_substr(replace(:v_lastline, ',', ',§'),
  '(§"[^"]*"|[^,]+)', 1, level) is not null;

       LVL ELEMENT                                
---------- ----------------------------------------
         1 29218368                                
         2 8062115                                 
         3 " Benedict Canyon Equities, Inc"        
         4 CLS                                     
         5                                         
         6 FAX                                     

It's an extension of a common method to split delimited strings, which is explained in detail here.

• replace(:v_lastline, ',', ',§') changes ...,CLS,,FAX to ...,§CLS,§,§FAX, where § is a character you'll never see.
• regexp_substr(..., '(§"[^"]*"|[^,]+)', 1, level) tokenises the updated value with a regex that looks for any double-quote-enclosed value (now preceded by the special character too) or a non-comma; the order of the evaluation means commas inside the quoted part are ignored.
• the level is part of the hierarchical query syntax, where:
• connect by regexp_substr(<same value and pattern>) is not null just figured out how many tokens there are.
• and finally replace(regexp_substr(...), , '§', null) removes the special character used in the first step.

You can then remove the double-quotes too with a further level of replace(), and trim whitespace, if you want/need to.

You have't said quite what you mean by an array, but you can run that query in PL/SQL and bulk-collect into a collection if that's what you intend to work with. For example, using the built-in ODCIVARCHAR2LIST collection type:

set serveroutput on
declare
  v_lastline varchar2(50);
  v_array sys.odcivarchar2list;
begin
  v_lastline := '29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX';

  select trim(replace(replace(
    regexp_substr(replace(:v_lastline, ',', ',§'),
      '(§"[^"]*"|[^,]+)', 1, level), '§', null), '"', null))
  bulk collect into v_array
  from dual
  connect by regexp_substr(replace(:v_lastline, ',', ',§'),
    '(§"[^"]*"|[^,]+)', 1, level) is not null;

  dbms_output.put_line('Number of elements: ' || v_array.count);
  for i in 1..v_array.count loop
    dbms_output.put_line('Index ' || i || ' has: ' || v_array(i));
  end loop;
end;
/

Number of elements: 6
Index 1 has: 29218368
Index 2 has: 8062115
Index 3 has: Benedict Canyon Equities, Inc
Index 4 has: CLS
Index 5 has: 
Index 6 has: FAX

---

With multiple empty elements this also (now) works:

exec :v_lastline := '29218368,8062115," Benedict Canyon Equities, Inc",,,,,,,CLS,,,,,FAX,,,,,,,,,,,,,,,,,,INVOICE';
select level as lvl,
  replace(regexp_substr(replace(:v_lastline, ',', ',§'),
    '(§"[^"]*"|[^,]+)', 1, level), '§', null) as element
from dual
connect by regexp_substr(replace(:v_lastline, ',', ',§'),
  '(§"[^"]*"|[^,]+)', 1, level) is not null;

       LVL ELEMENT                                
---------- ----------------------------------------
         1 29218368                                
         2 8062115                                 
         3 " Benedict Canyon Equities, Inc"        
         4                                         
...
         9                                         
        10 CLS                                     
        11                                         
...
        14                                         
        15 FAX                                     
        16                                         
...
        32                                         
        33 INVOICE                                 

Answer 2

Here is a solution without regular expressions, first of all create two helper functions

/* CAR   select car('hello,world,bla') from dual --> hello */
create or replace function car(PI_STR in varchar2,
                               PI_SEPARATOR in varchar2 default ',')
  return varchar2 is l_pos number;
begin
  l_pos := instr(PI_STR, PI_SEPARATOR);

  if l_pos > 0 then
    return substr(PI_STR, 1, l_pos - 1);
  end if;

  return PI_STR;
end;

/* CDR select cdr('hello,world,bla') from dual --> world,bla */
create or replace function cdr(PI_STR in varchar2,
                               PI_SEPARATOR in varchar2 default ',')
  return varchar2 is l_pos number;
begin
  l_pos := instr(PI_STR, PI_SEPARATOR);

  if l_pos > 0 then
    return substr(PI_STR, l_pos + length(PI_SEPARATOR));
  end if;

  return '';
end;

now: extract by ',' and for each result concat with next entry if escape character is found upto next escape character:

create or replace type csv_col is table of varchar2(4000);

create or replace function get_columns(PI_STR in varchar2,
                                       PI_SEPARATOR in varchar2,
                                       PI_ESC_CHAR in varchar2)
  return csv_col pipelined is l_car varchar2(4000);
l_cdr varchar2(4000);
l_car_esc varchar2(4000);
begin
  l_car := car(PI_STR, PI_SEPARATOR);
  l_cdr := cdr(PI_STR, PI_SEPARATOR);
  -- check for escape char
  l_car_esc := cdr(l_car, PI_ESC_CHAR);
  if l_car_esc is not null then
    l_car := l_car_esc || PI_SEPARATOR || car(l_cdr, PI_ESC_CHAR);
    l_cdr := cdr(cdr(l_cdr, PI_ESC_CHAR), PI_SEPARATOR);
  end if;
  loop
    if l_car is null and l_cdr is null then
      exit;
    end if;

    pipe row(l_car);
    l_car     := car(l_cdr, PI_SEPARATOR);
    l_cdr     := cdr(l_cdr, PI_SEPARATOR);
    l_car_esc := cdr(l_car, PI_ESC_CHAR);
    if l_car_esc is not null then
      l_car := l_car_esc || PI_SEPARATOR || car(l_cdr, PI_ESC_CHAR);
      l_cdr := cdr(cdr(l_cdr, PI_ESC_CHAR), PI_SEPARATOR);
      dbms_output.put_line(l_car);
      dbms_output.put_line(l_cdr);
    end if;
  end loop;
end;

call it like this:

select *
  from table(get_columns('29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX',
                         ',',
                         '"'));

--> result

29218368
8062115
 Benedict Canyon Equities, Inc
CLS

FAX

Answer 3

If the structure of your CSV if fixed, you can try with something like this:

with text(text) as ( select '29218368,8062115," Benedict Canyon Equities, Inc",CLS,,FAX' from dual)
select level,
       trim(',' from 
                       case
                        when level in (1,2) then
                            regexp_substr(text, '(.*??)\,', 1, level)
                        when level = 3 then
                            regexp_substr(text, '"(.*??)"', 1, 1)
                        when level in (4,5) then
                            regexp_substr(text, '(.*??)\,', instr(text, '"', 1, 2), level -2) 
                        when level = 6 then       
                            regexp_substr(text, '\,([^\,]*)', instr(text, '"', 1, 2), 3)
                       end
            )
from text
connect by level <= 6

This makes strong assumptions on the structure of CSV, by treating each part in a different way, but it seems to me difficult to find a really generic solution to the problem.