array function returning empty

Question

The aim here is to adapt this answer to return array instead of setof datatype.

CREATE FUNCTION split_csvline(
  line text,                 -- the input CSV string
  delim_char char(1) = ',',  -- can be also E'\t', ';', '|', etc.
  quote_char char(1) = '"'   -- field quotation
) RETURNS  text[] AS $f$
  import csv
  row = csv.reader(
      [line], 
      quotechar=quote_char, 
      delimiter=delim_char, 
      skipinitialspace=True, 
      escapechar='\\'
  )
  next(row)
$f$ IMMUTABLE language PLpythonU;

SELECT split_csvline('a,b');  -- empty!

EDIT

Notes

It is a question about "using Python with PostgreSQL".

I am using PLpythonU because the staff use Python and because CSV is complex and need reliable (years of test) algorithm.

Not need a workaround, because a simple workaround is in use:

CREATE FUNCTION split_csv_line(
  text, char(1) DEFAULT ',', char(1) DEFAULT '"'
) RETURNS text[] AS $f$
  SELECT x FROM split_csv($1,$2,$3) x LIMIT 1;
$f$ language SQL IMMUTABLE;

@BenH, CSV is complex, see this regex, perhaps using regexp_matches() ... — Peter Krauss
– Peter Krauss, Commented Mar 21, 2017 at 11:08
i know that csv is complex, but then you should consider a far better sample than 'a,b' — Ben H
– Ben H, Commented Mar 21, 2017 at 11:09
You didn't use a return statement. IMHO almost all procedural language requires that if you want to return something. — pozs
– pozs, Commented Mar 21, 2017 at 11:46

Clodoaldo Neto · Accepted Answer · 2017-03-21 14:15:24Z

2

The csv.reader function returns a reader object. next is a reader object method:

create or replace function split_csvline(
    _line text,
    _delim_char char(1) = ',',  
    _quote_char char(1) = '"'
) returns  text[] as $f$
    import csv
    reader = csv.reader(
        [_line], 
        quotechar = _quote_char, 
        delimiter = _delim_char, 
        skipinitialspace = True, 
        escapechar = '\\'
    )
    return reader.next()
$f$ immutable language plpythonu;

select split_csvline('a,b');
 split_csvline 
---------------
 {a,b}

edited Mar 21, 2017 at 14:15

answered Mar 21, 2017 at 14:08

Clodoaldo Neto

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Laurenz Albe · Accepted Answer · 2017-03-21 11:48:28Z

2

You could use PL/pgSQL:

CREATE OR REPLACE FUNCTION split_csvline(
   line text,                 -- the input CSV string
   delim_char char(1) = ',',  -- can be also E'\t', ';', '|', etc.
   quote_char char(1) = '"',  -- field quotation
   OUT result text[]
) LANGUAGE plpgsql AS
$$DECLARE
   i integer;
   t text := '';
   c char(1);
   /*
    * 0 means unquoted
    * 1 means quoted
    * 2 means quoted, and we just read a quote
    */
   q integer := 0;
BEGIN
   /* loop through the characters */
   FOR i IN 1..length(line) LOOP
      /* get i-th character */
      c := substring(line FROM i FOR 1);
      /* end of string is at an unquoted delimiter */
      IF c = delim_char AND q <> 1 THEN
         result := result || t;
         t := '';
         q := 0;
      ELSIF c = quote_char THEN
         CASE q
            WHEN 0 THEN
               q := 1;
            WHEN 1 THEN
               IF c = quote_char THEN
                  q := 2;
               ELSE
                  t := t || c;
               END IF;
            WHEN 2 THEN
               q := 1;
               t := t || quote_char;
         END CASE;
      ELSE
         IF q = 2 THEN
            q := 0;
         END IF;
         t := t || c;
      END IF;
   END LOOP;
   /* add the last string */
   result := result || t;
END;$$;

answered Mar 21, 2017 at 11:48

Laurenz Albe

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2 Comments

Peter Krauss Over a year ago

Thanks @LaurenzAble! Hum... This algorithm will work with any CSV case? even simple algorith as this perl one use regular expression... I was trying to use regexp_matches() but need to adapt ?:^|,)(?=[^"]|(")?)"?((?(1)[^"]*|[^,"]*))"?(?=,|$) to POSIX...

Laurenz Albe Over a year ago

It should work for everything, unless I made a programming error. Sometimes it is simpler to write a state machine than to specify the grammar.

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