How to put values starting from the right side into columns? [duplicate] - mysql

Question

I have a table with a single column in a Postgres 13.1 database. It consists of many rows with comma-separated values - around 20 elements at most.
I want to split the data into multiple columns. But I have only a limited number of columns say 5 and more than 5 CSV values in a single row, so excess values must be shifted to new/next row). How to do this?
Example:
a1, b1, c1
a2, b2, c2, d2, e2, f2
a3, b3, c3, d3, e3, f3, g3, h3, i3, j3
a4
a5, b5, c5
'
'
'
Columns are only 5, so the output would be like:
c1 c2 c3 c4 c5
---------------
a1 b1 c1
a2 b2 c2 d2 e2
f2
a3 b3 c3 d3 e3
f3 g3 h3 i3 j3
a4
a5 b5 c5
'
'
'

Answer 1

It is typically bad design to store CSV values in a single column. If at all possible, use an array or a properly normalized design instead.
While stuck with your current situation ...
For known small maximum number of elements
A simple solution without trickery or recursion will do:
SELECT id, 1 AS rnk
, split_part(csv, ', ', 1) AS c1
, split_part(csv, ', ', 2) AS c2
, split_part(csv, ', ', 3) AS c3
, split_part(csv, ', ', 4) AS c4
, split_part(csv, ', ', 5) AS c5
FROM tbl
WHERE split_part(csv, ', ', 1) <> '' -- skip empty rows
UNION ALL
SELECT id, 2
, split_part(csv, ', ', 6)
, split_part(csv, ', ', 7)
, split_part(csv, ', ', 8)
, split_part(csv, ', ', 9)
, split_part(csv, ', ', 10)
FROM tbl
WHERE split_part(csv, ', ', 6) <> '' -- skip empty rows
-- three more blocks to cover a maximum "around 20"
ORDER BY id, rnk;
db<>fiddle here
id being the PK of the original table.
This assumes ', ' as separator, obviously.
You can adapt easily.
Related:
Split comma separated column data into additional columns
For unknown number of elements
Various ways. One way use regexp_replace() to replace every fifth separator before unnesting ...
-- for any number of elements
SELECT t.id, c.rnk
, split_part(c.csv5, ', ', 1) AS c1
, split_part(c.csv5, ', ', 2) AS c2
, split_part(c.csv5, ', ', 3) AS c3
, split_part(c.csv5, ', ', 4) AS c4
, split_part(c.csv5, ', ', 5) AS c5
FROM tbl t
, unnest(string_to_array(regexp_replace(csv, '((?:.*?,){4}.*?),', '\1;', 'g'), '; ')) WITH ORDINALITY c(csv5, rnk)
ORDER BY t.id, c.rnk;
db<>fiddle here
This assumes that the chosen separator ; never appears in your strings. (Just like , can never appear.)
The regular expression pattern is the key: '((?:.*?,){4}.*?),'
(?:) ... “non-capturing” set of parentheses
() ... “capturing” set of parentheses
*? ... non-greedy quantifier
{4}? ... sequence of exactly 4 matches
The replacement '\1;' contains the back-reference \1.
'g' as fourth function parameter is required for repeated replacement.
Further reading:
PostgreSQL & regexp_split_to_array + unnest
Apply `trim()` and `regexp_replace()` on text array
PostgreSQL unnest() with element number
Other ways to solve this include a recursive CTE or a set-returning function ...
Fill from right to left
(Like you added in How to put values starting from the right side into columns?)
Simply count down numbers like:
SELECT t.id, c.rnk
, split_part(c.csv5, ', ', 5) AS c1
, split_part(c.csv5, ', ', 4) AS c2
, split_part(c.csv5, ', ', 3) AS c3
, split_part(c.csv5, ', ', 2) AS c4
, split_part(c.csv5, ', ', 1) AS c5
FROM ...
db<>fiddle here

Answer 2

CREATE UNLOGGED TABLE foo( x TEXT );
\copy foo FROM stdin
a1, b1, c1
a2, b2, c2, d2, e2, f2
a3, b3, c3, d3, e3, f3, g3, h3, i3, j3
a4
a5, b5, c5
\.
From lines to single column...
SELECT (ROW_NUMBER() OVER () - 1)/5 AS r, u FROM (SELECT unnest(string_to_array(x,', ')) u from foo) y;
r | u
---+----
0 | a1
0 | b1
0 | c1
0 | a2
0 | b2
1 | c2
1 | d2
...etc
...and back to lines of known length.
SELECT r,array_agg(u) a FROM (
SELECT (ROW_NUMBER() OVER () - 1)/5 AS r, u FROM (
SELECT unnest(string_to_array(x,', ')) u from foo) y) y1
GROUP BY r ORDER BY r;
r | a
---+------------------
0 | {a1,b1,c1,a2,b2}
1 | {c2,d2,e2,f2,a3}
2 | {b3,c3,d3,e3,f3}
3 | {g3,h3,i3,j3,a4}
4 | {a5,b5,c5}
After this you can insert it into a table using a[] for each column. What to do with the last line is left as an exercise to the reader...

Answer 3

Answer to related question: How to put values starting from the right side into columns?
The accepted great answer from #ErwinBrandstetter can be easily adapted to required right-to-left output.
You just need to change to order of the split parts. So you don't return split parts 1-5 and 6-10 but 5-1 and 10-6:
demo:db<>fiddle
SELECT id, 1 AS rnk
, split_part(csv, ', ', 5) AS c1
, split_part(csv, ', ', 4) AS c2
, split_part(csv, ', ', 3) AS c3
, split_part(csv, ', ', 2) AS c4
, split_part(csv, ', ', 1) AS c5
FROM tbl
WHERE split_part(csv, ', ', 1) <> '' -- skip empty rows
UNION ALL
SELECT id, 2
, split_part(csv, ', ', 10)
, split_part(csv, ', ', 9)
, split_part(csv, ', ', 8)
, split_part(csv, ', ', 7)
, split_part(csv, ', ', 6)
FROM tbl
WHERE split_part(csv, ', ', 6) <> '' -- skip empty rows
-- more?
ORDER BY id, rnk;

Answer 4

You need to do this in the whatever backend layer you are using.
First, convert the CSV rows to array of string
Then, use logic something like this to add Values to the database
int row = 0; // database row index - can be used to just have a count
final int MAX_COLUMNS = 5;
for(int i = 0; i<rows.length; i++) {
// Convert csv row string to array of each value.
String [] values = rows[i].split(",");
// Dividing whole row into chunks of size of number of columns
for(int j = 0; j < (values.length/(MAX_COLUMNS)) + 1; j++) {
Add Values [MAX_COLUMNS*j,MAX_COLUMNS*j+(MAX_COLUMNS - 1)] to the row [row + j]
row++;
}
}