to find new lines character in postgres - postgresql

Question

Having couple of entries in database table that have multiple line "names" data.
I try to find single newline character from it.
SELECT
id,
strpos ( NAME, E'\n' ) AS Position_of_substring
FROM
problems
WHERE
strpos ( NAME, E'\n' ) > 0;
But it fails for the data that have more than 1 new line character (\n).
ANy way to find "n" number of "\n" in names data.

Answer 1

regexp_matches will emit a row for each match. doc
SELECT
id,
strpos ( NAME, E'\n' ) AS Position_of_substring
FROM
problems p
WHERE
(select count(*) from regexp_matches(p.name,E'\n','g') ) = ?;

Answer 2

This one gives you a list of all indexes with \n in your string. I am not sure if you were expecting this result:
demo:db<>fiddle
SELECT
name,
array_remove( -- 5
(array_agg(sum))::int[], -- 4
length(name) + 1
)
FROM (
-- 3
SELECT
name,
SUM(length(lines) + 1) OVER (PARTITION BY name ORDER BY row_number)
FROM (
-- 2
SELECT
*,
row_number() OVER ()
FROM (
-- 1
SELECT
name,
regexp_split_to_table(name, '\n') as lines
FROM problems
)s
)s
) s
GROUP BY name
Splitting the string at the \n chars. Every split part is now one row in a temporary table.
Adding a row_count to assure the right order of the split parts
This counts the length of all single split parts. The (length + 1) gives the position of the \n. The SUM window function sums up all values within a group (your original text). That's why the order is relevant. For example: The first two parts of "abc\nde\nfgh" have the lengths of 3 and 2. So the breaks are at 4 (abc = 3, + 1) and 3 (de = 2, + 1). But the 3 of the second part is no real index, but if you sum up these values you get the right indexes: 4 and 7.
Aggregating these results
If (as in my example) the last char is always a \n and you are only interested in the \n chars the string you could remove the last entry of the aggregated array.
Changed problem in comments below:
Would like to replace \n with spaces. So I am thinking how above query
will look in the Update statement. – Pranav Unde
Replacing the \n by spaces is a quiet different problem then getting indexes for all occurances of a special character. And it's much simpler:
UPDATE problems
SET name = trim(regexp_replace(name, E'\n', ' ', 'g'));
regexp_replace(..., 'g') finds all occurances of \n and does the replacing
trim() removes the whitespaces before and after the string if necessary (maybe because there was a trailing \n as in my example - which was replaced by a space as well in the step before)
demo:db<>fiddle