How to remove backslash in redshift?
For example:
1\234
I want only:
1 234
I have tried regexp_replace function but it didn't work.
REPLACE(string1, old_chars, new_chars)
select replace('1\\2345','\\',' ')
http://docs.aws.amazon.com/redshift/latest/dg/r_REPLACE.html
Related
a have a column as below
mystring
AC1853551,AC1854125,AC1855220,AC188115,AC1884120,AC1884390,AC1885102
I need to transformm it to get this output
mystring
('AC1853551','AC1854125','AC1855220','AC188115','AC1884120','AC1884390','AC1885102')
Here is my query that i tried
select CONCAT('( , CONCAT (mystring, ')')) from mytablename
I'm getting an error when it comes to insert a single quote '
Then i thought about replacing the comma with a ','
How to get desired output
i'm using postgres 10
A literal quote is coded as a doubled quote:
select '(''' || replace(mycolumn, ',', ''',''') || ''')'
from mytable
See live demo.
I know that I can do a text search in Postgres with TextSearch and get some result with
select ts_headline('german',content, tq, 'MaxFragments=4, MinWords=5, MaxWords=12,
ShortWord=3, StartSel = <strong>, StopSel = </strong>') as highlight, ...
FROM to_tsquery('german', 'test') tq ...
Is there a similar way to apply to content the same limitations? i.e. to get directly up to 12 words from the column content.
You could use regular expressions:
SELECT (regexp_match(
regexp_replace(content, '[^\w\s]+', ' ', 'g'),
'^\s*((?:\w+\s+){9}\w+)'
))[1] FROM ...
That will first replace everything that is not a space or alphanumerical character with a space and then return the first 10 words.
I am trying to import JSON file where a value contains data like this
"\"Koslov Git †girish\" Ges"
The return should be like "Koslov Git †girish Ges"
The extra \" are the troubles
I tried REGEXP_REPLACE(values, '\\''+', '''', 'g') with no luck. How can I get rid of this. Note the first and last double quotes should not be replaced
with data(json_object) as (
values ('"\"Koslov Git †girish\" Ges"'::jsonb)
)
select regexp_replace(json_object::text, '\\"', '', 'g')
from data;
Db<>fiddle.
I'm using PostgreSQL and I need to truncate a text string, I need to show from an escape character (:) onwards.
I'm trying something like that:
SELECT SUBSTRING ('CATEGORIA DE TRABAJOS: EJECUTIVO' FROM '%#":#"%' FOR '#');
t=# select split_part('CATEGORIA DE TRABAJOS: EJECUTIVO',':',2);
split_part
------------
EJECUTIVO
(1 row)
just split by first found delimiter?..
I am wondering how I can remove all newline characters in Redshift from a field. I tried something like this:
replace(replace(body, '\n', ' '), '\r', ' ')
and
regexp_replace(body, '[\n\r]+', ' ')
But it didn't work. Please share if you know how to do this.
Use chr(10) instead of \n
example:
select replace(CONCAT('Text 1' , chr(10), 'Text 2'), chr(10), '-') as txt
This should help
regexp_replace(column, '\r|\n', '')
To remove line breaks:
SELECT REPLACE('This line has
a line break', CHR(10), '');
This gives output: This line hasa line break. You can see more ASCII or CHR() codes here: https://www.petefreitag.com/cheatsheets/ascii-codes/
To remove special characters like \r, \n, \t
Assuming col1 has text like This line has\r\n special characters.
Using replace()
SELECT REPLACE(REPLACE(col1, '\\r', ''), '\\n', '');
We need to escape \ because backslash is special character in SQL (used to escape double quotes, etc...)
Using regexp_replace()
SELECT REGEXP_REPLACE(col1, '(\\\\r|\\\\n)', '');
We need to escape \ because it is a special character in SQL and we need to escape the resulting backslashes again because backslash is a special character in regex as well.
Both replace() and regexp_replace() gives output: This line has special characters.