Currently looking into way to convert number like:
699 937,57
Into
699937.57
I'm looking into something like
SELECT to_number(column, 'FM99G999G999') from mytable;
But the last example will drop decimal numbers
It's probably easier to simply replace the comma with a dot and then convert that result to a number:
select to_number(replace(the_column, ',', '.'), 'FM999999.99')
from mytable
This however requires you to know the maximum number of digits before the decimal point. Another option would be to remove all whitespace from the string (after replacing the comma) and then cast that to a number:
select regexp_replace(replace(the_column, ',', '.'), '\s+', '', 'g')::numeric
from mytable;
Regular expressions are somewhat expensive, to the second solution is probably slower (but more robust) then the first one.
The following:
with mytable (the_column) as (
values ('699 937,57'), ('123,45'), ('456,789'), ('123 456 789,1234')
)
select regexp_replace(replace(the_column, ',', '.'), '\s+', '', 'g')::numeric
from mytable;
returns:
regexp_replace
--------------
699937.57
123.45
456.789
123456789.1234
Related
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.
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') ) = ?;
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
I have two columns, COL1 and COL2. COL1 has value like 'Birds sitting on $1 and enjoying' and COL2 has value like 'the.location_value[/tree,\building]'
I need to update third column COL3 with values like 'Birds sitting on /tree and enjoying'
i.e. $1 in 1st column is replaced with /tree
which is the 1st word from list of comma separated words with in square brackets [] in COL2 i.e. [/tree,\building]
I wanted to know the best suitable combination of string function in postgresql to use to achieve this.
You need to first extract the first element from the comma separated list, to do that, you can use split_part() but you first need to extract the actual list of values. This can be done using substring() with a regular expression:
substring(col2 from '\[(.*)\]')
will return /tree,\building
So the complete query would be:
select replace(col1, '$1', split_part(substring(col2 from '\[(.*)\]'), ',', 1))
from the_table;
Online example: http://rextester.com/CMFZMP1728
This one should work with any (int) number after $:
select t.*, c.col3
from t,
lateral (select string_agg(case
when o = 1 then s
else (string_to_array((select regexp_matches(t.col2, '\[(.*)\]'))[1], ','))[(select regexp_matches(s, '^\$(\d+)'))[1]::int] || substring(s from '^\$\d+(.*)')
end, '' order by o) col3
from regexp_split_to_table(t.col1, '(?=\$\d+)') with ordinality s(s, o)) c
http://rextester.com/OKZAG54145
Note:it is not the most efficient though. It splits col2's values (in the square brackets) each time for replacing $N.
Update: LATERAL and WITH ORDINALITY is not supported in older versions, but you could try a correlating subquery instead:
select t.*, (select array_to_string(array_agg(case
when s ~ E'^\\$(\\d+)'
then (string_to_array((select regexp_matches(t.col2, E'\\[(.*)\\]'))[1], ','))[(select regexp_matches(s, E'^\\$(\\d+)'))[1]::int] || substring(s from E'^\\$\\d+(.*)')
else s
end), '') col3
from regexp_split_to_table(t.col1, E'(?=\\$\\d+)') s) col3
from t
I want to convert a column of type "character varying" that has integers with commas to a regular integer column.
I want to support numbers from '1' to '10,000,000'.
I've tried to use: to_number(fieldname, '999G999G999'), but it only works if the format matches the exact length of the string.
Is there a way to do this that supports from '1' to '10,000,000'?
select replace(fieldname,',','')::numeric ;
To do it the way you originally attempted, which is not advised:
select to_number( fieldname,
regexp_replace( replace(fieldname,',','G') , '[0-9]' ,'9','g')
);
The inner replace changes commas to G. The outer replace changes numbers to 9. This does not factor in decimal or negative numbers.
You can just strip out the commas with the REPLACE() function:
CREATE TABLE Foo
(
Test NUMERIC
);
insert into Foo VALUES (REPLACE('1,234,567', ',', '')::numeric);
select * from Foo; -- Will show 1234567
You can replace the commas by an empty string as suggested, or you could use to_number with the FM prefix, so the query would look like this:
SELECT to_number(my_column, 'FM99G999G999')
There are things to take note:
When using function REPLACE("fieldName", ',', '') on a table, if there are VIEW using the TABLE, that function will not work properly. You must drop the view to use it.
In short, I am looking for a single recursive query that can perform multiple replaces over one string. I have a notion it can be done, but am failing to wrap my head around it.
Granted, I'd prefer the biz-layer of the application, or even the CLR, to do the replacing, but these are not options in this case.
More specifically, I want to replace the below mess - which is C&P in 8 different stored procedures - with a TVF.
SET #temp = REPLACE(RTRIM(#target), '~', '-')
SET #temp = REPLACE(#temp, '''', '-')
SET #temp = REPLACE(#temp, '!', '-')
SET #temp = REPLACE(#temp, '#', '-')
SET #temp = REPLACE(#temp, '#', '-')
-- 23 additional lines reducted
SET #target = #temp
Here is where I've started:
-- I have a split string TVF called tvf_SplitString that takes a string
-- and a splitter, and returns a table with one row for each element.
-- EDIT: tvf_SplitString returns a two-column table: pos, element, of which
-- pos is simply the row_number of the element.
SELECT REPLACE('A~B!C#D#C!B~A', MM.ELEMENT, '-') TGT
FROM dbo.tvf_SplitString('~-''-!-#-#', '-') MM
Notice I've joined all the offending characters into a single string separated by '-' (knowing that '-' will never be one of the offending characters), which is then split. The result from this query looks like:
TGT
------------
A-B!C#D#C!B-A
A~B!C#D#C!B~A
A~B-C#D#C-B~A
A~B!C-D-C!B~A
A~B!C#D#C!B~A
So, the replace clearly works, but now I want it to be recursive so I can pull the top 1 and eventually come out with:
TGT
------------
A-B-C-D-C-B-A
Any ideas on how to accomplish this with one query?
EDIT: Well, actual recursion isn't necessary if there's another way. I'm pondering the use of a table of numbers here, too.
You can use this in a scalar function. I use it to remove all control characters from some external input.
SELECT #target = REPLACE(#target, invalidChar, '-')
FROM (VALUES ('~'),(''''),('!'),('#'),('#')) AS T(invalidChar)
I figured it out. I failed to mention that the tvf_SplitString function returns a row number as "pos" (although a subquery assigning row_number could also have worked). With that fact, I could control cross join between the recursive call and the split.
-- the cast to varchar(max) matches the output of the TVF, otherwise error.
-- The iteration counter is joined to the row number value from the split string
-- function to ensure each iteration only replaces on one character.
WITH XX AS (SELECT CAST('A~B!C#D#C!B~A' AS VARCHAR(MAX)) TGT, 1 RN
UNION ALL
SELECT REPLACE(XX.TGT, MM.ELEMENT, '-'), RN + 1 RN
FROM XX, dbo.tvf_SplitString('~-''-!-#-#', '-') MM
WHERE XX.RN = MM.pos)
SELECT TOP 1 XX.TGT
FROM XX
ORDER BY RN DESC
Still, I'm open to other suggestions.
I want to search a column and get values where value containts \ .
I tried select * from "Values" where "ValueName" like '\'. But returns no value.
Also tried like "\" and like'\''%' etc. But no results.
See the DB2 Documentation on the LIKE predicate, in particular the parts about escape expressions.
What you want is
select * from Values where ValueName like '\\%' escape '\'
To give an example of usage:
create table backslash_escape_test
(
backslash_escape_test_column varchar(20)
);
insert into backslash_escape_test(backslash_escape_test_column)
values ('foo\');
insert into backslash_escape_test(backslash_escape_test_column)
values ('no slashes here');
insert into backslash_escape_test(backslash_escape_test_column)
values ('foo\bar');
insert into backslash_escape_test(backslash_escape_test_column)
values ('\bar');
select count(*) from backslash_escape_test where
backslash_escape_test_column like '%\\%' escape '\';
returns 3 (all 3 rows with \ in them).
select count(*) from backslash_escape_test where
backslash_escape_test_column like '\\%' escape '\';
returns 1 (the \bar row).
select * from Values where ValueName like '%\\%'
values is a not so good name because it may be confused with the values keyword
Don't escape it. You just need wildcards around it like this:
select count(*)
from escape_test
where test_column like '%\%'
But, suppose you really do need to escape the slash. Here's a simpler, more straightforward answer:
The escape-expression allows you to specify whatever character for escaping that you wish. So why use a character that you're looking for, thus requiring you to escape it? Use any other character instead. I'll use a plus sign as an example, but it could be a backslash, pound-sign, question-mark, anything other than a character you are looking for or one of the wildcard characters (% or _).
select count(*)
from escape_test
where test_column like '%\%' escape '+';
Now you don't have to add anything into your like-pattern.
To hold myself to the same standard of proof that #Michael demonstrated --
create table escape_test
( test_column varchar(20) );
insert into escape_test
(test_column)
values ('foo\'),
('no slashes here'),
('foo\bar'),
('\bar');
select 'test1' trial, count(*) result
from escape_test
where test_column like '%\%'
UNION
select 'test2', count(*)
from escape_test
where test_column like '%\\%' escape '\'
UNION
select 'test3', count(*)
from escape_test
where test_column like '%\%' escape '+'
;
Which returns the same number of rows for each method:
TRIAL RESULT
----- ------
test1 3
test2 3
test3 3