Is there a way i could convert an actual string to a number in postgresql
i have record stored in db and below is an example of unique identifier found in each record.
d895774d-7b91-4adc-bfba-f1264e5aae04
I want to store this record in another database but i would want to generate another unique number for each transaction/line
Is there a way i can convert this d895774d-7b91-4adc-bfba-f1264e5aae04 to some number.
like this actual numbers 13693660123082308270370273012321321312321
select 'd895774d-7b91-4adc-bfba-f1264e5aae04' as id
First convert the string into a table of single characters (the from clause);
Then select the digits 'as is', a as 10, b as 11 and so on (the case expression);
Finally aggregate into a string (string_agg)
skipping the - characters (where clause).
select
string_agg(case when c between '0' and '9' then c else (ascii(c) - 87)::text end, '')
from unnest(string_to_array('d895774d-7b91-4adc-bfba-f1264e5aae04', null)) c
where c <> '-';
Result: 13895774137119141013121115111015126414510101404
Edit
select
td.date,
(
select string_agg(case when c between '0' and '9' then c else (ascii(c) - 87)::text end, '')
from unnest(string_to_array(td.id, null)) c
where c <> '-'
) as id
from table_d td;
Related
I find in my sql database string whit weird whitespace which cannot be replace like REPLACE(string, ' ', '') RTRIM and cant it even find with string = '% %'. This space is even transfered to new table when using SELECT string INTO
If i select this string in managment studio and copy that is seems is normal space and when everything is works but cant do nothing directly from database. What else can i do? Its some kind of error or can i try some special character for this?
First, you must identify the character.
You can do that by using a tally table (or a cte) and the Unicode function:
The following script will return a table with two columns: one contains a char and the other it's unicode value:
DECLARE #Str nvarchar(100) = N'This is a string containing 1 number and some words.';
with Tally(n) as
(
SELECT TOP(LEN(#str)) ROW_NUMBER() OVER(ORDER BY ##SPID)
FROM sys.objects a
--CROSS JOIN sys.objects b -- (unremark if there are not enough rows in the tally cte)
)
SELECT SUBSTRING(#str, n, 1) As TheChar,
UNICODE(SUBSTRING(#str, n, 1)) As TheCode
FROM Tally
WHERE n <= LEN(#str)
You can also add a condition to the where clause to only include "special" chars:
AND SUBSTRING(#str, n, 1) NOT LIKE '[a-zA-Z0-9]'
Then you can replace it using it's unicode value using nchar (I've used 32 in this example since it's unicode "regular" space:
SELECT REPLACE(#str, NCHAR(32), '|')
Result:
This|is|a|string|containing|1|number|and|some|words.
How to return last n words using Postgres.
I have tried using LEFT method.
SELECT DISTINCT LEFT(name, -4) FROM my_table;
but it return last 4 characters ,i want to return last 3 words.
demo:db<>fiddle
You can do this using a the SUBSTRING() function and regular expressions:
SELECT
SUBSTRING(name FROM '((\S+\s+){0,3}\S+$)')
FROM my_table
This has been explained here: How can I match the last two words in a sentence in PostgreSQL?
\S+ is a string of non-whitespace characters
\s+ is a string of whitespace characters (e.g. one space)
(\S+\s+){0,3} Zero to three words separated by a space
\S+$ one word at the end of the text.
-> creates 4 words (or less if there are no more).
One way is to use regexp_split_to_array() to split the string into the words it contains and then put a string back together using the last 3 words in that array.
SELECT coalesce(w.words[array_length(w.words, 1) - 2] || ' ', '')
|| coalesce(w.words[array_length(w.words, 1) - 1] || ' ', '')
|| coalesce(w.words[array_length(w.words, 1)], '')
FROM mytable t
CROSS JOIN LATERAL (SELECT regexp_split_to_array(t."name", ' ') words) w;
db<>fiddle
RIGHT() should do
SELECT RIGHT('MYCOLUMN', 4); -- returns LUMN
UPD
You can convert to array and then back to string
SELECT array_to_string(sentence[(array_length(sentence,1)-3):(array_length(sentence,1))],' ','*')
FROM
(
SELECT regexp_split_to_array('this is the one of the way to get the last four words of the string', E'\\s+') AS sentence
) foo;
DEMO HERE
PostgreSql gives me this error when i try to cast a TEXT colum to a integer.
select pro_id::integer from mmp_promocjas_tmp limit 1;
This colum contains only digits, valid integer. How can "1" be invalid integer?
select pro_id, length(pro_id) ,length(trim(pro_id)) from mmp_promocjas_tmp limit 1;
outputs:
1 | 2 | 2
Query select pro_id from mmp_promocjas_tmp where trim(pro_id) = '1' shows nothing.
I tried to remove whitespaces, without no result:
select pro_id from mmp_promocjas_tmp where regexp_replace(trim(pro_id), '\s*', '', 'g')
There are probably spurious invisible contents in the column.
To make them visible, try a query like this:
select pro_id, c,lpad(to_hex(ascii(c)),4,'0') from (
select pro_id,regexp_split_to_table(pro_id,'') as c
from (select pro_id from mmp_promocjas_tmp limit 10) as s
) as g;
This will show the ID and each character its contains, both as a character and as its hexadecimal code in the repertoire.
I want to implement a stored procedure that extract letters from a varchar in firebird.
Example :
v_accountno' is of type varchar(50) and has the following values
accountno 1 - 000023208821
accountno 2 - 390026826850868140H
accountno 3 - 0700765001003267KAH
I want to extract the letters from v_accountno and output it in o_letter.
In my example: o_letter will store H for accountno 2 and KAH for accountno 3.
I tried the following stored procedure, which obviously won't work for accountno 3. (Please help).
CREATE OR ALTER PROCEDURE SP_EXTRACT_LETTER
returns (
o_letter varchar(50))
as
declare variable v_accountno varchar(50);
begin
v_accountno = '390026826850868140H';
if (not (:v_accountno similar to '[[:DIGIT:]]*')) then
begin
-- My SP won't work in for accountno 3 '0700765001003267KAH'
v_accountno = longsubstr(v_accountno, strlen(v_accountno), strlen(v_accountno));
o_letter = v_accountno;
end
suspend;
end
One solution would be to replace every digits with empty string like:
o_letter = REPLACE(v_accountno, '0', '')
o_letter = REPLACE(o_letter, '1', '')
o_letter = REPLACE(o_letter, '2', '')
...
Since Firebird 3, you can use substring for this, using its regex facility (using the similar clause):
substring(v_accountno similar '[[:digit:]]*#"[[:alpha:]]*#"' escape '#')
See also this dbfiddle.
I'm generating T-SQL SELECT statements for tables for which I have no data type information up-front. In these statements, I need to perform string manipulation operations that depend on the length of the original value of the tables' columns.
One example (but not the only one) is to insert some text at a specific position in a string, including the option to insert it at the end:
SELECT
CASE WHEN (LEN ([t0].[Product] = 8)
THEN [t0].[Product] + 'test'
ELSE STUFF ([t0].[Product], 8, 0, 'test')
END
FROM [OrderItem] [t0]
(The CASE WHEN + LEN is required because STUFF doesn't allow me to insert text at the end of a string.)
The problem is that LEN excludes trailing blanks, which will ruin the calculation.
I know I can use DATALENGTH, which does not exclude trailing blanks, but I can't convert the bytes returned by DATALENGTH to the characters required by STUFF because I don't know whether the Product column is of type varchar or nvarchar.
So, how can I generate a SQL statement that depends on the exact length of a string in characters without up-front information about the string data type being used?
Here's what I ended up using:
SELECT
CASE WHEN ((LEN ([t0].[Product] + '#') - 1) = 8)
THEN [t0].[Product] + 'test'
ELSE STUFF ([t0].[Product], 8, 0, 'test')
END
FROM [OrderItem] [t0]
Measurements indicate that the LEN (... + '#') - 1 trick is about the same speed as LEN (...) alone.
Thanks for all the good answers!
try this:
SELECT
CASE WHEN (LEN (REPLACE([t0].[Product],' ', '#') = 8)
THEN [t0].[Product] + 'test'
ELSE STUFF ([t0].[Product], 8, 0, 'test')
END
FROM [OrderItem] [t0]
Can't you look up the type information for the columns in the system tables?
If not then to determine whether or not a column is varchar or nvarchar this would do it.
create table #test
(
c varchar(50),
n nvarchar(50)
)
insert into #test values ('1,2,3,4 ',N'1,2,3,4,5 ')
SELECT
CASE
WHEN datalength(CAST(c AS nvarchar(MAX))) = datalength(c)
THEN 'c is nvarchar'
ELSE 'c is char'
END,
CASE
WHEN datalength(CAST(n AS nvarchar(MAX))) = datalength(n)
THEN 'n is nvarchar'
ELSE 'n is char'
END
FROM #test
Use DATALENGTH and SQL_VARIANT_PROPERTY:
SELECT
CASE
WHEN 8
= DATALENGTH([t0].[Product])
/ CASE SQL_VARIANT_PROPERTY([t0].[Product],'BaseType') WHEN 'nvarchar' THEN 2 ELSE 1 END
THEN [t0].[Product] + 'test'
ELSE STUFF ([t0].[Product], 8, 0, 'test')
END
FROM [OrderItem] [t0]
If there are no leading blanks, len(reverse(column_name)) will give you the column length.