Searched the postgresql docs http://www.postgresql.org/docs/8.4/interactive/functions-bitstring.html for information on converting bit varying to integer
But couldnt' find any info.
select '011111'::bit(4)::varbit(4)::integer as varbit
Appreciate your response.
One way:
SELECT b, lpad(b::text, 32, '0')::bit(32)::int
FROM (
VALUES
('01'::varbit)
,('011111')
,('111')
) t (b);
Result:
b | lpad
-------+------
01 | 1
011111 | 31
111 | 7
Related answer:
Convert hex in text representation to decimal number
Let's say our table has 3 columns and 3 rows. We can simulate it with:
select *
from (
values ('row1', 1::int, 12::bit(8)::varbit),
('row2', 2::int, 23::bit(8)::varbit),
('row3', 3::int, 34::bit(8)::varbit)
) as T(A,B,C);
As you can see, first columns is varchar , second is int and the third is varbit.
Let's convert third column to int:
select C::bit(8)::int
from (
values ('row1', 1::int, 12::bit(8)::varbit),
('row2', 2::int, 23::bit(8)::varbit),
('row3', 3::int, 34::bit(8)::varbit)
) as T(A,B,C);
==C==
12
23
34
The point is, you have to convert it to bit(n) first, then you can convert bit(n) to varbit. The same thing is also true for int to varbit.
Related
In my hive table I need to extract from column DATE(stored as bigint), 2 digits(month) in center.
For Example:
20220120 -> 01
20211201 -> 12
Is it possible to do directly without casting column DATE to string?
Thanks
substr should work.
select substr( 20220101,4,2)
New to postgres and unsure how to accomplish the following. I have a table as follows:
create table if not exists my_table (
id int GENERATED BY DEFAULT AS IDENTITY primary key,
key int default 0
)
What I am trying to do is take an integer value (my_key) and if it's >= 0 and < 10 then add a leading zero (0) to it and insert it into my_table.key
I have tried to_char(my_key::integer,'09')::integer where my_key = 0 and it doesn't insert 00 within the key column.
Any help would be great.
Leading zeros don't change the value of an integer, so this is a question about formatting numbers.
If you want to display the id column with leading zeros, you could do that like this:
SELECT to_char(id, '00'), key
FROM "table";
The format 00 formats the number as a two-digit string with leasing zeros. If id is greater than 99, the number cannot be formatted like this, and you will get ##.
See the documentation for details about to_char and the available formats.
I have a column tattoo that is varchar(20) type. I want this to be padded left with 3 zeros. N009 should become 000N009.
to_char(integer/double precsion/int/timestamp,text) but is there any way we could do this to varchar?
Casting varchar(20) to integer did not worked either. Some of my code added below:
select tattoo,to_char(tattoo,'0000FM')::varchar as tm from animals where soc_code = 'AURO' limit 100
select tattoo,to_char(tattoo::integer,'0000FM')::varchar as tm from animals where soc_code = 'AURO' limit 100
select tattoo,to_char(cast(tattoo as integer),'0000FM')::varchar as tm from animals where soc_code = 'AURO' limit 100
They both throw errors saying invalid input syntax for integer and function_to_char does not exist.
have you tried LPAD()
select LPAD( tattoo, 7, '0' )
I want this to be padded left with 3 zeros. N009 should become 000N009.
You may be overthinking this. Just concatenate:
SELECT '000' || tatoo;
If tatoo is NULL, so is the result.
I have the following query
Select * from Common.dbo.Zip4Lookup where
zipcode='76033' and
StreetName='PO BOX' and
'704' between AddressLow and AddressHigh and
(OddEven='B' or OddEven = 'E')
The AddressLow and AddressHigh columns are varchar(10) fields.
The records returned are
AddressLow AddressHigh
------------ ------------
1 79
701 711
The second is the desired record How do I get rid of the first record.
The problem is that SQL is using a string compare instead of a numeric compare. This is because AddressLow/High are varchar and not int.
As long as AddressLow/High contain numbers, this should work:
Select * from Common.dbo.Zip4Lookup where
zipcode='76033' and
StreetName='PO BOX' and
704 between
CAST(AddressLow as INT) and
CAST(AddressHigh as INT) and
(OddEven='B' or OddEven = 'E')
The problem is that your condition fits to the first record in 7 on the beginning of the 79 because it's the string value. The easist way is IMHO change the data type to some numeric one.
I have a varchar column in Postgres 8.3 that holds values like: '0100011101111000'
I need a function that would consider that string to be a number in base 2 and spits out the numeric in base 10. Makes sense?
So, for instance:
'000001' -> 1.0
'000010' -> 2.0
'000011' -> 3.0
Thanks!
Cast to a bit string then to an integer.
An example:
'1110'::bit(4)::integer -> 14
Though you had varying length examples, and were after bigint, so instead use bit(64) and pad the input with zeroes using the lpad function.
lpad('0100011101111000',64,'0')::bit(64)::bigint
Here's a complete example...
create temp table examples (val varchar(64));
insert into examples values('0100011101111000');
insert into examples values('000001');
insert into examples values('000010');
insert into examples values('000011');
select val,lpad(val,64,'0')::bit(64)::bigint as result from examples;
The result of the select is:
val | result
------------------+--------
0100011101111000 | 18296
000001 | 1
000010 | 2
000011 | 3
(4 rows)