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Extract date (yyyy/mm/dd) from a timestamp in PostgreSQL
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Closed 6 years ago.
I have date column on my postgres table as timestamp format i.e "2017-01-01 22:00:00". When I wrote in the queries
select date from table where date = '2017-01-01' it did not give me any result.
Should I always include the time information on those queries? Can I just put the yyyy-mm-dd only on my queries to search the date column?
regards
Cast to date:
select "date" from table where "date"::date = '2017-01-01'
Note that I enclosed references to the date column in double quotes, because date is a Postgres keyword. You should avoid naming your columns, tables, or schemas using keywords.
Use date() function, it will extract date from datetime like:
select DATE(my_field) from my_table;
Try this way:
select date("Date") from yourTable
Related
I am facing an issue extracting the month and year from the data (in Character varying type) >> InvoiceDate in SQL Postgres. I have seen the solution is relatively easy with MySQL function: DATEFROMPARTS as per the below Code which is not available in SQLpostgres. How can I get the same result DATA_PART function in Postgres SQL, but simultaneously I need to change the data type of the column "InvoiceDate" to the date
Select
CustomerID,
min(InvoiceDate) first_purchase_date,
DATEFROMPARTS(year(min(InvoiceDate)), month(min(InvoiceDate)), 1) Cohort_Date
into #cohort
from #online_retail_main
group by CustomerID
The output:
Customer ID| first_purchase_date |Cohort_Date|
-----------+-------------------------+-----------+
12345 | 2010-12-20 15:47:00:00 | 2010-12-01|
I am trying to make a date consits of Year and Month , while the day to be set as 1 for all
Assuming a valid Postgres timestamp:
select date_trunc('month', '2010-12-20 15:47:00.00'::timestamp)::date;
date_trunc
------------
12/01/2010
--or ISO format
set datestyle = 'ISO,MDY';
select date_trunc('month', '2010-12-20 15:47:00.00'::timestamp)::date;
date_trunc
------------
2010-12-01
Uses date_trunc to truncate the timestamp to a month which means the first of the month. Then cast(::date) to a date type. The DateStyle just deals with how the value is presented to the user. The value is not stored formatted.
To do something similar to what you did in MySQL:
select make_date(extract(year from '2010-12-20 15:47:00.00'::timestamp)::integer, extract(month from '2010-12-20 15:47:00.00'::timestamp)::integer, 1);
This uses make_date from here Date/time functions and extract to build a date.
Hi I have a date format in one table in Text 'YYYY/MM'. example 2018/01, 2020/08 etc.
I need to join it with another table where the date is in Number type( and DATETIME20 format attached it it) and convert it to month and compare.
Is there any way to do it in PROC SQL as the rest of my query is in PROC SQL?
eg. Table 1: Month= 2018/01; Table 2: Date =20.01.2018 10:48:17 . They should be joined in the PROC SQL query.
I would also like to calculate difference in Months between these two dates.
Thank you in advance.
Convert both to the same DATE value. To convert a datetime value to a date use the DATEPART() function. To move to the first day of the month use the INTNX() function with the month interval. To convert a string like '2018/01' to a date you could use INPUT() function with YYMMDD informat by appending '/01'.
proc sql ;
create table want as
select *
from table1,table2
where input(cats(table1.month,'/01'),yymmdd10.)=intnx('month',datepart(table2.date),0)
;
quit;
I have column 'jobstarttimeiso' and I want to create another column for Weeks of the year based on the date. How would I go about doing that? I am using Redash to query from redshift database. Please help! Thank you.
The extract function will return a week number e.g.
select extract(week from jobstarttimeiso) as weeknumber
In general:
EXTRACT ( datepart FROM { TIMESTAMP 'literal' | timestamp } )
See: https://docs.aws.amazon.com/redshift/latest/dg/r_EXTRACT_function.html
alternatively you can also use to_char()
TO_CHAR (timestamp_expression | numeric_expression , 'format')
with the parameter IYYY as the format for ISO 8601 week-numbering year (4 or more digits)
I have two columns (date character varying, time character varying) in table VM. Now I want to merge these two columns and change the type as a timestamp. can anyone help me out to do this?
Or else
I want to separate data based on the year (2017, 2018). I used substring(date, '2018') command but it replies with 2017 data showing NULL instead of 2017 along with 2018 data.
You didn't tell us what exactly the contents of that table is, but something like this should work:
Add a new timestamp column (and please find a better name than "timestamp" or ts_column for it)
alter table the_table add ts_column timestamp;
Then you need to update the data
update the_table
set ts_column = to_timestamp(concat("date", ' ', "time"), 'yyyy-mm-dd hh24:mi:ss')
where "date" is not null;
The above assumes that the date stored in the column named "date" and the time stored in the column named "time" are formatted as ISO date and time values. If they are not, you need to adjust the parameter to the to_timestamp() function. For details on the format mask, please see the manual
Then drop the old columns:
alter table the_table drop "date", drop "time";
To extract the year from a date or timestamp column, use the extract function:
where extract(year from ts_column) = 2018
Note that the above will not be able to use an index on ts_column to speed up the query.
An alternative way of writing is:
where ts_column >= date '2018-01-01'
and ts_column < date '2019-01-01';
which would be able to use an index on ts_column
I want to insert date and time in oracle database, I have created the table with columns
create table myadmin
( employe_id number(5),
supervisor Varchar2(20),
department Varchar2(20),
action Varchar2(20),
sdate date,
stime date)
While inserting the values below it gives an error. Please tell me how to insert the time ?
insert into myadmin
( employe_id,supervisor,department,action,sdate,stime) values
(83,'gaurav','helpdesk','pick','23-jan-2013','09:43:00');
You have to use keyword to_date for date insert in oracle like this.
to_date('23-01-2013','dd-mm-yyyy')
Basically you have to use keyword to_date('your date','your date format').
You can also add date and time together if you want and it would be something like this
to_date('23-01-2013 09:43:00','dd-mm-yyyy hh24:mi:ss')
A date in Oracle always has a date part and a time part. Having date and time in two separate columns only makes sense, if it can occur that date is null and time is not. (And still, you could set date to an improbable value like 1.1.0001 then.)
However, if you want to stick to those two separate fields, then make your string a datetime with the to_date function specifying the format used:
insert into myadmin
( employe_id,supervisor,department,action,sdate,stime) values
(83,'gaurav','helpdesk','pick',to_date('23-01-2013','dd-mm-yyyy'), to_date('09:43:00', 'hh24:mi:ss'));