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When I subtract timestamps, the interval is in form DD:HH:MM:SS. How can I convert it all to minutes without extracting days and hours and multiplication/addition? I'm looking for a single function which I can substitute for date_part in this query so that it returns 65:
select date_part('minutes', '65 minutes'::interval); Context: I need to know how many minutes have passed since given timestamp.
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In PostgreSQL the interval data type is used to store and manipulate a time period. It holds 16 bytes of space and ranging from -178, 000, 000 years to 178, 000, 000 years. It also has additional attribute called “precision (denoted by p)” that can be used to set the level of precision in the query results.
In PostgreSQL, the make_interval() function creates an interval from years, months, weeks, days, hours, minutes and seconds fields. You provide the years, months, weeks, days, hours, minutes and/or seconds fields, and it will return an interval in the interval data type.
Postgres DATE data type Postgres uses the DATE data type for storing different dates in YYYY-MM-DD format. It uses 4 bytes for storing a date value in a column. You can design a Postgres table with a DATE column and use the keyword DEFAULT CURRENT_DATE to use the current system date as the default value in this column.
In PostgreSQL, the Interval is another type of data type used to store and deploy Time in years, months, days, hours, minutes, seconds, etc. And the months and days values are integers values, whereas the second's field can be the fractions values.
SELECT EXTRACT(EPOCH FROM '2 months 3 days 12 hours 65 minutes'::INTERVAL)/60; seems to work.
WARNING: "seems" is the key word.
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