Presto SQL - Converting a date string to date format
SqlPrestoSql Problem Overview
I'm on presto and have a date formatted as varchar that looks like -
7/14/2015 8:22:39 AM
I've looked the presto docs and tried various things(cast, date_format, using split_part to parse and then cast) and am not getting this to convert to a date format that I can use with functions like date_diff.
I've tried:
cast(fieldname as timestamp)
date_format(fieldname, '%Y-%m-%d %T)
Both give me an error like this
'Value cannot be cast to timestamp: 3/31/2016 6:05:04 PM'
How do I convert this?
Sql Solutions
Solution 1 - Sql
I figured it out. The below works in converting it to a 24 hr date format.
select date_parse('7/22/2016 6:05:04 PM','%m/%d/%Y %h:%i:%s %p')
Solution 2 - Sql
You can also do something like this
date(cast('2016-03-22 15:19:34.0' as timestamp))
Solution 3 - Sql
Use: cast(date_parse(inv.date_created,'%Y-%m-%d %h24:%i:%s') as date)
Input: String timestamp
Output: date format 'yyyy-mm-dd'
Solution 4 - Sql
Converted DateID having date in Int format to date format: Presto Query
Select CAST(date_format(date_parse(cast(dateid as varchar(10)), '%Y%m%d'), '%Y/%m-%d') AS DATE)
from
Table_Name
limit 10;
Solution 5 - Sql
If your string is in ISO 8601 format, you can also use from_iso8601_timestamp
Solution 6 - Sql
select date_format(date_parse(t.payDate,'%Y-%m-%d %H:%i:%S'),'%Y-%m-%d') as payDate
from testTable t
where t.paydate is not null and t.paydate <> '';
Solution 7 - Sql
SQL 2003 standard defines the format as follows:
<unquoted timestamp string> ::= <unquoted date string> <space> <unquoted time string>
<date value> ::= <years value> <minus sign> <months value> <minus sign> <days value>
<time value> ::= <hours value> <colon> <minutes value> <colon> <seconds value>
There are some definitions in between that just link back to these, but in short YYYY-MM-DD HH:MM:SS with optional .mmm milliseconds is required to work on all SQL databases.
Solution 8 - Sql
date_format requires first argument as timestamp so not the best way to convert a string. Use date_parse instead.
Also, use %c for non zero-padded month, %e for non zero-padded day of the month and %Y for four digit year.
SELECT date_parse('7/22/2016 6:05:04 PM', '%c/%e/%Y %r')