In presto SQL, the date is saved as a string like '2020-06-10'. I want to convert into a date format (yyyy-mm-dd)
This is how I did it:
select date_format(date_parse('2020-06-10', '%Y-%m-%d'),'%Y-%m-%d')
First I convert string to a timestamp format, then convert the new timestamp to date_format.
My question is that is there a function such that I convert only once?
For example
date_parse(string, format, expect_out_put_time_format)
You can use date function, which is a shortcut for CAST(x AS date).
presto> SELECT date('2020-06-10');
_col0
------------
2020-06-10
Related
I have a string column which has date value in the below format
6/1/2020 12:00:00 AM
I want only the date part and convert the column to date
so when i use date_trunc and parse date like below it is giving error
select CAST(DATE_TRUNC(PARSE_DATE('%m/%d/%Y', datecolumn), AS DATE)) from table name
PLease let me know what is the best way to achieve this
Use below
select date(parse_datetime('%m/%d/%Y %I:%M:%S %p', datecolumn))
from your_table
Using derived column i am adding 3 columns -> 2 columns for date and 1 for timestamp. for the date columns i am passing a string as parameter. for eg: 21-11-2021 and timstamp i am using currenttimestamp fucntion.
i wrote expressions in derived columns to convert them as date and timestamp datatype and also in a format that target table needs which is dd-MM-yyyy and dd-MM-yyyy HH:mm:ss repectively
For date->
expression used: toDate($initialdate, 'dd-MM-yyyy')
data preview output: 2021-01-21 --(not in the format i want)
After pipline Debug Run, value in target DB(Azure sql database) column:
2021-01-21T00:00:00 -- in table it shows like this I dont understand why
For Timstamp conversion:
Expression used:
toTimestamp(toString(currentTimestamp(), 'dd-MM-yyyy HH:mm:ss', 'Europe/Amsterdam'), 'dd-MM-yyyy HH:mm:ss')
Data preview output: 2021-11-17 19:37:04 -- not in the format i want
After pipline Debug Run, value in target DB(Azure sql database) column:
2021-11-17T19:37:04:932 -in table it shows like this I dont understand why
question 1: I am NOT getting values in the format the target requires ???and it should be only in DATE And Datetime2 dataype respectively so no string conversions
question 2: after debug run i dont know why after insert the table values look different from Data preview???
Kinldy let me know if i have written any wrong expressions??
--apologies i am not able post pictures---
toDate() converts input date string to date with default format as yyyy-[M]M-[d]d. Accepted formats are :[ yyyy, yyyy-[M]M, yyyy-[M]M-[d]d, yyyy-[M]M-[d]dT* ].
Same goes with toTimestamp(), the default pattern is yyyy-[M]M-[d]d hh:mm:ss[.f...] when it is used.
In Azure SQL Database as well the default date and datetime2 formats are in YYYY-MM-DD and YYYY-MM-DD HH:mm:ss as shown below.
But if your column datatypes are in string (varchar) format, then you can change the output format of date and DateTime in azure data flow mappings.
When loaded to Azure SQL database, it is shown as below:
Note: This format results when datatype is varchar
If the datatype is the date in the Azure SQL database, you can convert them to the required format using date conversions as
select id, col1, date1, convert(varchar(10),date1,105) as 'dd-MM-YYYY' from test1
Azure SQL Database always follows the UTC time zone. Using “AT TIME ZONE” convert it another non-UTC time zone.
select getdate() as a, getdate() AT TIME ZONE 'UTC' AT TIME ZONE 'Central Standard Time' as b
You can also refer to sys.time_zone_info view to check current UTC offset information.
select * from sys.time_zone_info
I have a parquet file with a start_date and end_date columns
Formatted like this
01-Jan-2021
I've tried every combination conversion toDate, toString, toInterger functions but I still get Nulls returned when viewing the data (see image).
I would like to have see the result in two ways YYYYMMDD as integer column and YYYY-MM-DD as Date columns.
eg 01012021 and 01-01-2021
I'm sure the default format has caused this issue.
Thanks
First, for the Date formatter, you need to first tell ADF what each part of your string represents. Use dd-MMM-yyy for your format. Then, use a string formatter to manipulate the output as such: toString(toDate('01-Jan-2021', 'dd-MMM-yyyy'), 'yyyy-MM-dd')
For the integer representation: toInteger(toString(toDate('01-Jan-2021', 'dd-MMM-yyyy'), 'yyyyMMdd'))
Ah, you say *"I would like to have see the result in two ways YYYYMMDD as integer column and YYYY-MM-DD as Date columns. eg 01012021 and 01-01-2021"* Do you want in YYYYMMDD or dd-mm-yyy cause your example is in the later format.
Anyways, please see below expression you could use:
My source:
Use derived column:
Edit expression:
start_date_toInteger : toString(toDate(substring(start_date,1,11), 'dd-MMM-yyyy'), 'yyyymmdd')
start_date_toDate: toString(toDate(substring(start_date,1,11), 'dd-MMM-yyyy'), 'yyyy-mm-dd')
Final results:
I have a string associated with date in ‘Teradata’ tables
Var1=09OCT2017-EMRT
I need to extract the date from the above string in ‘mm/dd/yyyy’ format
I tried the following
Cast(cast(substr(var1,1,9) as char(20)) as date format ‘mm/dd/yyyy’) as date
I am getting error as ‘invalid date supplied for var1’
I would appreciate your help
You need to apply a format matching the input string:
To_Date(Substr(var1,1,9), 'ddmonyyyy')
returns a DATE.
If you want to cast it back to a string:
To_Char(To_Date(Substr(var1,1,9), 'ddmonyyyy'), 'mm/dd/yyyy')
I'm very new to sql/hive. At first, I loaded a txt file into hive using:
drop table if exists Tran_data;
create table Tran_data(tran_time string,
resort string, settled double)
ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t' LINES TERMINATED BY '\n';
Load data local inpath 'C:\Users\me\Documents\transaction_data.txt' into table Tran_Data;
The variable tran_time in the txt file is like this:10-APR-2014 15:01. After loading this Tran_data table, I tried to convert tran_time to a "standard" format so that I can join this table to another table using tran_time as the join key. The date format desired is 'yyyymmdd'. I searched online resources, and found this: unix_timestamp(substr(tran_time,1,11),'dd-MMM-yyyy')
So essentially, I'm doing this: unix_timestamp('10-APR-2014','dd-MMM-yyyy'). However, the output is "NULL".
So my question is: how to convert the date format to a "standard" format, and then further convert it to 'yyyymmdd' format?
from_unixtime(unix_timestamp('20150101' ,'yyyyMMdd'), 'yyyy-MM-dd')
My current Hive Version: Hive 0.12.0-cdh5.1.5
I converted datetime in first column to date in second column using the below hive date functions. Hope this helps!
select inp_dt, from_unixtime(unix_timestamp(substr(inp_dt,0,11),'dd-MMM-yyyy')) as todateformat from table;
inp_dt todateformat
12-Mar-2015 07:24:55 2015-03-12 00:00:00
unix_timestamp function will convert given string date format to unix timestamp in seconds , but not like this format dd-mm-yyyy.
You need to write your own custom udf to convert a given string date to the format that you need as present Hive do not have any predefined functions. We have to_date function to convert a timestamp to date , remaining all unix_timestamp functions won't help your problem.
select from_unixtime(unix_timestamp('01032018' ,'MMddyyyy'), 'yyyyMMdd');
input format: mmddyyyy
01032018
output after query: yyyymmdd
20180103
To help someone in the future:
The following function should work as it worked in my case
to_date(from_unixtime(UNIX_TIMESTAMP('10-APR-2014','dd-MMM-yyyy'))
unix_timestamp('2014-05-01','dd-mmm-yyyy') will work, your input string should be in this format for hive yyyy-mm-dd or yyyy-mm-dd hh:mm:ss
Where as you are trying with '01-MAY-2014' hive won't understand it as a date string