How can I extract the weekenddate date from a string date, e.g. '2021-12-01', in SingleStore Database?
It might help you:
Use to_date() to convert string to date and dayname() to catch weekends only in where clause.
Select TO_DATE(date,'YYYY-MM-DD') from table
Where dayname(to_date(date,'YYYY-MM-DD')) in ('Saturday','Sunday');
Related
Can someone show me how to extract the year from the String date in databricks SQL.
I am based in the UK and our date format is normally as follows:
dd/mm/yyyy
The field containing the dates is set as StringType()
I am trying to extract the year from the string as follows:
select year(cast(financials_0_accountsDate as Date)) from `financiallimited_csv`
I'm using the following the code to extract the quarter
select quarter(cast(financials_0_accountsDate as Date)) from `financiallimited_csv`
However, both result in NULL values.
Any thoughts on how to extract the year and quarter from dates with StringType() dd/mm/yyyy?
The table looks like the following:
Could you try the to_date function?
select year(to_date(financials_0_accountsDate, 'dd/MM/yyyy')) from `financiallimited_csv`
I have a csv file in which a date column has values like 01080600, basically MM-dd-HH-mm.
I want to add a column in dataframe which shows this in a more readable format.
I do :
spark.sql("SELECT date...")
.withColumn("readable date", to_date(col("date"), "MM:dd HH:mm"))
.show(10)
But readable date is returned null.
What am I missing here?
While formating or converting to date or timestamp you need to provide the date_format as is following your pattern , example in your case you need to modify your format as below and further which can be formatted depending on the final format you wish your date col to take using date_format
References to various patterns and parsing can be found here
To Timestamp
sql.sql("""
SELECT
TO_TIMESTAMP('01080600','ddMMhhmm') as date,
DATE_FORMAT(TO_TIMESTAMP('01080600','ddMMhhmm'),'MM/dd hh:mm') as formated_date
""").show()
+-------------------+-------------+
| date|formated_date|
+-------------------+-------------+
|1970-08-01 06:00:00| 08/01 06:00|
+-------------------+-------------+
Could you please guide with below query.
I need to convert below string column to date.
Input and expected output is provided in screenshot.
Input table: column maturity_date is in string datatype.
I tried below but not working as expected
to_date(from_unixtime(unix_timestamp(maturity_date,'MM/DD/YYYY H:mm:ss'),'yyyy-mm-dd')
Try using lower case letters. Upper case means another thing (day of year (D) and week-year (Y)).
to_date(from_unixtime(unix_timestamp(maturity_date,'MM/dd/yyyy H:mm:ss'),'yyyy-MM-dd')
Correct input format is 'MM/dd/yyyy H:mm:ss', not 'MM/DD/YYYY H:mm:ss'
Correct output format is yyyy-MM-dd, not yyyy-mm-dd. mm is minutes. MM is month
Read more about date format used in Hive here SimpleDateFormat
I have tried to insert date into the postgresql in 'dd.mm.yyyy' format.
First inserted the string type, then the date type
moment(new Date()).format('DD.MM.YYYY') but in database format always like 'yyyy-mm-dd'
What can I do?
Plesase, help
I have table which stores datetime as varchar
Format looks like this 2018-07-16 15:00:00.0 ,
I want to parse this to extract only date part so that I use date part to compare with date in string format such as '2018-07-20' in where clause. What is the best way to achieve this in presto?
This particular format (based on example value 2018-07-16 15:00:00.0 in the question) is understood by cast from varchar to timestamp. You then need to extract date part with another cast:
presto> SELECT CAST(CAST('2018-07-16 15:00:00.0' AS timestamp) AS date);
_col0
------------
2018-07-16
(1 row)