selecting the rows based on the distinct column values

Question

I have the following data in a table sample . I am using Oracle database.

---

column1      column2         column3
A             01           2017-07-27-04.24.58.851340
B             01           2017-07-27-06.43.19.654420
C             01           2017-08-10-08.26.47.633480

I need to pull out the count of distinct records based on SUBSTR(column3, 1,10) value.

For example:

In the first row, the SUBSTR(column3, 1,10) value of column3 is 2017-07-27 ,

In the second row, the SUBSTR(column3, 1,10) value of column3 is 2017-07-27 which is same as first row

In third row, the SUBSTR(column3, 1,10) value of column3 is 2017-08-10 which is different from first two rows.

So based on distinct SUBSTR(column3, 1,10) values, the count should be returned as 2 .

Can anybody help me out?

Answer 1

select count(distinct SUBSTR(column3, 1,10))
from sample

Answer 2

Since you appear to be storing timestamp values as strings (which you should not do, and ought to store them as timestamps) you can convert them to a timestamp and then truncate the time component to the start of the day:

SELECT   COUNT( DISTINCT TRUNC( TO_TIMESTAMP( column3, 'YYYY-MM-DD-HH24.MI.SS.FF6' ) ) )
           AS num_days
FROM     your_table

But you could also just use the string value:

SELECT   COUNT( DISTINCT SUBSTR( column3, 1, 10 ) ) AS num_days
FROM     your_table

If you convert your table to store them as timestamps then you can just use:

SELECT   COUNT( DISTINCT TRUNC( column3 ) ) AS num_days
FROM     your_table