Extract date from string and insert into field Microsoft SQL Server 2012

Question

Say I have a field UserAddedInfo with a string "User was added to the system and removed from list on 16/05/2016 by User Annon" and a field DateAdded in the same table.

Is there any way in SQL to extract that 16/05/2016 date from the string field and insert it into the DateAdded field as a datetime?

The date in the string is always going to be dd/MM/yyyy.

Thanks!

Answer 1

Use PATINDEX to get the start position of the date string in the column and extract 10 characters from that position. To convert the extracted string to date , use CONVERT with format 103 .

103 = dd/mm/yyyy

select 
convert(date,
substring(UserAddedInfo,patindex('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%',UserAddedInfo),10)
      ,103)
from table_name
where patindex('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%',UserAddedInfo) > 0

To update the dateadded field in the table, use

update table_name
set dateadded = convert(date,
substring(UserAddedInfo,patindex('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%',UserAddedInfo),10)
      ,103)
where patindex('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%',UserAddedInfo) > 0

Use try_cast or try_convert to return null when the substring returns invalid dates.

select 
try_cast(
substring(UserAddedInfo,patindex('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%',UserAddedInfo),10) 
as date) 
--or
--try_convert(date,
--substring(UserAddedInfo,patindex('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%',UserAddedInfo),10) 
--) 
from table_name
where patindex('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%',UserAddedInfo) > 0

Answer 2

You can use patindex to find date string, and use cast to convert it to datetime

select 
    cast(substring('User was added to the system and removed from list on 27/08/2014 by User Annon', 
        patindex('%[0-9][0-9]/[0-9][0-9]/[0-9][0-9][0-9][0-9]%', 
                 'User was added to the system and removed from list on 27/08/2014 by User Annon'), 10) as datetime)