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Former Member
Apr 16, 2015 at 12:10 AM

'Like' statement

72 Views

So for illustration of my problem I have created a simple view consisting of 1 column.

I apply filter thus : Views.ZZ_List.NormalizedName = 'Java 6 Update 11'

and wonderfully this results in 1 row which shows the value 'Java 6 Update 11'

However, in reality, I want this filter to return all rows with the string 'Java' in it.

So in traditional SQL : Like '%Java%' (or Like '*Java*')

In Information Steward, I see that there is no 'Like' statement.

Hence, reading online, I tried this one : match_regex(Views.ZZ_List.NormalizedName, 'Java', null)

Yet after applying the above, I get zero rows returned.

Also tried :

match_pattern(Views.ZZ_List.NormalizedName, 'Java')

match_pattern(Views.ZZ_List.NormalizedName, '*Java*')

.... and may I say, many more variations.

Please can someone point out what the proper way of achieving a 'like' functionality is ?

Many Thanks

😊