EJB-QL, underscores

[Jeremy Wright]

I received this question in a Whizlabs test, and I'm not sure I understand this right...

Select all pattern values of a LIKE expression that would identify a word starting with a 'J' and having 'on' as the last two characters.
[A] 'J%on_'
[B] 'J*on%'
[C] 'J%on*'
[D] 'J_on%'
[E] 'J%_on_'
They give the correct answer as A,E.
Quote: "The only way to identify words having 'on' as the last two characters is by having a string that contains 'on' and an underscore, as in 'on_'.

However, I understood that '_' stands for any single character. Can it also stand for nothing?

Thanks,

Jeremy

[Sankar Subbiramaniam]

I do have the same understanding.
Quote from EJB specs page230:

The pattern-value is a string literal in
which an underscore (_) stands for any single character, a percent (%) character stands for any sequence
of characters (including the empty sequence), and all other characters stand for themselves.

[Devender Thareja]

Originally posted by Jeremy Wright:
I received this question in a Whizlabs test, and I'm not sure I understand this right...

Select all pattern values of a LIKE expression that would identify a word starting with a 'J' and having 'on' as the last two characters.
[A] 'J%on_'
[B] 'J*on%'
[C] 'J%on*'
[D] 'J_on%'
[E] 'J%_on_'
They give the correct answer as A,E.
Quote: "The only way to identify words having 'on' as the last two characters is by having a string that contains 'on' and an underscore, as in 'on_'.

However, I understood that '_' stands for any single character. Can it also stand for nothing?

Thanks,

Jeremy

I have the same understanding too. It seems that the answer shoud be J%on which is not in the options.
Probable they mean that It should start with J and two charecters before last character should be 'on'. That would make the answer "J%on_". Still E wouldn't be right.