That works. Thank you!!
On Fri, Apr 29, 2022 at 10:43 AM David G. Johnston <
david(dot)g(dot)johnston(at)gmail(dot)com> wrote:
> On Friday, April 29, 2022, Blake McBride <blake1024(at)gmail(dot)com> wrote:
>
>> I am wanting it to do the same as: select * from se_user where user_id
>> in (3, 4, 5)
>>
>
> Then rewrite the query to use the “val = any(array)” expression like I
> said.
>
> David J.
>
>