Generate newid() uniqueidentifier as a char(16) field

I need to generate a 16 digits universal unique alphanumeric number.
I tried to do it using newid() function  and save it in a char(16) field but it doesn't work.
Any other ideas?
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Gustavo Perez BuenrostroCommented:

declare @sMyOwnId char(16)
select @sMyOwnId = substring(replace(cast(newid() as varchar(36)),'-',''),1,16)
select @sMyOwnId

PD: I think you could create your own unique-value generator routine based on GETDATE and RAND functions. Let me know if you need a better approach for creating unique values.

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mordiAuthor Commented:
To use the full universal capability of the newid function, i have to convert every 2 digits(hex) of the 32 that the function returns, to one digit, and thus converting to a 16 digit guid.
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