Function inside quotes 1

[chrissparkle]

I have this line of code in a proc:

 CASE WHEN DATEDIFF(n, mbx.date_received, GETDATE()) BETWEEN 0 AND 10 THEN 'DATEDIFF(n, mbx.date_received, GETDATE()) min ago' END as USE_DATE

How can I get the DateDiff function to output the actual value when inside quotes? At the moment it just outputs the raw sql??

 

[gmmastros]

Try this...

[COLOR=blue]CASE[/color] [COLOR=blue]WHEN[/color] [COLOR=#FF00FF]DATEDIFF[/color](n, mbx.date_received, [COLOR=#FF00FF]GETDATE[/color]()) BETWEEN 0 AND 10 
     [COLOR=blue]THEN[/color] [COLOR=#FF00FF]Convert[/color]([COLOR=blue]VarChar[/color](20), [COLOR=#FF00FF]DATEDIFF[/color](n, mbx.date_received, [COLOR=#FF00FF]GETDATE[/color]())) + [COLOR=red]' min ago'[/color]
     [COLOR=blue]END[/color] [COLOR=blue]as[/color] USE_DATE

DateDiff always returns an integer, so you need to convert that integer to a varchar and append it to your ' min ago' string.

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