MYSQL if

Sveiki,

iskilo problema su mysql if, googleje ir bandziau ieskot, bet nelabai sekesi.

sql kodas:

ORDER BY  if( `lift` > '".today()."' , ( `vip2`  , `lift` ) , ( `lift` DESC) )

kai vygdo sql uzklausa ismeta err:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'DESC) ) LIMIT 0,25' at line 2

Pakeitus koda i (nors ir desc yra butinas):

ORDER BY  if( `lift` > '".today()."' , ( `vip2`  , `lift` ) , ( `lift`) )

Meta: Operand should contain 1 column(s)

lift -> timestampas

vip2 - > [0-9]

today() - > siandienos data

dar bandziau su : ORDER BY concat(`vip2`,lift) DESC

,bet man reikia kad desc nebutu jei `lift` > '".today()."'

Gal padesit su sia sintakse, aciu

Pabandyk taip:

ORDER BY if(`lift` > '".today()."', '`vip2`  , `lift`' , '`lift` DESC')

Pabandyk taip:

ORDER BY if(`lift` > '".today()."', '`vip2`  , `lift`' , '`lift` DESC')

Bandziau, bet sql nereguoja i tai kas parasyta tarp ' ' ,nes nenuskaito tada reiksmiu.

(tapati rodo parasius ORDER BY 'lift' arba ORDER BY 'lift desc')

gal dar ka pasiulysit?