SQL access “IIF” 问题

来源:百度知道 编辑:UC知道 时间:2024/06/21 09:20:29
TCOO: IIf(COST_US_CURRENCY ="U","USA",
IIf[COST_US_CURRENCY ="L","MAL","Other"])

COST_US_CURRENCY , COST_US_CURRENCY = 是一个table

可是一直处一个警告~我哪里写错呢?
IIf(COST_US_CURRENCY ="U","USA",IIf(COST_US_CURRENCY ="L","MAL","Other"))
这个试过了~还是一样的error出现~~

IIf(COST_US_CURRENCY ="U","USA",IIf(COST_US_CURRENCY ="L","MAL","Other"))

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那么请楼主写清你的表结构,主要是表名和你想用作判断的字段名。

COST_US_CURRENCY , COST_US_CURRENCY = 是一个table,那么table怎么可能是一个值呢?你这个iif函数的判断出问题了,肯定报错。
如果改成字段就没问题,如IIf(字段1="U","USA", IIf[字段2="L","MAL","Other"])

=(IIf[COST_US_CURRENCY ="L","MAL","Other"])
要加“=”

那是一个表