错误的参数数量SQL MSACCESS

Question

运行以下查询时出现错误，指示错误的参数数量：

SELECT
population_postcodes.*, 
target_postcodes.*, 
SQR( EXP(population_postcodes.longitude- target_postcodes.longitude, 2) + EXP(population_postcodes.latitude-target_postcodes.latitude, 2) ) as distance
FROM population_postcodes INNER JOIN target_postcodes on Population_postcodes.Population_postcode = Target_postcodes.Target_postcode;

有人可以建议我如何解决此问题吗？

我还尝试了以下代码：

SELECT Population_postcodes.*, Target_postcodes.* 

FROM population_postcodes
INNER JOIN target_postcodes
ON Population_postcodes.Population_postcode = Target_postcodes.Target_postcode
SQR( (population_postcodes.longitude- target_postcodes.longitude)^2 + (population_postcodes.latitude-target_postcodes.latitude)^2 ) as distance;

这段代码：

     SELECT Population_postcodes.*, Target_postcodes.*, SQR( (population_postcodes.longitude- target_postcodes.longitude)^2 + (population_postcodes.latitude-target_postcodes.latitude)^2 ) as distance
FROM population_postcodes
INNER JOIN target_postcodes
ON Population_postcodes.Population_postcode = Target_postcodes.Target_postcode;

Answer 1

Exp需要一个参数，您需要两个。

旧版： EXP(population_postcodes.longitude- target_postcodes.longitude, 2)

新增内容： (population_postcodes.longitude- target_postcodes.longitude)*(population_postcodes.longitude- target_postcodes.longitude)

Answer 2

尝试更换...

EXP(<expression>, 2)

...至...

<expression>^2

在Access中，EXP函数将e（自然对数的底数）返回为幂。 要将表达式提升为幂，请使用^运算符。

根据您的情况，请小心在表达式两边加上方括号，例如...

(population_postcodes.longitude- target_postcodes.longitude)^2

...强制使用最后的电源。 默认情况下，^运算符在-运算符之前进行求值。