Question

我尝试计算两点之间的距离（给定十进制格式的那些点的纬度/经度）。

VBA代码：

Const pi = 3.14159265358979

Function distance(lat1, lon1, lat2, lon2)
Dim theta, dist
theta = lon1 - lon2
dist = Sin(deg2rad(lat1)) * Sin(deg2rad(lat2)) + Cos(deg2rad(lat1)) * Cos(deg2rad(lat2)) * Cos(deg2rad(theta))
dist = acos(dist)
dist = rad2deg(dist)
distance = (dist * 60 * 1.1515) * 1.609344
End Function

Function acos(rad)
If Abs(rad) <> 1 Then
acos = pi / 2 - Atn(rad / Sqr(1 - rad * rad))
ElseIf rad = -1 Then
acos = pi
End If
End Function

Function deg2rad(deg)
deg2rad = CDbl(deg * pi / 180)
End Function

Function rad2deg(rad)
rad2deg = CDbl(rad * 180 / pi)
End Function

我得到错误运行时错误'94'：在此行“deg2rad = CDbl（deg * pi / 180）”中无效使用Null。我写了一个查询：

SELECT DISTINCT 

([band].E_laip+([band].E_min*(1/60))+([band].E_sec*(1/3600))) AS Band_E_dec, 
([band2].E_laip+([band2].E_min*(1/60))+([band2].E_sec*(1/3600))) AS Band2_E_dec,
([band].N_laip+([band].N_min*(1/60))+([band].N_sec*(1/3600))) AS Band_N_dec, 
([band2].N_laip+([band2].N_min*(1/60))+([band2].N_sec*(1/3600))) AS Band2_N_dec,

distance([Band_N_dec],[Band_E_dec],[Band2_N_dec],[Band2_E_dec]) AS Atstumas

FROM [band] LEFT JOIN band2 ON [band].Stotis = band2.Stotis;

也许有人有想法？提前谢谢。

Answer 1

如果您没有匹配的行，则在LEFT JOIN ed表中使用JOIN，允许空值。我猜你的band表格中至少有一条记录band2字段stotis中没有匹配的行。

要了解，请执行：

SELECT *
FROM band
LEFT JOIN band2
ON [band].Stotis = band2.Stotis
WHERE band2.stotis IS NULL

如果您有任何点击，则会出现参考数据问题。

您的解决方案是使用LEFT JOIN执行WHERE Band2.stotis IS NOT NULL，或者使用相同的条件执行INNER JOIN，这只会返回两个表中的匹配行。

访问VBA：运行时错误'94'：无效使用Null

1 个答案: