请教关于多维数组变量类型转换赋值的问题

[aojin54]

dim url(100,4) as String
dim zxd(100) as Integer
dim z as integer
..........
..........
给url(100,4)赋值
..........
For z = 0 To 100
zxd(z) = CInt(Url(z, 0))

Next z
程序编译执行时总是说：实施错误13，类型不匹配
有那位大虾知道怎么样正确的赋值，请告诉我

[SoHo_Andy]

我试过了，没问题，是不是你的Url初始化有问题

测试通过的程序
Private Sub Command1_Click()
Dim url(100, 4) As String
Dim zxd(100) As Integer
Dim z As Integer
For z = 0 To 100
url(z, 0) = z
Next
For z = 0 To 100
zxd(z) = CInt(url(z, 0))
Next z
End Sub

测试不通过的，报类型不匹配错误
Private Sub Command1_Click()
Dim url(100, 4) As String
Dim zxd(100) As Integer
Dim z As Integer
For z = 0 To 100
url(z, 0) = z & "和"
Next
For z = 0 To 100
zxd(z) = CInt(url(z, 0))
Next z
End Sub

区别就在于url初始化的值要是能够转化为Int类型的值
你的程序出错应该也是这个原因

[TechnoFantasy]

使用Val，不要用CInt。或者将dim url(100,4) as String改为dim url(100,4) as Variant

[TechnoFantasy]

Val的作用在于即时你的字符串并不是一个数字字符串也不会抱错 :=)