为下一步添加经典asp
For next to add with classic asp
下一个 "A to Z" 得到结果。
为了进入; 5 和数字 6,我想补充一下。
怎么做。
for k = asc("A") to asc("Z")
response.write chr(k)
next
结果:
一个
乙
C
..
Z
我要
一个
乙
C
..
Z
5个
6
例如(
k = asc("A") 到 asc("Z") 添加“5”并添加“6”)
谢谢。
strString = "ABCDEFGHIJKLMNOPQRSTUVWXYZ56"
For i=1 To Len(strString)
Response.Write Mid(strString,i,1)
Next
你不能真的有一个循环,只需添加单独的命令:
for k = asc("A") to asc("Z")
response.write chr(k)
next
response.write "5"
response.write "6"
另一种选择是将所有 ASCII 数字存储在数组中,然后循环该数组:
Dim arrLetters(), x
ReDim arrLetters(-1)
For k=Asc("A") To Asc("Z")
ReDim Preserve arrLetters(UBound(arrLetters) + 1)
arrLetters(UBound(arrLetters)) = k
Next
ReDim Preserve arrLetters(UBound(arrLetters) + 1)
arrLetters(UBound(arrLetters)) = Asc("5")
ReDim Preserve arrLetters(UBound(arrLetters) + 1)
arrLetters(UBound(arrLetters)) = Asc("6")
For x=0 To UBound(arrLetters)
k = arrLetters(x)
response.write chr(k)
Next
Erase arrLetters
下一个 "A to Z" 得到结果。 为了进入; 5 和数字 6,我想补充一下。 怎么做。
for k = asc("A") to asc("Z")
response.write chr(k)
next
结果:
一个 乙 C .. Z
我要 一个 乙 C .. Z 5个 6
例如( k = asc("A") 到 asc("Z") 添加“5”并添加“6”)
谢谢。
strString = "ABCDEFGHIJKLMNOPQRSTUVWXYZ56"
For i=1 To Len(strString)
Response.Write Mid(strString,i,1)
Next
你不能真的有一个循环,只需添加单独的命令:
for k = asc("A") to asc("Z")
response.write chr(k)
next
response.write "5"
response.write "6"
另一种选择是将所有 ASCII 数字存储在数组中,然后循环该数组:
Dim arrLetters(), x
ReDim arrLetters(-1)
For k=Asc("A") To Asc("Z")
ReDim Preserve arrLetters(UBound(arrLetters) + 1)
arrLetters(UBound(arrLetters)) = k
Next
ReDim Preserve arrLetters(UBound(arrLetters) + 1)
arrLetters(UBound(arrLetters)) = Asc("5")
ReDim Preserve arrLetters(UBound(arrLetters) + 1)
arrLetters(UBound(arrLetters)) = Asc("6")
For x=0 To UBound(arrLetters)
k = arrLetters(x)
response.write chr(k)
Next
Erase arrLetters