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#1
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a string to an integer
I have a string:
string(3) = "543" I try to change it to an integer: int = cint(string(3)) It gives me a 'Type Mismatch' Error. Maybe there may be spaces before of after the numbers but does that matter? How do I change from a string to an integer? |
#2
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a string to an integer
Spaces do matter. Try int = cint(trim(string(3)) "Philosophaie" wrote: I have a string: string(3) = "543" I try to change it to an integer: int = cint(string(3)) It gives me a 'Type Mismatch' Error. Maybe there may be spaces before of after the numbers but does that matter? How do I change from a string to an integer? |
#3
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a string to an integer
There were no spaces. I checked. Why else am I getting 'Type Mismatch' Error
and not letting me use cint to change from a string? |
#4
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a string to an integer
I've just done this:
Sub test() Dim s As String s = "543" Dim i As Integer i = CInt(Trim(s)) MsgBox i End Sub and it works. Have you ot an apostrophe/single quote ' preceeding the 5? "Philosophaie" wrote: There were no spaces. I checked. Why else am I getting 'Type Mismatch' Error and not letting me use cint to change from a string? |
#5
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a string to an integer
Don't use String and Int as variable names.
String is a VBA function and data type. Int is a VBA function. Dim myString(1 To 3) As String Dim myInt As Long myString(3) = "543" myInt = CInt(myString(3)) Personally, I wouldn't use "as integer" or cInt(). I'd use "As long" and clng(). Philosophaie wrote: I have a string: string(3) = "543" I try to change it to an integer: int = cint(string(3)) It gives me a 'Type Mismatch' Error. Maybe there may be spaces before of after the numbers but does that matter? How do I change from a string to an integer? -- Dave Peterson |
#6
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a string to an integer
I am using what is already in the cell from "Left and Right" operations to a
bunch of similar data strings. I have ran my own trimming operation thru if left(a,1)="-" then a=Right(a,2) but it will still not let me take the cint or cdbl of any of the data. So the quote sign is irrelevant. |
#7
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a string to an integer
Ok... another approach - try
msgbox len(str) * " """ & str & """" which should display 3 "543" to make sure there are no extra characters in there. Sam "Philosophaie" wrote: I am using what is already in the cell from "Left and Right" operations to a bunch of similar data strings. I have ran my own trimming operation thru if left(a,1)="-" then a=Right(a,2) but it will still not let me take the cint or cdbl of any of the data. So the quote sign is irrelevant. |
#8
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a string to an integer
I typed verbatum:
msgbox len(str) * " """ & str & """" even that gave me a 'Type Mismatch error' "Sam Wilson" wrote: Ok... another approach - try msgbox len(str) * " """ & str & """" which should display 3 "543" to make sure there are no extra characters in there. Sam "Philosophaie" wrote: I am using what is already in the cell from "Left and Right" operations to a bunch of similar data strings. I have ran my own trimming operation thru if left(a,1)="-" then a=Right(a,2) but it will still not let me take the cint or cdbl of any of the data. So the quote sign is irrelevant. |
#9
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a string to an integer
msgbox len(str) & " """ & str & """"
Sorry, my mistake. Change str for whatever string it is you're trying to convert. Sam "Philosophaie" wrote: I typed verbatum: msgbox len(str) * " """ & str & """" even that gave me a 'Type Mismatch error' "Sam Wilson" wrote: Ok... another approach - try msgbox len(str) * " """ & str & """" which should display 3 "543" to make sure there are no extra characters in there. Sam "Philosophaie" wrote: I am using what is already in the cell from "Left and Right" operations to a bunch of similar data strings. I have ran my own trimming operation thru if left(a,1)="-" then a=Right(a,2) but it will still not let me take the cint or cdbl of any of the data. So the quote sign is irrelevant. |
#10
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a string to an integer
they all had len=2 and nothing looked out of the ordinary.
"Sam Wilson" wrote: msgbox len(str) & " """ & str & """" Sorry, my mistake. Change str for whatever string it is you're trying to convert. |
#11
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a string to an integer
Also String is an intrinsic function in VB!
Use Option Explicit at the top of your module and make sure you dim all of your variables using names that are not intrinsic functions. "Philosophaie" wrote: I have a string: string(3) = "543" I try to change it to an integer: int = cint(string(3)) It gives me a 'Type Mismatch' Error. Maybe there may be spaces before of after the numbers but does that matter? How do I change from a string to an integer? |
#12
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a string to an integer
Type mismatch probably comes from the fact you are using "int"
int = CInt() Int is an intrinsic function in VB. You shouldn't name a variable "int" As for the other problem use Mid not Right to get the remaining portion of the string if left(a,1)="-" then a=mid(a,2) "Philosophaie" wrote: I have a string: string(3) = "543" I try to change it to an integer: int = cint(string(3)) It gives me a 'Type Mismatch' Error. Maybe there may be spaces before of after the numbers but does that matter? How do I change from a string to an integer? |
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