vba convert string to int with string not a number
0
I need to convert a string into a numerical value.
However, my string has not a number in it. So I define the numerical value of this string in my program.
This is the test I am performing:
-
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
The main issue is that I don't understand why the value b and d are not taken into account.
Moreover a message error : "type mismatch" appear when I run the program.
Does anyone can explain me where is my mistake?
Thanks a lot
excel string excel-vba type-conversion integer
edited Nov 12 '18 at 14:34
Davesexcel
5,07921936
asked Nov 12 '18 at 13:06
A.MNIERA.MNIER
1
add a comment |
0
I need to convert a string into a numerical value.
However, my string has not a number in it. So I define the numerical value of this string in my program.
This is the test I am performing:
-
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
The main issue is that I don't understand why the value b and d are not taken into account.
Moreover a message error : "type mismatch" appear when I run the program.
Does anyone can explain me where is my mistake?
Thanks a lot
excel string excel-vba type-conversion integer
edited Nov 12 '18 at 14:34
Davesexcel
5,07921936
asked Nov 12 '18 at 13:06
A.MNIERA.MNIER
1
1
You are trying to do calculations on the letter "b".
– SJR
Nov 12 '18 at 13:16
add a comment |
0
0
0
I need to convert a string into a numerical value.
However, my string has not a number in it. So I define the numerical value of this string in my program.
This is the test I am performing:
-
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
The main issue is that I don't understand why the value b and d are not taken into account.
Moreover a message error : "type mismatch" appear when I run the program.
Does anyone can explain me where is my mistake?
Thanks a lot
excel string excel-vba type-conversion integer
edited Nov 12 '18 at 14:34
Davesexcel
5,07921936
asked Nov 12 '18 at 13:06
A.MNIERA.MNIER
1
I need to convert a string into a numerical value.
However, my string has not a number in it. So I define the numerical value of this string in my program.
This is the test I am performing:
-
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
The main issue is that I don't understand why the value b and d are not taken into account.
Moreover a message error : "type mismatch" appear when I run the program.
Does anyone can explain me where is my mistake?
Thanks a lot
excel string excel-vba type-conversion integer
excel string excel-vba type-conversion integer
edited Nov 12 '18 at 14:34
Davesexcel
5,07921936
asked Nov 12 '18 at 13:06
A.MNIERA.MNIER
1
edited Nov 12 '18 at 14:34
Davesexcel
5,07921936
asked Nov 12 '18 at 13:06
A.MNIERA.MNIER
1
edited Nov 12 '18 at 14:34
Davesexcel
5,07921936
edited Nov 12 '18 at 14:34
Davesexcel
5,07921936
edited Nov 12 '18 at 14:34
Davesexcel
5,07921936
5,07921936
asked Nov 12 '18 at 13:06
A.MNIERA.MNIER
1
asked Nov 12 '18 at 13:06
A.MNIERA.MNIER
1
asked Nov 12 '18 at 13:06
A.MNIERA.MNIER
1
1
1
You are trying to do calculations on the letter "b".
– SJR
Nov 12 '18 at 13:16
add a comment |
1
You are trying to do calculations on the letter "b".
– SJR
Nov 12 '18 at 13:16
1
1
You are trying to do calculations on the letter "b".
– SJR
Nov 12 '18 at 13:16
You are trying to do calculations on the letter "b".
– SJR
Nov 12 '18 at 13:16
add a comment |
2 Answers
2
active
oldest
votes
0
Convert them back to the variables.
ValCellY = IIf(Cells(StartNumber, 2).Value = "b", b, d)
ValCellZ = IIf(Cells(StartNumber, 3).Value = "b", b, d)
answered Nov 12 '18 at 13:23
DavesexcelDavesexcel
5,07921936
Hello, I am not sure to understand the purpose of this. Can you explain me? Thanks
– A.MNIER
Nov 12 '18 at 15:04
It makes ValCellY & ValCellZ equal to the variables in your code, did you not try it?
– Davesexcel
Nov 12 '18 at 15:40
Yes I tried and it works! Thanks you so much
– A.MNIER
Nov 13 '18 at 12:14
You need to click this as the answer.
– Davesexcel
Nov 13 '18 at 16:02
add a comment |
0
You are using strings to calculate instead of actual values. You might use if, or select case but if your actual problem is more complex then I suggest using dictionary.
Option Explicit
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
If ValCellY = "b" Then
ValCellY = b
ElseIf ValCellY = "d" Then
ValCellY = d
End If
If ValCellZ = "b" Then
ValCellZ = b
ElseIf ValCellZ = "d" Then
ValCellZ = d
End If
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
answered Nov 12 '18 at 13:25
Pawel CzyzPawel Czyz
6502517
Thanks a lot it is working!
– A.MNIER
Nov 12 '18 at 14:57
Can you explain me what do you mean by dictionary?
– A.MNIER
Nov 12 '18 at 15:17
A dictionary stores unique keys and values assigned to keys. You can create dictionary and later refer to it by the key. Here is a nice article about dictionaries in VBA: excelmacromastery.com/vba-dictionary If my reply was helpful please approve the answer :)
– Pawel Czyz
Nov 12 '18 at 15:20
Thanks a lot, I will study it. It seems very interesting.
– A.MNIER
Nov 13 '18 at 11:27
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
Convert them back to the variables.
ValCellY = IIf(Cells(StartNumber, 2).Value = "b", b, d)
ValCellZ = IIf(Cells(StartNumber, 3).Value = "b", b, d)
answered Nov 12 '18 at 13:23
DavesexcelDavesexcel
5,07921936
Hello, I am not sure to understand the purpose of this. Can you explain me? Thanks
– A.MNIER
Nov 12 '18 at 15:04
It makes ValCellY & ValCellZ equal to the variables in your code, did you not try it?
– Davesexcel
Nov 12 '18 at 15:40
Yes I tried and it works! Thanks you so much
– A.MNIER
Nov 13 '18 at 12:14
You need to click this as the answer.
– Davesexcel
Nov 13 '18 at 16:02
add a comment |
0
Convert them back to the variables.
ValCellY = IIf(Cells(StartNumber, 2).Value = "b", b, d)
ValCellZ = IIf(Cells(StartNumber, 3).Value = "b", b, d)
answered Nov 12 '18 at 13:23
DavesexcelDavesexcel
5,07921936
Hello, I am not sure to understand the purpose of this. Can you explain me? Thanks
– A.MNIER
Nov 12 '18 at 15:04
It makes ValCellY & ValCellZ equal to the variables in your code, did you not try it?
– Davesexcel
Nov 12 '18 at 15:40
Yes I tried and it works! Thanks you so much
– A.MNIER
Nov 13 '18 at 12:14
You need to click this as the answer.
– Davesexcel
Nov 13 '18 at 16:02
add a comment |
0
0
0
Convert them back to the variables.
ValCellY = IIf(Cells(StartNumber, 2).Value = "b", b, d)
ValCellZ = IIf(Cells(StartNumber, 3).Value = "b", b, d)
answered Nov 12 '18 at 13:23
DavesexcelDavesexcel
5,07921936
Convert them back to the variables.
ValCellY = IIf(Cells(StartNumber, 2).Value = "b", b, d)
ValCellZ = IIf(Cells(StartNumber, 3).Value = "b", b, d)
answered Nov 12 '18 at 13:23
DavesexcelDavesexcel
5,07921936
answered Nov 12 '18 at 13:23
DavesexcelDavesexcel
5,07921936
answered Nov 12 '18 at 13:23
DavesexcelDavesexcel
5,07921936
answered Nov 12 '18 at 13:23
DavesexcelDavesexcel
5,07921936
5,07921936
Hello, I am not sure to understand the purpose of this. Can you explain me? Thanks
– A.MNIER
Nov 12 '18 at 15:04
It makes ValCellY & ValCellZ equal to the variables in your code, did you not try it?
– Davesexcel
Nov 12 '18 at 15:40
Yes I tried and it works! Thanks you so much
– A.MNIER
Nov 13 '18 at 12:14
You need to click this as the answer.
– Davesexcel
Nov 13 '18 at 16:02
add a comment |
Hello, I am not sure to understand the purpose of this. Can you explain me? Thanks
– A.MNIER
Nov 12 '18 at 15:04
It makes ValCellY & ValCellZ equal to the variables in your code, did you not try it?
– Davesexcel
Nov 12 '18 at 15:40
Yes I tried and it works! Thanks you so much
– A.MNIER
Nov 13 '18 at 12:14
You need to click this as the answer.
– Davesexcel
Nov 13 '18 at 16:02
Hello, I am not sure to understand the purpose of this. Can you explain me? Thanks
– A.MNIER
Nov 12 '18 at 15:04
Hello, I am not sure to understand the purpose of this. Can you explain me? Thanks
– A.MNIER
Nov 12 '18 at 15:04
It makes ValCellY & ValCellZ equal to the variables in your code, did you not try it?
– Davesexcel
Nov 12 '18 at 15:40
It makes ValCellY & ValCellZ equal to the variables in your code, did you not try it?
– Davesexcel
Nov 12 '18 at 15:40
Yes I tried and it works! Thanks you so much
– A.MNIER
Nov 13 '18 at 12:14
Yes I tried and it works! Thanks you so much
– A.MNIER
Nov 13 '18 at 12:14
You need to click this as the answer.
– Davesexcel
Nov 13 '18 at 16:02
You need to click this as the answer.
– Davesexcel
Nov 13 '18 at 16:02
add a comment |
0
You are using strings to calculate instead of actual values. You might use if, or select case but if your actual problem is more complex then I suggest using dictionary.
Option Explicit
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
If ValCellY = "b" Then
ValCellY = b
ElseIf ValCellY = "d" Then
ValCellY = d
End If
If ValCellZ = "b" Then
ValCellZ = b
ElseIf ValCellZ = "d" Then
ValCellZ = d
End If
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
answered Nov 12 '18 at 13:25
Pawel CzyzPawel Czyz
6502517
Thanks a lot it is working!
– A.MNIER
Nov 12 '18 at 14:57
Can you explain me what do you mean by dictionary?
– A.MNIER
Nov 12 '18 at 15:17
A dictionary stores unique keys and values assigned to keys. You can create dictionary and later refer to it by the key. Here is a nice article about dictionaries in VBA: excelmacromastery.com/vba-dictionary If my reply was helpful please approve the answer :)
– Pawel Czyz
Nov 12 '18 at 15:20
Thanks a lot, I will study it. It seems very interesting.
– A.MNIER
Nov 13 '18 at 11:27
add a comment |
0
You are using strings to calculate instead of actual values. You might use if, or select case but if your actual problem is more complex then I suggest using dictionary.
Option Explicit
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
If ValCellY = "b" Then
ValCellY = b
ElseIf ValCellY = "d" Then
ValCellY = d
End If
If ValCellZ = "b" Then
ValCellZ = b
ElseIf ValCellZ = "d" Then
ValCellZ = d
End If
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
answered Nov 12 '18 at 13:25
Pawel CzyzPawel Czyz
6502517
Thanks a lot it is working!
– A.MNIER
Nov 12 '18 at 14:57
Can you explain me what do you mean by dictionary?
– A.MNIER
Nov 12 '18 at 15:17
A dictionary stores unique keys and values assigned to keys. You can create dictionary and later refer to it by the key. Here is a nice article about dictionaries in VBA: excelmacromastery.com/vba-dictionary If my reply was helpful please approve the answer :)
– Pawel Czyz
Nov 12 '18 at 15:20
Thanks a lot, I will study it. It seems very interesting.
– A.MNIER
Nov 13 '18 at 11:27
add a comment |
0
0
0
You are using strings to calculate instead of actual values. You might use if, or select case but if your actual problem is more complex then I suggest using dictionary.
Option Explicit
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
If ValCellY = "b" Then
ValCellY = b
ElseIf ValCellY = "d" Then
ValCellY = d
End If
If ValCellZ = "b" Then
ValCellZ = b
ElseIf ValCellZ = "d" Then
ValCellZ = d
End If
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
answered Nov 12 '18 at 13:25
Pawel CzyzPawel Czyz
6502517
You are using strings to calculate instead of actual values. You might use if, or select case but if your actual problem is more complex then I suggest using dictionary.
Option Explicit
Sub TestFinal()
Dim Result As Integer
Dim StartNumber As Integer
Dim EndNumber As Integer
Dim b As Variant
Dim d As Variant
Dim ValCellY As String
Dim ValCellZ As String
StartNumber = 2 'First line value of my table
EndNumber = 3 'Last line value of my table
b = 80
d = 20
For StartNumber = 2 To EndNumber 'Begining of my loops
ValCellY = Cells(StartNumber, 2).Value
ValCellZ = Cells(StartNumber, 3).Value
If Cells(StartNumber, 1) = "Yes" Then 'First if condition
Result = 100
End If
If Cells(StartNumber, 1) = "No" Then 'Second if condition
If ValCellY = "b" Then
ValCellY = b
ElseIf ValCellY = "d" Then
ValCellY = d
End If
If ValCellZ = "b" Then
ValCellZ = b
ElseIf ValCellZ = "d" Then
ValCellZ = d
End If
Result = 75 * CInt(ValCellY) + 25 * CInt(ValCellZ) 'I use CInt function to convert my string into a integer
End If
Cells(StartNumber, 4).Value = Result 'I associate the varaible Result to the column D
Next StartNumber 'End of my loops
End Sub
answered Nov 12 '18 at 13:25
Pawel CzyzPawel Czyz
6502517
answered Nov 12 '18 at 13:25
Pawel CzyzPawel Czyz
6502517
answered Nov 12 '18 at 13:25
Pawel CzyzPawel Czyz
6502517
answered Nov 12 '18 at 13:25
Pawel CzyzPawel Czyz
6502517
6502517
Thanks a lot it is working!
– A.MNIER
Nov 12 '18 at 14:57
Can you explain me what do you mean by dictionary?
– A.MNIER
Nov 12 '18 at 15:17
A dictionary stores unique keys and values assigned to keys. You can create dictionary and later refer to it by the key. Here is a nice article about dictionaries in VBA: excelmacromastery.com/vba-dictionary If my reply was helpful please approve the answer :)
– Pawel Czyz
Nov 12 '18 at 15:20
Thanks a lot, I will study it. It seems very interesting.
– A.MNIER
Nov 13 '18 at 11:27
add a comment |
Thanks a lot it is working!
– A.MNIER
Nov 12 '18 at 14:57
Can you explain me what do you mean by dictionary?
– A.MNIER
Nov 12 '18 at 15:17
A dictionary stores unique keys and values assigned to keys. You can create dictionary and later refer to it by the key. Here is a nice article about dictionaries in VBA: excelmacromastery.com/vba-dictionary If my reply was helpful please approve the answer :)
– Pawel Czyz
Nov 12 '18 at 15:20
Thanks a lot, I will study it. It seems very interesting.
– A.MNIER
Nov 13 '18 at 11:27
Thanks a lot it is working!
– A.MNIER
Nov 12 '18 at 14:57
Thanks a lot it is working!
– A.MNIER
Nov 12 '18 at 14:57
Can you explain me what do you mean by dictionary?
– A.MNIER
Nov 12 '18 at 15:17
Can you explain me what do you mean by dictionary?
– A.MNIER
Nov 12 '18 at 15:17
A dictionary stores unique keys and values assigned to keys. You can create dictionary and later refer to it by the key. Here is a nice article about dictionaries in VBA: excelmacromastery.com/vba-dictionary If my reply was helpful please approve the answer :)
– Pawel Czyz
Nov 12 '18 at 15:20
A dictionary stores unique keys and values assigned to keys. You can create dictionary and later refer to it by the key. Here is a nice article about dictionaries in VBA: excelmacromastery.com/vba-dictionary If my reply was helpful please approve the answer :)
– Pawel Czyz
Nov 12 '18 at 15:20
Thanks a lot, I will study it. It seems very interesting.
– A.MNIER
Nov 13 '18 at 11:27
Thanks a lot, I will study it. It seems very interesting.
– A.MNIER
Nov 13 '18 at 11:27
add a comment |
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1
You are trying to do calculations on the letter "b".
– SJR
Nov 12 '18 at 13:16