Count Nos in string

[billa856]

Hi,
I have string like below and i wanna count nos in that string

1st string ;0;80;0;0;39;0;0;0;81;42;5;100;0;0 in this string there are 6 Nos.

2nd string ;11;0;5;2;0;0;0;0;89;65;0;26;0;2;5 in this line there are 8 Nos.

3rd string ;0;0;0;0;50;41;20;30;0;50;0 in this line there are 5 Nos.

I can count total nos in string

1st string -14 Nos.
2nd string -15 Nos.
3rd String - 11 Nos.

can anyone help please?
Thank You.

[missinglinq]

Assuming you're running Acc2000 or later version:

If your “Nos” are not preceded by a semi-colon, i.e.

0;80;0;0;39;0;0;0;81;42;5;100;0;0

NosCount = (Len([YourString]) - Len(Replace([YourString], ","; ""))) + 1
 

If your “Nos” are preceded by a semi-colon, i.e

;0;80;0;0;39;0;0;0;81;42;5;100;0;0

NosCount =  Len([YourString]) - Len(Replace([YourString], ","; ""))
 

Linq ;0)>

[billa856]

Assuming you're running Acc2000 or later version:

If your “Nos” are not preceded by a semi-colon, i.e.

0;80;0;0;39;0;0;0;81;42;5;100;0;0

NosCount = (Len([YourString]) - Len(Replace([YourString], ","; ""))) + 1
 

If your “Nos” are preceded by a semi-colon, i.e

;0;80;0;0;39;0;0;0;81;42;5;100;0;0

NosCount =  Len([YourString]) - Len(Replace([YourString], ","; ""))
 

Linq ;0)>

There is an error missing list operator.can u plz check its format?

[billa856]

I used this fuction and its working fine.

Public Function test(strToTest As String) As Integer
Dim strTest As String
Dim ara() As String
Dim I As Integer
Dim Count As Integer
strTest = strToTest
ara = Split(strTest, ";")
Count = 0
For I = 0 To UBound(ara)
If ara(I) = "0 " Or ara(I) = ";0 " Or ara(I) = "" Then
Else
Count = Count + 1
End If
Next I

test = Count
End Function

By the way thanks for ur reply.

[ADezii]

I used this fuction and its working fine.
Here is another option for you:
Public Function test(strToTest As String) As Integer
Dim strTest As String
Dim ara() As String
Dim I As Integer
Dim Count As Integer
strTest = strToTest
ara = Split(strTest, ";")
Count = 0
For I = 0 To UBound(ara)
If ara(I) = "0 " Or ara(I) = ";0 " Or ara(I) = "" Then
Else
Count = Count + 1
End If
Next I

test = Count
By the way thanks for ur reply.

Here is another option for you:

Public Function test(strToTest As String) As Integer
Dim ara As Variant
Dim I As Integer
Dim Count As Integer
 
ara = Split(strToTest, ";")
 
Count = 0
 
For I = LBound(ara) To UBound(ara)
  If IsNumeric(ara(I)) And ara(I) > 0 Then
    Count = Count + 1
  Else
  End If
Next I
 
test = Count
End Function

[missinglinq]

Sorry I am late getting back to you. The code I posted worked fine in testing, and is, in fact, something I've used for this kind of thing a number of times. My guess is that you either have a typo when you use the code or all of your data was formatted the same as your example.

At any rate, I'm glad you found something that works!

Linq ;0)>