Vista - RE: searching for a pattern in a string!!

Hi,

I am using for example:

str = "04234329"
pattern= "40"

Code:
result=instr(str,pattern)

to search. So if I am searching for 40 and it finds 04 and says it
found 40 which is incorrect as 04 and 40 are not the same. Is there a
way around this problem. Thanks for helping

SA

On Nov 14, 1:18*pm, sa <agarwasa2...@xxxxxx> wrote:Actually the str is not correctly defined earlier. It is

str="030502012715290406281"
pattern="30"

So it correctly identifies 30 using the instr code. The problem is
that the str is a combination of 2 chars as shown below:
03
05
02
01
27
15
29
04 etc

So as can be seen its 03 and then 05 and its combined into 0305 and
the instr function finds 30 which is not what I want. Is there a way
around this ehre it can search in 2's only for a search pattern

SA

If the search string is always composed of 2-character
pairs then you can do it like this:

s1 = "03040608"
s2 = "40"

i = InStr(s1, s2)
If i Mod 2 = 1 Then
MsgBox s2 & " found at offset " & i
End If

The Mod operator returns the remainder of division.
So x Mod 2 will return 0 for even numbers and 1 for
odd numbers. Since your pair always starts at an odd
number offset you can check a valid match by using
i Mod 2.

If there's a possibility of finding more than one
instance then you can use a loop. For example,
in "04011440" there's an invalid "40" at offset 2
and a valid "40" at offset 7. To find the first valid
match where there might also be invalid matches,
you can use a function like so:

'------------------------------------

s1 = "04060508"
s2 = "40"

MsgBox GetOffset(s1, s2)

Function GetOffset(s1, sToFind)
Dim Pt1, Pt2
Pt1 = 1
Pt2 = 0
GetOffset = 0
Do
Pt2 = InStr(Pt1, s1, sToFind)
If Pt2 = 0 Then Exit Function
If (Pt2 Mod 2 = 1) Then
GetOffset = Pt2
Exit Function
End If
Pt1 = Pt2 + 1
If Pt1 >= Len(s1) Then Exit Function
Loop
End Function

'-------------------------------

Apparently you can also use RegExp. Personally I
find RegExp to be very tedious, and overkill in the vast
majority of cases. (Such as the case at hand.) But it
seems to be a love/hate kind of thing. Some people hate
RegExp while other people like to use them in any function
where they can be shoehorned in. I suspect the RegExp
lovers are the same people who like to work out 85-character
Perl expressions that will format a disk, take out the
rubbish and wash the dishes, all in a matter of milliseconds.

Actually the str is not correctly defined earlier. It is

str="030502012715290406281"
pattern="30"

So it correctly identifies 30 using the instr code. The problem is
that the str is a combination of 2 chars as shown below:
03
05
02
01
27
15
29
04 etc

So as can be seen its 03 and then 05 and its combined into 0305 and
the instr function finds 30 which is not what I want. Is there a way
around this ehre it can search in 2's only for a search pattern

SA