Left - Right - Mid Functions

I am having a lot of trouble
understanding the Left - Right and Mid functions and how they work. If i gave you an example of one could you break it down and explain to me what each piece of code is doing?

Lets say that FirstString = < and SecondString = > Lets say TheWholeString = <L>

We will use the Mid Function As the example.

Text1.Text = Mid$(TheWholeString,_ InStr(TheWholeString,_
FirstString) + Len(FirstString),_ InStr(TheWholeString,_
SecondString) - (Len(FirstString) +_
InStr(TheWholeString, FirstString)))

I know after running this code, text1.text would = L
but i cant figure out how it does this. Can you explain
to me what each piece of code in the brackets is doing.
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Where did you come up with this?  Okay, let's break it down.  You have the Mid function with parameters a, b, c. thus Mid(a, b, c).

a is the string named TheWholeString or <L>

b is InStr(TheWholeString, FirstString) + Len(FirstString)  OR

    InStr(<L>, <) equals 1 since the character first appears in the first position of the target string <L>.  If the target string had been &&<<<L>> then the value of InStr(<L>, <) would be 3.

   Len(<) equals 1 or the length of the target string.

   Therefore b = 1 + 1 = 2

c is InStr(TheWholeString, SecondString) - (Len(FirstString) +                  InStr(TheWholeString, FirstString)))

   InStr(<L>, >) = 3

   Len(<) = 1

   InStr(<L>, <) = 1

So c = 3 - (1 + 1) = 1

Now back to our original formula

Mid(<L>, 2, 1) means start at position number 2 in the string <L> and take 1 charachter.  Thus the answer is L.

More examples:   Mid("Hello", 2, 3) = ell

                            Mid("Good Morning", 3, 6) = od Mor

                            InStr("Hello", l) = 3

                            InStr("Hello", o) = 5

                            Len("Good Morning") = 12

Hope this helps.

You need to look at the function in  algebraic terms... Always resolve things in parenthesis to get to the lowest absolute, which in this case is the Mid$ function:

Mid$ returns a substring of a string.  There are 3 arguments that are passed to Mid$:

      mid$(String_to_use, Start_of_substring, Length_of_substring)

Here is an algebraic breakdown of the formula. Paste the following into VB so it will make sense.  It will not show up properly here...

Dim FirstString, SecondString, TheWholeString
FirstString = "<"
SecondString = ">"
TheWholeString = "<L>"

text1.Text = Mid$(TheWholeString, InStr(TheWholeString, FirstString) + Len(FirstString), InStr(TheWholeString, SecondString) - (Len(FirstString) + InStr(TheWholeString, FirstString)))
'                                 \                                /   \              /  \                                 /    \              /   \                               /
'                                  \           returns            /     \   returns  /    \            returns            /      \   returns  /     \           returns           /
'                                                 1                  +         1       ,                  3                  -  (       1        +                 1                 )
'  Plug the numbers into the mid$ function, so it looks like this:
'       text1.Text = Mid$(TheWholeString, 1 + 1, 3 - (1 + 1))
'  Resolve the numbers again so it looks like this:
'       text1.Text = Mid$(TheWholeString, 2, 3 - 2)
'  Resolve the numbers again so it looks like this:
'       text1.Text = Mid$(TheWholeString, 2, 1)

Hope this helps!


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poorboyAuthor Commented:
I just made the formula up in case
you were wondering. Thanks for the help.
You made the formula up?? Ok...

Thanks for the points! Glad I could help!

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