[RESOLVED] Mid function

[Academy]

Hi,

I hope somebody in the forum knows VB.6. I am looking for a function that does the samething as Mid function in VB.6. Please see the below sample.

Dim MyString, FirstWord
MyString = "Mid Function Demo" ' Create text string.
FirstWord = Mid(MyString, 1, 3) ' Returns "Mid".


Thanks.

[ComputerJy]

What does it do? your example couldn't be more vague

[CornedBee]

It's like String.substring, except I think Mid's last argument is the length of the substring, not the end index like in Java.

[ComputerJy]

I don't think so, I found this example:

' Creates text string.
Dim TestString As String = "Mid Function Demo"
' Returns "Mid".
Dim FirstWord As String = Mid(TestString, 1, 3)
' Returns "Demo".
Dim LastWord As String = Mid(TestString, 14, 4)
' Returns "Function Demo".
Dim MidWords As String = Mid(TestString, 5)

[CornedBee]

Oh yeah, Mid is 1-based, while String.substring is 0-based. But aside from that, you just proved my point.

[ComputerJy]

Oh yeah, Mid is 1-based, while String.substring is 0-based. But aside from that, you just proved my point.

err....I don't know
Why do they call it MID if it has nothing to do with Middle!!

[CornedBee]

Well, it has. It takes a part in the middle of the string. As opposed to Beg and End, or whatever they're called, which take a part from the beginning and end of the string, respectively.
Why the designers of VB thought it necessary to have 3 functions is beyond me.

In Java, you do it all with String.substring(), String.length() and a bit of math that a 7-year-old can do.

[Academy]

MyString = "Mid Function Demo" ' Create text string.
FirstWord = Mid(MyString, 1, 3) ' Returns "Mid".


It returns a string from MyString, starts from postion 1 and 3 characters long.

[ComputerJy]

This is how its done in Java

    String MyString, FirstWord ;
    MyString = "Mid Function Demo" ;
    FirstWord = MyString.substring( 0, 3 ) ;

[Academy]

Thanks

[CornedBee]

And my posts didn't tell you this because ...?

[AlnavPlatinum]

it actually does not do what you have talked (above). With the below code, it should start at a position 2 and returns 5 characters.

System.out.println(strFirstBigIn.substring(2,5)+"");

See the out put for more explaination:

Please enter your first BigInt

123456789
345

[|2eM!x]

it actually does not do what you have talked (above). With the below code, it should start at a position 2 and returns 5 characters.

System.out.println(strFirstBigIn.substring(2,5)+"");

See the out put for more explaination:

Please enter your first BigInt

123456789
345

Im not seeing that anywhere in the thread..?

It goes from position 2 to position 5 which is 3 characters..where is the problem?