VB 6 - Useful Vb Functions

VB Functions

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String Functions

• InStr
• Left and Right
• Mid
• Split
• InStrRev
• Ucase And LCase
• Join
• LTrim and RTrim
• Replace

Conversion Functions

• CInt
• CLng
• CDate
• CStr
• CBool
• CDbl

Dates And Time

• Date
• Time
• Now
  

Maths Functions

• IsNumeric
• Val

Other Functions

• TypeName

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InStr


Use -

VB Code:

instr([Start],[String 1],[String 2],[Compare Method])

Its not as bad as it looks firstly to keep this simple we can ignore the [Compare Method] and we will always set the [start] value to 1,

[String 1] = Is the string we will be looking in

[String 2] = Is the string we will be looking for


Instr will return the position of [String 2] in [String 1]


Example

VB Code:

MsgBox InStr(1, "abcdefghi", "d")

Will return 4 because that is the position of [String2] ('d') within [String1]

Left And Right Functions


Left


Use –

Code:

---

[String2] = Left([String1],[Length])

---

This will put the the number of characters specified from [String1] (starting from the left) into [String2]


Example

Code:

---

mystring = "abcdefghijk"
    newstring = Left(mystring, 4)

---

newstring would now be “abcd”

Right

Is exactly the same as above but works from the right hand side

Example

Code:

---

mystring = "abcdefghijk"
    newstring = Right(mystring, 4)

---

newstring would now be “hijk”

Mid


The mid function is slightly different

Use –

Code:

---

String2 = Mid([String 1],[Start], [Length])

---

[String 1] is the string we will be looking in

[Start]Is the start position of where we will begin (See example)

[Length] Is the number of characters we will be copying


Example

Code:

---

mystring = "abcdefghijk"
    newstring = Mid(mystring, 5, 3)

---

So here we are starting from the 5th character along and taking 3 characters from that position so, newString would now be “efg”

Split

This is a very useful function, it separates certain parts of a string using a delimiter for this one the results must be stored in an array (Assume SplitString is an array)

Use -

Code:

---

SplitString = Split([Expression],[Delimiter])
    ‘ For now this is all we need to worry about

---

[Expression] is the string we will be splitting up

[Delimiter] is the character that we are using to break up each ‘part’ of the string

Example

Code:

---

mystring = "ab,cd,ef,gh,ijk"
    newstring = Split(mystring, ",")

---

Now how do we go about getting the values? Well because its an array Vb has indexed each ‘section’ in order so

Code:

---

newstring(0) = “ab”
   
        newstring(1) = “cd”
   
        newstring(2) = “ef”
   
        newstring(3) = “gh”
   
        newstring(4) = “ijk”

---

InStrRev

Use

Code:

---

result = InStrRev([StringCheck],[StringMatch])

---

This function returns 1 if [StringMatch] is contained in [StringCheck]

Example

Code:

---

mystring = "abcdefghijk"
 otherstring = "p"
 newstring = InStrRev(mystring, otherstring)

---

newString here would return 0 because [otherString] is not in [Mystring]

where as ...

Code:

---

mystring = "abcdefghijk"
  otherstring = "abc"
  newstring = InStrRev(mystring, otherstring)

---

newString here would return 1 because [otherString] is not in [Mystring]

UCase and LCase

These functions make a chosen string UpperCase (Capital letters) or LowerCase (Small Letters)

UCase

Use -

Code:

---

[NewString] = Ucase([String])

---

Example -

Code:

---

mystring = "abcdefghijk"
  otherstring = UCase(mystring)

---

[Otherstring] would now be "ABCDEFGHIJK"

LCase

Use -

Code:

---

[NewString] = Lcase([String])

---

Example -

Code:

---

mystring = "ABCDEFGHIJK"
  otherstring = LCase(mystring)

---

[OtherString] would now be "abcdefghijk"

Join

Used to join a number of strings from an array seperated by a character (Dilemeter)

Use -

Code:

---

Join([Array Souce],[Dilemeter]) as string

---

Example

Code:

---

myarray(0) = "1"
 myarray(1) = "2"
 myarray(2) = "3"
 myarray(3) = "4"
 myarray(4) = "5"
 
 newstring = Join(myarray, ",")
 
 MsgBox newstring

---

newstring would now be "1,2,3,4,5"

Len

Use -

VB Code:

Len([Expression])

[Expression] - Could be a varible, a textbox, "String of text", anything really

Example

VB Code:

Len(Text1.Text) 'Would return the length of text1.text
 
Len(myVarible) 'Would return the length of myVarible
 
Len("Hello") 'Would return the length of that string which is 5

LTrim\RTrim

LTrim Removes the spaces from the beginning of a string, (Left to right)
RTrim Removes the spaces from the end of a string, (Right to Left)

Use -

VB Code:

LTrim([String]) 
RTrim([String])

Example -

VB Code:

Msgbox LTrim("       Hello) 'Would remove the spaces at the start and display "Hello"
Msgbox RTrim("Hello         ") 'Would do the same but remove the sapces from the end

Date And Time (3 Functions)

Date, Now, Time

Date - Returns the current date
Now - Returns the current date and time
Time - Returns the current time

There isnt a specific way to use these functions you can assign them to varibles display them in msgbox whatever :)

Examples

VB Code:

Msgbox Date ' Would show current date according to system clock :)
Msgbox Time ' Would show current time according to system clock (Note it will be in the same format as the system clock)
Msgbox Now ' Would show the Date & time together

Replace

Use -

VB Code:

[i][b][string][/b][/i] = Replace([b][i][expresion][/i][/b],[b][i][find][/i][/b],[b][i][Replace With][/i][/b])

[string] - This will hold the result of our Replace() function[expresion] - This is the expresion we we be changing (a text box or label or "23.78.09" etc[find] - This is what we will be changing[Replace With] - and this is what we will be chnaging it to

Example -

Lets say we have a textbox (Named text1) and it contains (12.7.90.50) and for whatever reason we need to change the '.' (period) to a '/' (forward slash)

VB Code:

myString = Replace(Text1.Text, ".", "/")
 msgbox MyString

MyString would be 12/7/90/50,

Please note it would not change the contents of the textbox, to change the contents of the textBox you would need to do this.

VB Code:

text1.text = Replace(Text1.Text, ".", "/")

TypeName()

This returns the variable type of any given varible name could be usefull for debugging purpose.

Use -

VB Code:

[color=#00007f]String[/color] = [color=#00007f]TypeName[/color]([Variable Name])

[Variable Name] Would as it suggest be the name of the variable ;)

Example

So lets say you did,

VB Code:

Dim V1 As String
Dim V2 As Long
Dim V3 As Boolean
'etc
 
MsgBox TypeName(V1) ' Would Return string
MsgBox TypeName(V2) ' Would Return Long
MsgBox TypeName(V3) ' Would Return Boolean

IsNumeric()

This function returns true if the selected item is Numberic (Numbers) and false if it is not.

Use -

VB Code:

IsNumeric([Expression]) as boolean

[Expression] Would be the item you were looking to see if it was numeric.

Example -

VB Code:

If IsNumeric("78906") = True Then
    MsgBox ("True")
End If
 
'The above statement would return the true msgbox
 
If IsNumeric("5566pp578") = True Then
      MsgBox ("true")
End If
 
' This statement would not because the string we specified contains characters "pp"

So this function returns true if a specified expresion is numerical and flase if it is not

Note - It does also work for decimals so 0.8950 would return true :)

CLng, CDbl, CInt, CBool, CDate, CStr

All these functions work in a similar way they all convert a specified expression to the respective type.

Use -

VB Code:

Clng([Expression]) as long
 CDbl([Expression]) as Double
 CInt([Expression]) as Integer
 CBool([Expression]) as Boolean
 CDate([Expression]) as Date
 CStr([Expression]) as String

[Expression] being the thing we wish to convert, the functions explain themselves but i'll just go through what each does,

CLng - Converts a specified expression so it will fit into a long variable type
CDbl - Converts a specified expression so it will fit into a Double variable type
CInt - Converts a specified expression so it will fit into a Integer variable type
CBool - Converts a specified expression so it will fit into a Boolean variable type
CDate - Converts a specified expression so it will fit into a date variable type
CStr - Converts a specified expression so it will fit into a string variable type

Example

To use these functions is very simple, (myLong, myString etc are defined varibles as there name suggest so myString will be a string etc)

VB Code:

myLong = CLng(Text1.text)
 'Text1 must contain digits
 myDouble = CDbl(Text1.text)
 'Text1 must contain digits
 myInteger = CInt(Text1.text)
 'Text1 must contain digits (Whole numbers no decimals)
 myBoolean = CBool(Text1.text)
 'Text1.text must contain True Or False or 0 or 1
 myDate = CDate(text1.text)
 'Text1 must contain a valid date ie (26/10/1926)
 myString = CStr(Text1.text)
 'You wouldnt need to do this since text1.text is allready a string :)

Note - trying to convert impossible expressions will result in a Type Mismatch Error ie

VB Code:

myBoolean = CBool("Ya Its Not Gonna Work")

Val

The val function take the Value of a given epression.

Use

VB Code:

Val([Expression]) as Double

Expression is the thing we wish to take the value of

Example

VB Code:

'This would return the value of text1.text
 
Msgbox Val(Text1.Text)

If text1.text contained "456789" then thats what the msgbox would say, So whats the differnace between this and CDbl (Remember the val function is a double)

Well lets say in the textbox you have "99Chickens"
The code used above would now output "99"

But if the textbox contained "Chickens 99"
The code used above would output 0

So it will output all the leading numbers