Dim strnumber
strnumber = "0.3"
Dim add
add = 0.1
Dim result
result = strnumber + add
MsgBox result
I want to get 0.4 as result, but get 3.1.
I tried clng(strnumber) and int(strnumber), nothing works. There is a simple solution for sure but I won't find it.
EDIT: Solution
result = CDbl(Replace(strnumber,".",",") + add
Has to do with your locale settings. Automatic conversion (as well as explicit one) observes it in the same manner as in CStr() function.
E.g. in my locale CStr( 0.3) results to 0,3 that is invert to CDbl("0,3") while CDbl("0.3") results to an error.
BTW: always use option explicit and, for debugging purposes, On Error Goto 0
Following below procedures can help:
Replacing the dot(".") with comma (",") in the string
change the string to double by Cdbl
example:
dim a,b,c
a="10.12"
b="5.05"
a=Replace(a,".",",")
b= Replace(b,".",",")
c=Cdbl(a)+Cdbl(b)
msgbox c
You want to add two numbers. So you should use numbers (and not a string (strnumber) and a number (add):
>> n1 = 0.3
>> n2 = 0.1
>> r = n1 + n2
>> WScript.Echo r
>>
0,4
As you can see from the output (0,4), I'm using a locale (German) that uses "," as decimal 'point'. But literals always use ".". So by using the proper data types you can write your scripts in a language/locale independent fashion as long as you don't need to process external string data (from a file or user input). Then you have to modify the input before you feed it to a conversion function like CDbl(). For simple cases that can be done with Replace(inp, badmarker, goodmarker).
P.S. You said you " tried clng(strnumber) and int(strnumber)". You should have tried CDbl(). CLng() tries to get a long integer (cf. CInt()), Int() removes/rounds the fraction from number.
Related
I am trying to add a random set of numbers to the end of a string. I'm still learning the basics of VBS but this has really tricked me and I can't seem to find anything online.
I've tried:
string2 = "hello" + (Rnd() * Len(VALID_TEXT)) + 1
And:
x = rnd*10
string2 = "hello" + x
What am I doing wrong?
All random number generators rely on an underlying algorithm, usually fed by what’s called a seed number. You can use the Randomize statement to create a new seed number to ensure your random numbers don’t follow a predictable pattern.
To get the random numbers, using rnd alone is not sufficient as you will keep on getting the same random number again and again. You have to use randomize to achieve the task as shown below:
Dim strTest:strTest = "Hello"
Dim intNoOfDigitsToAppend:intNoOfDigitsToAppend = 5
Randomize
Msgbox "String before appending: " & strTest
strTest = fn_appendRandomNumbers(strTest,intNoOfDigitsToAppend)
Msgbox "String before appending: " & strTest
function fn_appendRandomNumbers(strToAppend,intNoOfRandomDigits)
Dim i
for i=1 to intNoOfRandomDigits
strToAppend= strToAppend & int(rnd*10) 'rnd gives a random number between 0 and 1, something like 0.8765341. Then, we multiply it by 10 so that the number comes in the range of 0 to 9. In this case, it becomes 8.765341. After that, we use the int method to truncate the decimal part so that we are only left with the Integer part. In this case, 765341 is truncated and we are left with only the integer 8
next
fn_appendRandomNumbers = strToAppend
end function
Reference 1
Reference 2
& is the string concatenation character. + is an old compatability concat character and will error if you mix text and numbers. Use + for maths only.
What is the Value of the below in vbscript
1)x=1+"1"
2)x="1"+"1"
3)x=1+"mulla"
Note:In the all above three cases I am using first variable as either string or integer and second on as always as string.
Case 1:Acting as a numeric and auto conversion to numeric during operation
enter code here
y=inputbox("Enter a numeric value","") Rem I am using 1 as input
x=1
msgbox x+y Rem value is 2
msgbox x*y Rem value is 1
Case 2:Acting as a String and no conversion to numeric during operation it fails
enter code here
y=inputbox("Enter a numeric value","") Rem I am using 1 as input
x=1
if y= x then
msgbox "pass"
else
msgbox "fail"
end if
Case 3:Acting as a String and explicit conversion to numeric during operation it passes
enter code here
y=inputbox("Enter a numeric value","") Rem I am using 1 as input
x=1
if Cint(y) = x then
msgbox "pass"
else
msgbox "fail"
end if
I need a logic reason for the different behaviors. but in other language it is straight forward and will work as expected
Reference: Addition Operator (+) (VBScript)
Although you can also use the + operator to concatenate two character strings, you should use the & operator for concatenation to eliminate ambiguity. When you use the + operator, you may not be able to determine whether addition or string concatenation will occur.
The type of the expressions determines the behavior of the + operator in the following way:
If Both expressions are numeric then the result is the addition of both numbers.
If Both expressions are strings then the result is the concatenation of both strings.
If One expression is numeric and the other is a string then an Error: type mismatch will be thrown.
When working with mixed data types it is best to cast your variables into a common data type using a Type Conversion Function.
I agree with most of what #thomas-inzina has said but the OP has asked for a more detailed explanation so here goes.
As #thomas-inzina point's out using + is dangerous when working with strings and can lead to ambiguity depending on how you combine different values.
VBScript is a scripting language and unlike it's big brothers (VB, VBA and VB.Net) it's typeless only (some debate about VB and VBA also being able to be typeless but that's another topic entirely) which means it uses one data type known as Variant. Variant can infer other data types such as Integer, String, DateTime etc which is where the ambiguity can arise.
This means that you can get some unexpected behaviour when using + instead of & as + is not only a concatenation operator when being used with strings but also a addition operator when working with numeric data types.
Dim x: x = 1
Dim y: y = "1"
WScript.Echo x + y
Output:
2
Dim x: x = "1"
Dim y: y = "1"
WScript.Echo x + y
Output:
11
Dim x: x = 1
Dim y: y = 1
WScript.Echo x + y
Output:
2
Dim x: x = 1
Dim y: y = "a"
WScript.Echo x + y
Output:
Microsoft VBScript runtime error (4, 5) : Type mismatch: '[string: "a"]'
I am using in my sql server reports a simple vbscript command to format some numeric fields:
FormatNumber(value,,-1,0,-1)
Important for me is the second parameter NumDigAfterDec that is set to default, meaning
that "the computer's regional settings are used" (http://www.w3schools.com/vbscript/func_formatnumber.asp). This is exactly what I prefer. But when the current number has more digits after decimal, it is rounded. In this case I would like to see all places.
e.g. for two places after decimal, I would like:
0 0.00
90.7 90.70
1.2345 1.2345 (instead of 1.23)
Is this possible without writing code in my reports?
If I understand correctly, you want a condition stating if length of decimal is greater than 2 places, then display full decimal number, else 2 decimal places?
A simple way to do this is:
Dim value : value = "100.54367"
Dim GetDecCount : GetDecCount = Len(Mid(Value, instr(value, ".")+1, len(value)))
if GetDecCount>2 then
msgbox "This number exceeds the regional decimal points of 2, with a length of " & GetDecCount & " decimal points."
else
FormatNumber(value,,-1,0,-1)
end if
Try something like this:
Set re = New RegExp
re.Pattern = ".\d{3,}$"
if re.Test(value) Then
fnum = CStr(value)
Else
fnum = FormatNumber(value, , -1, 0, -1)
End If
I am trying to convert letters to numbers.
I have a sub which ensures only numbers are put into the textbox.
My questions is will the following code work. I have a textbox(for numbers) and combobbox(for letters)
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Mid(1, cmbEngletter.Text, 1))
MsgBox "stringposition"
EngNumber = (txtManuNo.Text * 10) + stringposition
My only question above would be will the multiplication work with a .text. I believe it won't because it is a string. Please advise then on how to deal with a situation.
You can use CLng() to convert a string to a Long variable
CLng() will throw an error though if it doesn't like the contents of the string (for example if it contains a non-numeric character), so only use it when you are certain your string will only contain numbers
More forgiving is it to use Val() to convert a string into a numeric variable (a Double by default)
I also suggest you look into the following functions:
Asc() : returns the ASCII value of a character
Chr$() : coverts an ASCII value into a character
Left$() : returns the first characters of a string
CStr() : convert a number into a string
I think in your code you mean to show the contents of your variable stringposition instead of the word "stringposition", so you should remove the ""
I do wonder though what you are trying to accomplish with your code, but applying the above to your code gives:
Dim sha As String
Dim stringposition As Long
Dim EngNumber As Long
sha = "abcdefghifjklmnopqrstuvwxyz"
stringposition = InStr(1, sha, Left$(cmbEngletter.Text, 1))
MsgBox CStr(stringposition)
EngNumber = (Val(txtManuNo.Text) * 10) + stringposition
I used Val() because I am not certain your txtManuNo will contain only numbers
To ensure an user can only enter numbers you can use the following code:
Private Sub txtManuNo_KeyPress(KeyAscii As Integer)
Select Case KeyAscii
Case vbKeyBack
'allowe backspace
Case vbKey0 To vbKey9
'allow numbers
Case Else
'refuse any other input
KeyAscii = 0
End Select
End Sub
An user can still input non-numeric charcters with other methods though, like copy-paste via mouse actions, but it is a quick and easy first filter
I am trying to replace the double quotes in a string with a single quote, got the following code but get error message saying "Object Required strLocation"
Sub UpdateAdvancedDecisions(strLocation)
Dim d
Dim strLLength
strLLength = Len(strLocation) - 1
For d = 0 To strLLength
alert strLocation
strValue = strLocation.Substring(2,3)
If strLocation.substring(d,d+1)=" " " Then
strLLength = strLLength.substring(0, d) + "'" + strLLength.substring(d + 1,strLLength.length)
Next
End Sub
As Helen said, you want to use Replace, but her example assigned the result to your weird strLLength variable. Try this instead:
strLocation = Replace(strLocation, """", "'")
This one line does the job you asked about and avoids all the code currently in your given subroutine.
Other things that are problems in the code you posted:
a variable holding a number like the length of a string would not have a "str" prefix, so strLLength is misleading
strings in VBScript are indexed from 1 through length, not 0 through length-1
there is no "alert" keyword in VBScript
you assign a value to strValue, then never use it again
you need to use Mid to get a substring, there is no "substring" string method in VBScript
c = Mid(strLocation, d, 1) ' gets one character at position d
The more I look at this, the more clear it is that its some JavaScript that you're trying to run as VBScript but are not translating at all correctly.
Use a reference for VBScript like one of the following:
MSDN Library: VBScript Language Reference
W3Schools VBScript Tutorial