Is there any way to get the value of a variable that has been named in a string.
For instance:
Dim number As Integer = 12
Dim numberCopy As Integer = Convert("number")
' numberCopy will now equal 12
' Convert is the function to convert the string to the variable
Something like that can be done only with Introspection, and only if the variable holding the value is a property of a class. Your example, where the variable is only temporarily known inside a function, cannot be made work as you like.
To learn more about Introspection, see the documentation and examples of the same name.
You can also use a Dictionary instead of straight properties and variables.
For instance
dim d as new dictionary
d.value("number") = 12
Then when you need the value, you can do
Dim numberCopy As Integer = d.value("number")
Nice thing about Dictionary is since keys as well as values are variant, you can throw at it pretty much anything, all variable types, as well as objects.
See Language Reference
Related
How can I convert a string to a variable name in VBscript? I found similar Q/A for JavaScript, C#, Pyton, Perl, Ruby, Java, but not for VBscript.
This is why I can not work directly with variables instead of strings:
There is a list of comma separated permission names which I get from a web service. The project manager may update this list any time and expects that I check if a user of software is granted to these permissions. User's permissions are set in run time by complex methods as well as profile data, history of activities in the software etc. e.g. if a user has bought 1000 products, the variable SuperCustomerPermission will set for him in header of page using SuperCustomerPermission="yes" (not as cookies or session nor stored in a databse).
To check the list of permission I want to pass the list of permission names to a function and loop through the names:
Permission names which I get from a web service:
permissionNames = "adminPermission,deletePermission,configPermission,SuperCustomerPermission"
I try to pass these strings as variables to the subroutine:
Sub checkPermission(PermissionNames)
permissionNamesArray = split(permissionNames,",",-1,1)
For Each x In permissionNamesArray
'How to convert x to its variable equivalent and check if it equals "yes"
Next
End Sub
Call checkPermission(permissionNames)
If it is possible to put it into application or session you could do something like session(variableName) = 5.
If not, the next easiest way is using a dictionary:
Dim myDict
Set myDict= CreateObject("Scripting.Dictionary")
myDict.Add (variableName, value)
Using a dictionary should work fine with your situation and it's much cleaner.
That being said, since you asked for an Execute solution, you can do something like this:
Sub checkPermission(PermissionNames)
Dim permissionNamesArray, x
permissionNamesArray = split(permissionNames,",",-1,1)
For Each x In permissionNamesArray
Dim valueOfX
Execute "valueOfX = " & x
WScript.Echo valueOfX
Next
End Sub
Example of usage:
' Adding some values to the variables
Dim adminPermission, deletePermission, configPermission, SuperCustomerPermission
adminPermission = "yes"
deletePermission = "no"
configPermission = "yes"
SuperCustomerPermission = "no"
' Storing names of variables
Dim permissionNames
permissionNames = "adminPermission,deletePermission,configPermission,SuperCustomerPermission"
' Passing variables by names
checkPermission(permissionNames)
Output:
yes
no
yes
no
The following code always outputs "not":
print "input a number please. "
TestNumber = gets
if TestNumber % 2 == 0
print "The number is even"
else
print "The number is not even"
end
What is going wrong with my code?
The gets() method returns an object of type String.
When you call %() on a String object, the return value is a new String object (usually it changes the text. You can read more about string formatting here).
Since there are no String objects that == 0, the if/else will always take the same path.
If you want to use the return value of gets() like a number, you will need to transform it into one first. The simplest approach is probably to use the to_i() method on String objects, which returns a new 'Integer' object. If you're doing something where the user input will not always be an integer (e.g. 3.14 or 1.5), you might need to use a different approach.
One last thing: in your example the result of gets() is saved into a constant called TestNumber. Constants are different to normal variables, and they will probably cause problems if you're not using them intentionally. Normal variables don't start with capital letters. (You can read more about ruby variables here). In ruby you need to write you variable names like this: test_number.
I suspect your Testnumber variable might be interpreted as a string during the operation. make sure the testnum is converted to an integer first even if you put in say 100 it could be its being interpreted as the stirng "100" and not the integer 100.
A similar issue can be found here: Ruby Modulo Division
You have to convert TestNumber from string to integer, as your input has linefeed and/or other unwanted characters that do not match an integer.
Use TestNumber = gets.to_i to convert to integer before testing.
I'm creating a function that displays a lot of variables with the format Variable + Variable Name.
Define LibPub out(list)=
Func
Local X
for x,1,dim(list)
list[x]->name // How can I get the variable name here?
Disp name+list[x]
EndFor
Return 1
EndFunc
Given a list value, there is no way to find its name.
Consider this example:
a:={1,2,3,4}
b:=a ; this stores {1,2,3,4} in b
out(b)
Line 1: First the value {1,2,3,4} is created. Then an variable with name a is created and its value is set to {1,2,3,4}.
Line 2: The expression a is evaluated; the result is {1,2,3,4}. A new variable with the name b is created and its value is set to `{1,2,3,4}.
Line 3: The expression b is evaluated. The variable reference looks up what value is stored in b. The result is {1,2,3,4}. This value is then passed to the function out.
The function out receives the value {1,2,3,4}. Given the value, there is no way of knowing whether the value happened to be stored in a variable. Here the value is stored in both a and b.
However we can also look at out({1,1,1,1}+{0,2,3,4}).
The system will evaluate {1,1,1,1}+{0,2,3,4} and get {1,2,3,4}. Then out is called. The value out received the result of an expression, but an equivalent value happens to be stored in a and b. This means that the values doesn't have a name.
In general: Variables have a name and a value. Values don't have names.
If you need to print a name, then look into strings.
This will be memory intensive, but you could keep a string of variable names, and separate each name by some number of characters and get a substring based on the index of the variable in the list that you want to get. For instance, say you want to access index zero, then you take a substring starting at (index of variable * length of variable name, indexofvariable *length + length+1).
The string will be like this: say you had the variables foo, bas, random, impetus
the string will be stored like so: "foo bas random impetus "
I'm teaching myself VBS and I decided to write a message encryption program. It uses the left and right functions in a loop to read every individual character.
DO
wscript.sleep 100
if Az=0 then
EXIT DO
end if
CR=right(message,aZ)
DEL=left(CR,1)
aZ=aZ-1
zZ=zZ+1
supra=""
supra="supra"
CALL KEYCOUNT
CD=left(keyword,zZ)
TAC=right(CD,1)
....
From there, it sets every character equal to a different letter based on an encryption keyword and moves onto the next character. My problem is I don't know how to deal with spaces in the message. Is there a way to make a variable have the value of a space? I've tried:
set var=space(1)
set var="&"" ""&"
set var=""
set var=" "
set var=""" """
I'm certain there are things I'm not thinking of
Thanks
Joseph
set statement assigns an object reference to a variable or property, or associates a procedure reference with an event. That isn't our case.
var=space(1) ' var contains one space character
var="&"" ""&" ' var contains &" "&
var="" ' var contains a string of zero length
var=" " ' var contains one space character
var=""" """ ' var contains " "
Or, if you would like, declare a constant for use in place of literal value of space (anywhere in your script) as follows:
Const vbSp=" "
Constants are public by default. Within procedures, constants are always private; their visibility can't be changed. Within a script, the default visibility of a script-level constant can be changed using the Private keyword. There are variations:
Private Const vbSp=" "
Public Const vbSp=" "
The problem is that you use Set (cf. here) when assigning a simple/non-object value to a variable. If you loose it, your experimental statements will 'work' (compile & run without error). Look at the output and you'll see that
var = " "
is the correct and most efficient way to assign a (string containing one) space to a variable.
I'm extremely unfamiliar with ancient VB, and I'm trying to figure out the proper commands to concatenate an array (I am assuming it is in array form) into a string with comma separated values.
The values are being provided by a multiselect box, which is being assigned to the areas variable, which is grabbed from the areas select box.
dim name
dim from
dim company
dim phone
dim zip
dim message
dim areas
name = Request.Form("name")
from = Request.Form("from")
company = Request.Form("company")
phone = Request.Form("phone")
zip = Request.Form("zip")
areas = Request.Form("areas")
message = Request.Form("message")
I want to take areas, and implode it into a string.
What's the command in very old VB to do this?
The Join function does also exist in VB6, the syntax is like this:
myString = Join(myArray, ",")
EDIT: Note that the array goes before the delimiter. The delimiter is optional, it'll be a space if left empty.
you want to use the Join function String.Join(",",array)
If you are using VB6, try this:
Join(New String() {name, from, company}, ", ")
Join Function (Visual Basic) # MSDN
EDIT: I know it's a link to VB.NET, but it's the old VB6 function call, it should be compatible with VB6, syntax wise.