is if x <> empty the same as if not isempty(x) - if-statement

Are these two snippets equivalent in VBScript?
If x <> empty
'do something
End if
If not isempty(x)
'dosomething
End if
Context:
I want to rewrite a condition from
If ((Not IsEmpty( Var ) ) And ( Var <> "" ) ) Then
to the less verbose
If Var <> Empty And Var <> "" Then

No; the operator approach would fail for a variant typed as object
dim woof
set woof = nothing
msgbox isempty(woof) '// ok -> false
msgbox woof = empty '// object reference not set error
(Because the = approach would attempt to read the value of the default property of the object)

Related

VBA Function Returning Type Mismatch

I have a function calling another function. This was all one function but it got too big.
Function Wave(SubjectLine As Range)
Dim SubV
SubV = SubjectLine.Value
Dim regEx, Match, Matches ' Create variable.
Set regEx = New RegExp ' Create a regular expression.
regEx.Pattern = "[0-9][0-9][0-9][0-9][0-9][-][a-z][a-z][a-z]?[a-z]?[a-z]?[a-z]"
regEx.IgnoreCase = True ' Set case insensitivity.
regEx.Global = True ' Set global applicability.
Set Matches = regEx.Execute(SubV) ' Execute search.
For Each Match In Matches ' Iterate Matches collection.
RetStr = RetStr & Match.Value
Next
If RetStr = 0 Then
WaveSort = "Not Found"
Exit Function
End If
DivR = DivResult(Left(RetStr, 5))
...then something else
Function DivResult(PL As Integer)
Select Case PL
Case 1
DivResult = "Wave1"
...
Case Else
DivResult = "NO"
End Select
End Function
My result is alway Data Type Mismatch. I have tried changing variable types and renaming and everything I can think of. Left turns the variable to an integer so i just embraced it but nothing seems to work.

Parse a String in Excel Vba

I have a macro that send an XMLHTTP request to a server and it gets as response a plain text string, not a JSON format string or other standard formats (at least for what I know).
I would like to parse the output string in order to access the data in an structured approach in the same fashion as the parseJson subroutine in this link
My problem is I am not good with regular expressions and I am not able to modify the routine for my needs.
The string that I need to parse has the following structure:
The string is a single line
Each single parameter is defined by its parameter name the equal simbol, its value and ending with; "NID=3;" or "SID=Test;"
Parameter can be collected in "structures" starts and end with the symbol | and they are identified with their name followed by ; such as |STEST;NID=3;SID=Test;|
A structure can contain also other structures
An example of a output string is the following
|KC;|AD;PE=5;PF=3;|CD;PE=5;HP=test;|CD;PE=3;HP=abc;|
In this case there is a macro structure KC which contains a structure AD. The structure AD is composed by the parameters PE, PF and 2 structures CD. And finaly the structures CD have the parameters PE and HP
So I would like to parse the string to obtain an Object/Dictionary that reflects this structure, can you help me?
Adds after the first answers
Hi all, thank you for your help, but I think I should make more clear the output that I would like to get.
For the example string that I have, I would like to have an object with the following structure:
<KC>
<AD>
<PE>5</PE>
<PF>3</PF>
<CD>
<PE>5</PE>
<HP>test</HP>
</CD>
<CD>
<PE>3</PE>
<HP>abc</HP>
</CD>
</AD>
</KC>
So I started to wrote a possible working code base on some hint from #Nvj answer and the answer in this link
Option Explicit
Option Base 1
Sub Test()
Dim strContent As String
Dim strState As String
Dim varOutput As Variant
strContent = "|KC;|AD;PE=5;PF=3;|CD;PE=5;HP=test;|CD;PE=3;HP=abc;|"
Call ParseString(strContent, varOutput, strState)
End Sub
Sub ParseString(ByVal strContent As String, varOutput As Variant, strState As String)
' strContent - source string
' varOutput - created object or array to be returned as result
' strState - Object|Array|Error depending on processing to be returned as state
Dim objTokens As Object
Dim lngTokenId As Long
Dim objRegEx As Object
Dim bMatched As Boolean
Set objTokens = CreateObject("Scripting.Dictionary")
lngTokenId = 0
Set objRegEx = CreateObject("VBScript.RegExp")
With objRegEx
.Global = True
.MultiLine = True
.IgnoreCase = True
.Pattern = "\|[A-Z]{2};" 'Pattern for the name of structures
Tokenize objTokens, objRegEx, strContent, lngTokenId, bMatched, "str"
.Pattern = "[A-Z]{2}=[^\|=;]+;" 'Pattern for parameters name and values
Tokenize objTokens, objRegEx, strContent, lngTokenId, bMatched, "par"
End With
End Sub
Sub Tokenize(objTokens, objRegEx, strContent, lngTokenId, bMatched, strType)
Dim strKey As String
Dim strKeyPar As String
Dim strKeyVal As String
Dim strWork As String
Dim strPar As String
Dim strVal As String
Dim strLevel As String
Dim strRes As String
Dim lngCopyIndex As Long
Dim objMatch As Object
strRes = ""
lngCopyIndex = 1
With objRegEx
For Each objMatch In .Execute(strContent)
If strType = "str" Then
bMatched = True
With objMatch
strWork = Replace(.Value, "|", "")
strWork = Replace(strWork, ";", "")
strLevel = get_Level(strWork)
strKey = "<" & lngTokenId & strLevel & strType & ">"
objTokens(strKey) = strWork
strRes = strRes & Mid(strContent, lngCopyIndex, .FirstIndex - lngCopyIndex + 1) & strKey
lngCopyIndex = .FirstIndex + .Length + 1
End With
lngTokenId = lngTokenId + 1
ElseIf strType = "par" Then
strKeyPar = "<" & lngTokenId & "par>"
strKeyVal = "<" & lngTokenId & "val>"
strKey = strKeyPar & strKeyVal
bMatched = True
With objMatch
strWork = Replace(.Value, ";", "")
strPar = Split(strWork, "=")(0)
strVal = Split(strWork, "=")(1)
objTokens(strKeyPar) = strPar
objTokens(strKeyVal) = strVal
strRes = strRes & Mid(strContent, lngCopyIndex, .FirstIndex - lngCopyIndex + 1) & strKey
lngCopyIndex = .FirstIndex + .Length + 1
End With
lngTokenId = lngTokenId + 2
End If
Next
strContent = strRes & Mid(strContent, lngCopyIndex, Len(strContent) - lngCopyIndex + 1)
End With
End Sub
Function get_Level(strInput As String) As String
Select Case strInput
Case "KC"
get_Level = "L1"
Case "AD"
get_Level = "L2"
Case "CD"
get_Level = "L3"
Case Else
MsgBox ("Error")
End
End Select
End Function
This function creates a dictionary with an item for each structure name, parameter name and parameter value as shown in the figure
Thanks to the function get_Level the items associated to structures have a level that should help to preserve the original hierarchy of the data.
So what I am missing is a function to create an object that has the original structure of the input string. This is what the Retrieve function do in this answer link, but I do not know how to adapt it to my case
This looks like a simple nested delimited string. A couple of Split() functions will do the trick:
Option Explicit
Function parseString(str As String) As Collection
Dim a1() As String, i1 As Long, c1 As Collection
Dim a2() As String, i2 As Long, c2 As Collection
Dim a3() As String
a1 = Split(str, "|")
Set c1 = New Collection
For i1 = LBound(a1) To UBound(a1)
If a1(i1) <> "" Then
Set c2 = New Collection
a2 = Split(a1(i1), ";")
For i2 = LBound(a2) To UBound(a2)
If a2(i2) <> "" Then
a3 = Split(a2(i2), "=")
If UBound(a3) > 0 Then
c2.Add a3(1), a3(0)
ElseIf UBound(a3) = 0 Then
c2.Add a3(0)
End If
End If
Next i2
c1.Add c2
End If
Next i1
Set parseString = c1
End Function
Sub testParseString()
Dim c As Collection
Set c = parseString("|KC;|AD;PE=5;PF=3;|CD;PE=5;HP=test;|CD;PE=3;HP=abc;|")
Debug.Assert c(1)(1) = "KC"
Debug.Assert c(2)("PE") = "5"
Debug.Assert c(3)(1) = "CD"
Debug.Assert c(4)("HP") = "abc"
Debug.Assert c(4)(3) = "abc"
End Sub
Note that you can address values by both, index and key (if key existed in the input). If key was not provided you can only access the value by its index. You can also iterate collection recursively to get all the values in a tree structure.
Food for thought: since your structures may have repeated names (in your case "CD" structure happens twice) Collections / Dictionaries would find it problematic to store this elegantly (due to key collisions). Another good way to approach this is to create an XML structure with DOMDocument and use XPath to access its elements. See Program with DOM in Visual Basic
UPDATE: I've added XML example below as well. Have a look.
Here is another take on your string parsing issue using DOMDocument XML parser. You need to include Microsoft XML, v.6.0 in your VBA references.
Function parseStringToDom(str As String) As DOMDocument60
Dim a1() As String, i1 As Long
Dim a2() As String, i2 As Long
Dim a3() As String
Dim dom As DOMDocument60
Dim rt As IXMLDOMNode
Dim nd As IXMLDOMNode
Set dom = New DOMDocument60
dom.async = False
dom.validateOnParse = False
dom.resolveExternals = False
dom.preserveWhiteSpace = True
Set rt = dom.createElement("root")
dom.appendChild rt
a1 = Split(str, "|")
For i1 = LBound(a1) To UBound(a1)
If a1(i1) <> "" Then
a2 = Split(a1(i1), ";")
Set nd = dom.createElement(a2(0))
For i2 = LBound(a2) To UBound(a2)
If a2(i2) <> "" Then
a3 = Split(a2(i2), "=")
If UBound(a3) > 0 Then
nd.appendChild dom.createElement(a3(0))
nd.LastChild.Text = a3(1)
End If
End If
Next i2
rt.appendChild nd
End If
Next i1
Set parseStringToDom = dom
End Function
Sub testParseStringToDom()
Dim dom As DOMDocument60
Set dom = parseStringToDom("|KC;|AD;PE=5;PF=3;|CD;PE=5;HP=test;|CD;PE=3;HP=abc;|")
Debug.Assert Not dom.SelectSingleNode("/root/KC") Is Nothing
Debug.Assert dom.SelectSingleNode("/root/AD/PE").Text = "5"
Debug.Assert dom.SelectSingleNode("/root/CD[1]/HP").Text = "test"
Debug.Assert dom.SelectSingleNode("/root/CD[2]/HP").Text = "abc"
Debug.Print dom.XML
End Sub
As you can see this converts your text into an XML DOM document preserving all the structures and allowing for duplicates in naming. You can then use XPath to access any node or value. This can also be extended to have more nesting levels and further structures.
This is the XML document it creates behind the scenes:
<root>
<KC/>
<AD>
<PE>5</PE>
<PF>3</PF>
</AD>
<CD>
<PE>5</PE>
<HP>test</HP>
</CD>
<CD>
<PE>3</PE>
<HP>abc</HP>
</CD>
</root>
I've started to write a parser in VBA for the string structure specified by you, and it's not complete, but I'll post it anyways. Maybe you can pick up some ideas from it.
Sub ParseString()
Dim str As String
str = "|KC;|AD;PE=5;PF=3;|CD;PE=5;HP=test;|CD;PE=3;HP=abc;|"
' Declare an object dictionary
' Make a reference to Microsoft Scripting Runtime in order for this to work
Dim dict As New Dictionary
' If the bars are present in the first and last character of the string, replace them
str = Replace(str, "|", "", 1, 1)
If (Mid(str, Len(str), 1) = "|") Then
str = Mid(str, 1, Len(str) - 1)
End If
' Split the string by bars
Dim substring_array() As String
substring_array = Split(str, "|")
' Declare a regex object
' Check the reference to Microsoft VBScript Regular Expressions 5.5 in order for this to work
Dim regex As New RegExp
With regex
.Global = True
.IgnoreCase = True
.MultiLine = True
End With
' Object to store the regex matches
Dim matches As MatchCollection
Dim param_name_matches As MatchCollection
Dim parameter_value_matches As MatchCollection
' Define some regex patterns
pattern_for_structure_name = "^[^=;]+;"
pattern_for_parameters = "[^=;]+=[^=;]+;"
pattern_for_parameter_name = "[^=;]="
pattern_for_parameter_val = "[^=;];"
' Loop through the elements of the array
Dim i As Integer
For i = 0 To UBound(substring_array) - LBound(substring_array)
' Get the array element in a string
str1 = substring_array(i)
' Check if it contains a structure name
regex.Pattern = pattern_for_structure_name
Set matches = regex.Execute(str1)
If matches.Count = 0 Then
' This substring does not contain a structure name
' Check if it contains parameters
regex.Pattern = pattern_for_parameter
Set matches = regex.Execute(matches(0).Value)
If matches.Count = 0 Then
' There are no parameters as well as no structure name
' This means the string had || - invalid string
MsgBox ("Invalid string")
Else
' The string contains parameter names
' Add each parameter name to the dictionary
Dim my_match As match
For Each my_match In matches
' Get the name of the parameter
regex.Pattern = pattern_for_parameter_name
Set parameter_name_matches = regex.Execute(my_match.Value)
' Check if the above returned any matches
If parameter_name_matches.Count = 1 Then
' Remove = sign from the parameter name
parameter_name = Replace(parameter_name_matches(0).Value, "=", "")
' Get the value of the parameter
regex.Pattern = pattern_for_parameter_value
Set parameter_value_matches = regex.Execute(my_match.Value)
' Check if the above returned any matches
If parameter_value_matches.Count = 1 Then
' Get the value
parameter_value = Replace(parameter_value_matches(0).Value, ";", "")
' Add the parameter name and value as a key pair to the Dictionary object
dict.Item(parameter_name) = parameter_value
Else
' Number of matches is either 0 or greater than 1 - in both cases the string is invalid
MsgBox ("Invalid string")
End If
Else
' Parameter name did not match - invalid string
MsgBox ("Invalid string")
End If
Next
End If
ElseIf matches.Count = 1 Then
' This substring contains a single structure name
' Check if it has parameter names
Else
' This substring contains more than one structure name - the original string is invalid
MsgBox ("Invalid string")
End If
Next i
End Sub

In vbscript can you use two possible conditions in an if statement? (OR)

An example would be:
If filter_purchase = 0 Or "" Then
SetDocVar "filter_purchase", "0"
Else
SetDocVar "filter_purchase", CStr(filter_purchase)
End If
But I get a 'Type Mismatch'. Would there be an easier way than doing Else IFs?
you have to explicitly state the condition for each OR. Please see below
If filter_purchase = 0 Or filter_purchase = "" Then
SetDocVar "filter_purchase", "0"
Else
SetDocVar "filter_purchase", CStr(filter_purchase)
End If
This should be the condition you want
If ((filter_purchase = 0) Or (filter_purchase = "")) Then
#agamike, I believe a single = is used for comparison in a vbs if not and not == link here
If you're just trying to test for an uninitialized variable then your If expression is actually redundant. All variables in VBScript are variants and all variants start out with a default value of 0/False/"". For example:
Dim v
If v = "" Then MsgBox "Empty string"
If v = 0 Then MsgBox "Zero"
If v = False Then MsgBox "False"
All three of these tests will pass. Note how you can compare a single variable against string, numeric, and boolean literals. Uninitialized variables have no type yet, so these kinds of comparisons are completely fine.
However, once you assign a value to the variable, you need to consider its type when making comparisons. For example:
Dim v
v = ""
If v = "" Then MsgBox "Empty String" ' Pass. "" = "".
If v = 0 Then MsgBox "Zero" ' Fail! Illegal comparison.
If v = False Then MsgBox "False" ' Fail! "" <> False.
Now that the variant has been defined as holding a string, it will need to be compared against other string types (literals or variables) or values that can be cast (either implicitly or explicitly) to a string.

Dynamic Depending Lists in Separated WorkSheets in VBA

I'm not really expert in VBA and I have a problem with my code and I don't know how to solve it. (The code is from: http://siddharthrout.wordpress.com/2011/07/29/excel-data-validationcreate-dynamic-dependent-lists-vba/)
I'm working with 8 dynamic dependent lists, and I thought the best way to automate the process and avoid to modify the macro in a future if I modify the lists was a VBA code.
Trying to find the correct code, I'm just working with to lists. For after, apply it for all lists.
I've checked the code and I discovered that there's an error (method 'intersect' of object '_global' failed) because I'm comparing two ranges from a different worksheets.
My code is:
Private Sub Worksheet_Change(ByVal Target As Range)
Dim i As Long, LastRow As Long, n As Long
Dim MyCol As Collection
Dim SearchString As String, Templist As String
Application.EnableEvents = False
On Error GoTo Whoa
' Find LastRow in Col A
LastRow = Sheet2.Range("A" & Rows.Count).End(xlUp).Row
If Not Intersect(Target, Sheet2.Columns(1)) Is Nothing Then
Set MyCol = New Collection
' Get the data from Col A into a collection
For i = 2 To LastRow
If Len(Trim(Sheet2.Range("A" & i).Value)) <> 0 Then
On Error Resume Next
MyCol.Add CStr(Sheet2.Range("A" & i).Value), CStr(Sheet2.Range("A" & i).Value)
On Error GoTo 0
End If
Next i
' Create a list for the Data Validation List
For n = 1 To MyCol.Count
Templist = Templist & "," & MyCol(n)
Next
Templist = Mid(Templist, 2)
Range("A2").ClearContents: Range("A2").Validation.Delete
' Create the Data Validation List
If Len(Trim(Templist)) <> 0 Then
With Range("A2").Validation
.Add Type:=xlValidateList, AlertStyle:=xlValidAlertStop, Operator:=xlBetween, Formula1:=Templist
.IgnoreBlank = True
.InCellDropdown = True
.InputTitle = ""
.ErrorTitle = ""
.InputMessage = ""
.ErrorMessage = ""
.ShowInput = True
.ShowError = True
End With
End If
' Capturing change in cell A2
ElseIf Not Intersect(Target, Range("A2")) Is Nothing Then
SearchString = Range("A2").Value
Templist = FindRange(Sheet2.Range("A2:A" & LastRow), SearchString)
Range("B2").ClearContents: Range("B2").Validation.Delete
If Len(Trim(Templist)) <> 0 Then
' Create the DV List
With Range("B2").Validation
.Add Type:=xlValidateList, AlertStyle:=xlValidAlertStop, Operator:=xlBetween, Formula1:=Templist
.IgnoreBlank = True
.InCellDropdown = True
.InputTitle = ""
.ErrorTitle = ""
.InputMessage = ""
.ErrorMessage = ""
.ShowInput = True
.ShowError = True
End With
End If
End If
LetsContinue:
Application.EnableEvents = True
Exit Sub
Whoa:
MsgBox Err.Description
Resume LetsContinue
End Sub
' Function required to find the list from Col B
Function FindRange(FirstRange As Range, StrSearch As String) As String
Dim aCell As Range, bCell As Range, oRange As Range
Dim ExitLoop As Boolean
Dim strTemp As String
Set aCell = FirstRange.Find(what:=StrSearch, LookIn:=xlValues, lookat:=xlWhole, SearchOrder:= _
xlByRows, SearchDirection:=xlNext, MatchCase:=False, SearchFormat:=False)
ExitLoop = False
If Not aCell Is Nothing Then
Set bCell = aCell
strTemp = strTemp & "," & aCell.Offset(, 1).Value
Do While ExitLoop = False
Set aCell = FirstRange.FindNext(After:=aCell)
If Not aCell Is Nothing Then
If aCell.Address = bCell.Address Then Exit Do
strTemp = strTemp & "," & aCell.Offset(, 1).Value
Else
ExitLoop = True
End If
Loop
FindRange = Mid(strTemp, 2)
End If
End Function
Into the Sheet1, I just want the cells to select the list options and into the Sheet2, I want the all dynamic and dependent lists.
Is there any possibility to compare two ranges from a different worksheets using these algorithm? Or an alternative code to create a pick list for 8 depending and dynamic lists?
I am going to turn you to this page that describes dynamic dependent list usage very well.
Dynamic Dependent Lists
Perhaps you don't need VBA at all, unless you have to alter these on the fly, or based on some other variable. It's always best to use Excel's built-in functionality first, and code 2nd.
In case you are wandering, you can get around having lists on two different sheets by setting the named range scope to the entire workbook.
Edit: Adding answer to direct VBA error.
Since you didn't say, not sure if your Intersect is breaking here:
If Not Intersect(Target, Sheet2.Columns(1)) Is Nothing Then
but I think it is. Try this:
If Not Intersect(Target, Columns(1).EntireColumn) Is Nothing Then

Scala Regex enable Multiline option

I'm learning Scala, so this is probably pretty noob-irific.
I want to have a multiline regular expression.
In Ruby it would be:
MY_REGEX = /com:Node/m
My Scala looks like:
val ScriptNode = new Regex("""<com:Node>""")
Here's my match function:
def matchNode( value : String ) : Boolean = value match
{
case ScriptNode() => System.out.println( "found" + value ); true
case _ => System.out.println("not found: " + value ) ; false
}
And I'm calling it like so:
matchNode( "<root>\n<com:Node>\n</root>" ) // doesn't work
matchNode( "<com:Node>" ) // works
I've tried:
val ScriptNode = new Regex("""<com:Node>?m""")
And I'd really like to avoid having to use java.util.regex.Pattern. Any tips greatly appreciated.
This is a very common problem when first using Scala Regex.
When you use pattern matching in Scala, it tries to match the whole string, as if you were using "^" and "$" (and did not activate multi-line parsing, which matches \n to ^ and $).
The way to do what you want would be one of the following:
def matchNode( value : String ) : Boolean =
(ScriptNode findFirstIn value) match {
case Some(v) => println( "found" + v ); true
case None => println("not found: " + value ) ; false
}
Which would find find the first instance of ScriptNode inside value, and return that instance as v (if you want the whole string, just print value). Or else:
val ScriptNode = new Regex("""(?s).*<com:Node>.*""")
def matchNode( value : String ) : Boolean =
value match {
case ScriptNode() => println( "found" + value ); true
case _ => println("not found: " + value ) ; false
}
Which would print all all value. In this example, (?s) activates dotall matching (ie, matching "." to new lines), and the .* before and after the searched-for pattern ensures it will match any string. If you wanted "v" as in the first example, you could do this:
val ScriptNode = new Regex("""(?s).*(<com:Node>).*""")
def matchNode( value : String ) : Boolean =
value match {
case ScriptNode(v) => println( "found" + v ); true
case _ => println("not found: " + value ) ; false
}
Just a quick and dirty addendum: the .r method on RichString converts all strings to scala.util.matching.Regex, so you can do something like this:
"""(?s)a.*b""".r replaceAllIn ( "a\nb\nc\n", "A\nB" )
And that will return
A
B
c
I use this all the time for quick and dirty regex-scripting in the scala console.
Or in this case:
def matchNode( value : String ) : Boolean = {
"""(?s).*(<com:Node>).*""".r.findAllIn( text ) match {
case ScriptNode(v) => System.out.println( "found" + v ); true
case _ => System.out.println("not found: " + value ) ; false
}
}
Just my attempt to reduce the use of the word new in code worldwide. ;)
Just a small addition, use tried to use the (?m) (Multiline) flag (although it might not be suitable here) but here is the right way to use it:
e.g. instead of
val ScriptNode = new Regex("""<com:Node>?m""")
use
val ScriptNode = new Regex("""(?m)<com:Node>""")
But again the (?s) flag is more suitable in this question (adding this answer only because the title is "Scala Regex enable Multiline option")