Excel Error 2015
Contents |
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring excel macro error 2015 developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question
Excel Error 2029
x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; vba excel error 400 it only takes a minute: Sign up Excel VBA find string : Error 2015 up vote 1 down vote favorite I have to following code snippet ... Public Sub FindText(path As String, file As String) Dim Found As Range vba error 2042 myText = "test(" MacroBook = ActiveWorkbook.Name ' Open the File Workbooks.Open path & file, ReadOnly:=True, UpdateLinks:=False For Each ws In Workbooks(file).Worksheets With ws Set Found = .UsedRange.Find(What:=myText, LookIn:=xlFormulas, _ LookAt:=xlPart, MatchCase:=False) If Not Found Is Nothing Then ' do stuff ' ... I see in the debugger that Found contains Error 2015! The sheet contains the text I want in the formula. Any ideas why I'm getting the error? Thanks excel vba excel-vba find share|improve this question edited Feb 25
Vba Error 2402
'14 at 13:15 asked Feb 25 '14 at 12:22 Rueful Rabbit 3127 1 it's because your formula in the sheet returns #VALUE! error. You can handle it using IsError: If Not IsError(Found) Then –simoco Feb 25 '14 at 12:36 is this a sub or function? please, show complete code. –KazimierzJawor Feb 25 '14 at 12:43 1 Bravo Simoco, nice catch! –Rueful Rabbit Feb 25 '14 at 13:21 add a comment| 2 Answers 2 active oldest votes up vote 2 down vote accepted As follow up from comments to the Q, Error 2015 occurs because your formula in the sheet returns #VALUE! error. You can handle it using IsError: If Not Found Is Nothing Then If Not IsError(Found) Then ' do sth End If End If share|improve this answer edited Feb 25 '14 at 16:41 answered Feb 25 '14 at 13:41 simoco 26.7k93552 is there a way of finding which formula is giving the error? –user3540466 Dec 30 '15 at 16:30 add a comment| up vote 0 down vote You don't need to use 'Set' in your code. You only use this to assign a reference to an object. Try:- For Each ws In Workbooks(file).Worksheets With ws Found = .UsedRange.Find(What:=myText, LookIn:=xlFormulas, _ LookAt:=xlPart, MatchCase:=False) If Not Found Is Nothing Then ' do stuff ' ... Hopefully this should work. share|improve this answer answered Feb 25 '14 at 12:28 Mat Richardson 2
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about excel error 2015 vlookup hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask excel vba error 2015 evaluate Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join
Vba Error 2015 Vlookup
them; it only takes a minute: Sign up Excel VBA - Array Formula returns #VALUE! (Error 2015) in case a cell contains more than 255 characters up vote 2 down vote favorite 1 Dear all I am using an http://stackoverflow.com/questions/22014093/excel-vba-find-string-error-2015 array formula and pass the result to a VBA macro, which means the result of the formula is passed as array to my macro. In some cases I get in the array a #VALUE (error 2015) and I have no clue why. When I play arround with the affected cell where it gets the data from then it is happening when the cell has quite an amount of characters (~3000 chars). Has anyone a clue what I can http://stackoverflow.com/questions/35436777/excel-vba-array-formula-returns-value-error-2015-in-case-a-cell-contains-m try to solve the isuse? formula: {=arrayToCSV(removeElementsFrom2DimArray(IF('sheet'!$G$1:$G$2000=A1;'sheet'!$F$1:$F$2000)))} the IF('sheet'!$G$1:$G$2000=A1;'sheet'!$F$1:$F$2000) part returns an array like {FALSE, FALSE, FALSE... Value, Value, FALSE , ...}. This one I pass to my mVBA function removeElementsFrom2DimArray to remove the FALSE elements Now it seems that the IF('sheet'!$G$1:$G$2000=A1;'sheet'!$F$1:$F$2000) return in some cases #VALUE(Error 2015) for some of the elements i.e. {FALSE, FALSE, FALSE... #VALUE!, Value, FALSE , ...} Function removeElementsFrom2DimArray(ByRef arr() As Variant, Optional value As Variant = False) As String() On Error GoTo ErrorHandler Dim coll As New Collection Dim i As Integer If (IsArray(arr)) Then For i = 1 To UBound(arr, 1) If (arr(i, 1) <> value) Then --> Here happens the error that arr(i, 1) return #VALUE! Error 2015. Why? coll.Add (arr(i, 1)) End If Next i End If removeElementsFrom2DimArray = collectionToArray(coll) Exit Function ErrorHandler: MsgBox Err.Description Resume End Function UPDATES: I tried to modify to pass over arr as Variant or ByVal. It did not made any difference. I tried to modify to pass over arr as array arr() with and without the (). It did not made any difference. If I reduce the number of chars in the affected cell 'sheet'!$F$1:$F$2000 to below 256 chars then it works! If I use just the formula {=IF(sheet!$G$1:$G$2000=A1;sheet!$F$1:$F$2000)} (array formula) then it works too! so I assume there is a limitation in passing arrays from native formulas to own functions? So that values in the array with mor
be down. Please try the request again. Your cache administrator is webmaster. Generated Sat, 15 Oct 2016 09:02:42 GMT by s_ac15 (squid/3.5.20)
Forums Excel Questions Using Excel SEARCH in VBA (Error 2015) Results 1 to 6 of 6 Using Excel SEARCH in VBA (Error 2015)This is a discussion on Using Excel SEARCH in VBA (Error 2015) within the Excel Questions forums, part of the Question Forums category; Hello All, I have this macro... Code: Sub Indent() Dim value As Variant Dim cell As range Dim range As ... LinkBack LinkBack URL About LinkBacks Bookmark & Share Digg this Thread!Add Thread to del.icio.usBookmark in TechnoratiTweet this thread Thread Tools Show Printable Version Display Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Sep 7th, 2010,03:39 PM #1 dehinson New Member Join Date Sep 2010 Posts 2 Using Excel SEARCH in VBA (Error 2015) Hello All, I have this macro... Code: Sub Indent() Dim value As Variant Dim cell As range Dim range As range Dim source As range Dim sourcecell As range Set source = ActiveSheet.range("A:A") ActiveCell.EntireColumn.Select Set range = Selection Dim SearchText As String Dim SearchCell As String SearchText = "." For Each cell In range Set sourcecell = source.Cells(cell.Row) SearchCell = sourcecell.Address SearchCell = Replace(SearchCell, """", "") searchformula = "=SEARCH(""" & SearchText & """," & SearchCell & ",1)" value = Application.Evaluate(searchformula) cell.IndentLevel = cell.Offset(0, -1) Next End Sub It is not done yet. However, as I debug it, I am receiving an error 2015 on the Application.Evaluate line. The value of the searchformula is: "SEARCH(".",$A$1,1)". This looks like the same formula I would use in an Excel cell. Why am I getting this error? Share Share this post on Digg Del.icio.us Technorati Twitter Reply With Quote Sep 7th, 2010,05:02 PM #2 Scott Huish MrExcel MVPModerator Join Date Mar 2004 Location Oregon Posts 18,424 Re: Using Exc