Application.evaluate Error 2015

formulas as strings and have Excel VBA compute them for you. Only recently I discovered that some of the cells in a workbook contained #VALUE? instead of a sound value. After some investigation I found out that, without raising

Error 2015 Vba

any error at application level (in VBA), the Application.Evaluate function failed consistently with this error vba error 2015 vlookup whenever a particular condition occurred. This condition is related to limits in the underlying software layer of Excel, the one that runs excel error 2015 evaluate Application.Evaluate. Apparently this particular error occurs when the parsed function returns a string value having a length above 255 characters. As long as the length of the string is less than or equal to 255 characters, the

Error 2029 Vba Evaluate

return value is the desired string. Above 255 characters, the function "crashes". In my case, I was not immediately attended on this error because I used a Variant to store the returned value from Application.Evaluate (sample code): Dim vValue As Variant vValue = Application.Evaluate("GetData()") ' Error 2015 saved in vValue if GetData returns 255+ characters ActiveCell.Value = vValue If I would have used a String to store the value, I would probably have gotten

Excel Vba Error 2029

a "Runtime Error 13 - Type Mismatch" with Excel booming out of my routine without any debugging possibility (sample code): Dim sValue As String sValue = Application.Evaluate("GetData()") ' Run-time error if GetData() returns 255+ characters ActiveCell.Value = sValue To catch problems, I now completed code with an If statement (sample code): Dim vValue As Variant vValue = Application.Evaluate("GetData()") If (VBA.VarType(vValue) = vbError) Then ActiveCell.Value = "String Overflow on GetData()" Else ActiveCell.Value = vValue End If There seem to be more Excel functions that suffer from this 255 character limit, but could not establish which. For the moment, I will apply some workarounds until it gets solved by Microsoft. Dutch Like this:Like Loading... Related Comments (3) 3 Comments » Thanks for this, I was passing through a #value error as I was resetting a combo control Cell link to 0, by defining the field as String trapped the error allowing the "on error goto XXXX" to work as required Comment by John Allott -- June 21, 2012 @ 11:42 pm | Reply The error trapping in your post was very helpful for me, although for a different error type. This line of code was my aid: If (VBA.VarType(vValue) = vbError) Then I wanted to identify with VBA the first cell in a column that returns "" constant (double quotes) by using Application

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Excel Vba Vlookup

Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users vba error handling Badges Ask 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 them; it only takes a https://dutchgemini.wordpress.com/2009/08/07/error-2015-using-application-evaluate-in-excel-vba/ 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 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, http://stackoverflow.com/questions/22014093/excel-vba-find-string-error-2015 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 '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|impro

Forum Microsoft Office Application Help - Excel Help forum Excel Programming / VBA / Macros Error 2015 with Application.Evaluate To get replies by our experts at nominal charges, follow this link to buy points and post your thread in our Commercial Services forum! Here is http://www.excelforum.com/showthread.php?t=548937 the FAQ for this forum. + Reply to Thread Results 1 to 4 of 4 Error 2015 with Application.Evaluate Thread Tools Show Printable Version Subscribe to this Thread… Mark this thread as solved… Rate This Thread Current Rating http://learn-to-learn.854150.n3.nabble.com/Error-2015-using-Application-Evaluate-in-Excel-VBA-td922376.html ‎ Excellent ‎ Good ‎ Average ‎ Bad ‎ Terrible Display Linear Mode Switch to Hybrid Mode Switch to Threaded Mode 06-06-2006,09:10 AM #1 Jeff Guest Error 2015 with Application.Evaluate Hello, If someone could help me correct my VBA error 2015 procedure. I'm lost and I don't know how to fix it. I get #value error. Here's an example of my spreadsheet: A B C D E F G 20 6 F E Escada 1,940 495,866 Here's my VBA procedure: Dim mtchValue As Variant Dim getvalue As Variant Dim sh As String Dim fname As Variant fname = Application.GetOpenFilename If fname = False Then Exit Sub 'cancel End If sh = MakeReferenceNicer(fname) Workbooks.Open Filename:=fname mtchValue = ("MATCH(1,(" & sh application.evaluate error 2015 & "A1:A10000=20)*" & _ "(" & sh & "B1:B10000=6)*" & "(" & sh & _ "C1:C10000=""F"")*" & "(" & sh & _ "E1:E10000=""Escada""),0)") If Not IsError(mtchValue) Then getvalue = Application.Evaluate("Index(" & sh & "F1:F10000" & mtchValue & ")") End If Range("S1").Select ActiveCell.Value = getvalue End Sub Function MakeReferenceNicer(fname) As String 'taking '"C:\My Documents\excel\book5.xls" 'shooting for: '"'C:\My Documents\excel\[book5.xls]May'!" Dim iCtr As Long Dim myStr As String For iCtr = Len(fname) To 1 Step -1 If Mid(fname, iCtr, 1) = "\" Then 'found that last backslash myStr = "'" & Left(fname, iCtr) & _ "[" & Mid(fname, iCtr + 1) & "]May'!" Exit For End If Next iCtr MakeReferenceNicer = myStr End Function -- Regards, Jeff Register To Reply 06-06-2006,10:40 AM #2 Dave Peterson Guest Re: Error 2015 with Application.Evaluate Without testing.... getvalue _ = Application.Evaluate("Index(" & sh & "F1:F10000" & mtchValue & ")") needs an extra comma: getvalue _ = Application.Evaluate("Index(" & sh & "F1:F10000" & "," & mtchValue & ")") Jeff wrote: > > Hello, > > If someone could help me correct my VBA procedure. I'm lost and I don't know > how to fix it. I get #value error. > > Here's an example of my spreadsheet: > > A B C D E F G > 20 6 F E Escada 1,940 495,866 > Here's my VBA procedure: > Dim mtchValue As Variant > Dim getvalue As Varian

does not seem so perfect, which I just found after I had submitted my Excel VBA homework to my teacher already! Attention should be paid to this unconspicuous problem. [] and application.evaluate are two ways for us to invoke operation just like in the worksheet and sometime this practice is much more efficient. what's more, the latter way is much powerful than the first one for we can use string characters including variables for code. For example, application.evaluate("sum(" & your_array &")") (while ["sum(" & your_array &")"] doesn't work well). However there is a limit for this potential usage: as you can see from Excel help, while for a general string ·A variable-length string can contain up to approximately 2 billion (2^31) characters. ·A fixed-length string can contain 1 to approximately 64K (2^16) characters. For codes for String characters the length ranges from 0-255! means that you cannot include string code with length larger than 255 as parameters of application.evaluate else what you get is just Error 2015!! As for the "solution" for this problem I refer you to Butchgemini's blog####################################################################### Here is my homework(It's a problem about transfer probability matrix in Markov chain and we want to find out the converged probability matrix. I try to use array object to complete my task in this work) which remind me of this problem. (In my case I think we can deal with it by control the length of string within 255 but I did not do it in the workbook now) %E6%9C%B1%E5%85%83%E4%BC%9F0618%E7%AC%AC%E4%B8%80%E9%A2%981.xlsm Add a new comment « Return to Learn to Learn | 1 view|%1 views Loading... Free forum by Nabble Edit this page