Error Vba 2042
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Excel Vba Error 2042
a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Why am I getting Error 2042 in VBA Match? up vote 3 down vote
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favorite 1 I have Column A: +--+--------+ | | A | +--+--------+ | 1|123456 | |--+--------+ | 2|Order_No| |--+--------+ | 3| 7 | +--+--------+ Now if I enter: =Match(7,A1:A5,0) into a cell on the sheet I get 3 As a result. (This is desired) But when I enter this line: Dim CurrentShipment As Integer CurrentShipment = 7 CurrentRow = Application.Match(CurrentShipment, Range("A1:A5"), 0) CurrentRow gets a value of "Error 2042" My first vba error 2015 instinct was to make sure that the value 7 was in fact in the range, and it was. My next was maybe the Match function required a string so I tried Dim CurrentShipment As Integer CurrentShipment = 7 CurrentRow = Application.Match(Cstr(CurrentShipment), Range("A1:A5"), 0) to no avail. excel excel-vba excel-2007 share|improve this question asked Mar 20 '13 at 14:45 user2140261 5,21161939 Your code works for me as is. The error message you are getting is the equivalent of #N/A. Perhaps it is focusing on the wrong sheet sometimes? You might try actively selecting the right sheet before running the code. –techturtle Mar 20 '13 at 15:08 add a comment| 5 Answers 5 active oldest votes up vote 3 down vote accepted Try this : CurrentRow = Application.Match(CLng(CurrentShipment), Range("A1:A5"), 0) share|improve this answer answered Mar 20 '13 at 14:50 Vincent MAURY 994 This worked, but why? –user2140261 Mar 20 '13 at 14:52 And it will work for whole numbers, but not for decimal and not for string, so it feels a bit dodgy... –K_B Mar 20 '13 at 15:07 add a comment| up vote 3 down vote As a side note to this and for anyone who gets this error in future, with any fun
VBA Code & Other Help Excel Help [SOLVED] Solved: vlookup error 2042 If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link error 2023 vba above to proceed. To start viewing messages, select the forum that you want to visit vba error handling from the selection below. Results 1 to 13 of 13 Thread: Solved: vlookup error 2042 Thread Tools Show Printable Version Subscribe to vba error 2042 vlookup this Thread… Display Linear Mode Switch to Hybrid Mode Switch to Threaded Mode 02-18-2009,11:27 AM #1 Ago View Profile View Forum Posts View Blog Entries View Articles VBAX Tutor Joined Jan 2008 Posts 266 Location Solved: http://stackoverflow.com/questions/15526784/why-am-i-getting-error-2042-in-vba-match vlookup error 2042 i use a vlookup to find a match on a different sheet with this code: result = Application.VLookup(str, Sheets(3).Range("A1:B" & Cells(Rows.Count, 1).End(xlUp).Row), 2, False) MsgBox result Formatting tags added by mark007 and that works just fine while it finds something, but in 99,99% of the times it will not find anything. that makes the result become "error 2042" and causing it to crash. how can i avoid the msgbox-code when http://www.vbaexpress.com/forum/showthread.php?25189-Solved-vlookup-error-2042 result is 2042? i have tried If result = "error 2042" Formatting tags added by mark007 On Error Goto 1: msgbox result 1: Formatting tags added by mark007 If Is Not err.number = 2042 Formatting tags added by mark007 its probably simple, what did i do wrong? Reply With Quote 02-18-2009,11:44 AM #2 mdmackillop View Profile View Forum Posts View Blog Entries View Articles AdministratorVP-Knowledge Base VBAX Grand Master Joined May 2004 Location Scotland Posts 12,873 Location On Error Resume Next Result = Application.VLookup(MyStr, Sheets(3).Range("A1:B" & Cells(Rows.Count, 1).End(xlUp).Row), 2, False) If Result = "Error 2042" Then Result = "Nothing Found" MsgBox Result On Error GoTo 0 Formatting tags added by mark007 MVP (Excel 2008-2010) "Provide sample data and layout if you want a quicker solution." - MD To help indent your macros try Smart Indent Please remember to mark threads 'Solved' Reply With Quote 02-18-2009,11:45 AM #3 nst1107 View Profile View Forum Posts View Blog Entries View Articles VBAX Tutor Joined Nov 2008 Location Monticello Posts 245 Location Try On Error Resume Next Formatting tags added by mark007 Reply With Quote 02-18-2009,11:50 AM #4 lucas View Profile View Forum Posts View Blog Entries View Articles Moderator VBAX Wizard Joined Jun 2004 Location Tulsa, Oklahoma Posts 7,323 Location There are times when it is acceptable to use On
page describes how to return errors from VBA User Defined Functions. Returning Errors From VBA Functions If you use VBA or another COM language to create User Defined Functions http://www.cpearson.com/excel/ReturningErrors.aspx (functions that are called directly from worksheet cells) in a module https://www.reddit.com/r/excel/comments/3o56v3/vba_getting_rid_of_na_aka_error_2042/ or add-in, you likely will need to return an error value under some circumstances. For example, if a function requires a positive number as a parameter and the user passes in a negative number, you should return a #VALUE error. You might be tempted to return vba error a text string that looks like an error value, but this is not a good idea. Excel will not recognize the text string, for example #VALUE, as a real error, so many functions and formulas may misbehave, especially ISERROR, ISERR, and IFERROR, and ISNA. These functions require a real error value. VBA provides a function called CVErr that vba error 2042 takes a numeric input parameter specifying the error and returns a real error value that Excel will recognize as an error. The values of the input parameter to CVErr are in the XLCVError Enum and are as follows: xlErrDiv0 (= 2007) returns a #DIV/0! error. xlErrNA (= 2042) returns a #N/A error. xlErrName (= 2029) returns a #NAME? error. xlErrNull (= 2000) returns a #NULL! error. xlErrNum (= 2036) returns a #NUM! error. xlErrRef (= 2023) returns a #REF! error. xlErrValue (= 2015) returns a #VALUE! error. The only legal values of the input parameter to CVErr function are those listed above. Any other value causes CVErr to return a #VALUE. This means, unfortunately, that you cannot create your own custom error values. In order to return an error value, the function's return data type must be a Variant. If the return type is any other data type, the CVErr function will terminate VBA execution and Excel will report a #VALUE error in the cell. Note that these errors are meaning
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