I have this code:
If cells(3,11).value="c:users\flap-fis.txt"
This condition is getting failed every time either we have file location of or not.
How can I write the code how to check specific cell value with static string?
Do you know how to debug VBA projects? Best thing to do for such a case is putting a breakpoint on that particular line and use the "immediate" window, where you can ask ? <var> in order to know the value of a variable, as in these examples:
? Cells(3, 11).Value = "c:users\flap-fis.txt"
False
? Cells(3, 11).Value
c:\users\flap-fis.txt
=> Mind the extra backslash, as already mentioned by PEH.
Related
From title you can already tell what is my problem.
Somethink from my side:
I know there is for sure work around to run the macro once with creating boolean or check and hadling the error (with I still dont know how).
I want to know why the macro runs twice. My guess is that it takes mouse movement or somethink like that.
How to fix it?
Or use better alternative of capturing click on text (I want to evoid selection change/change and I am not big fan of using FollowHyperlink with linking on same cell).
Function I am using : =HYPERTEXTOVÝ.ODKAZ("#LinkClick()";"CLICK")
Eng version : =HYPERLINK("#LinkClick()";"CLICK")
Function LinkClick()
Range("A1").Value = Range("A1").Value + 1
End Function
It should be the same function. It is just different in my language :
https://support.office.com/cs-cz/article/hypertextov%C3%BD-odkaz-funkce-333c7ce6-c5ae-4164-9c47-7de9b76f577f
https://support.office.com/en-us/article/hyperlink-function-333c7ce6-c5ae-4164-9c47-7de9b76f577f?omkt=en-US&ui=en-US&rs=en-US&ad=US
PS: My first post and my english isn't best. Thanks for any answers.
You need to Set LinkClick = Selection so you return a cell with your function otherwise the link is invalid.
According to the documentation your formula =HYPERLINK("#LinkClick()";"CLICK") needs a link_location as first parameter HYPERLINK(link_location, [friendly_name]). But because you have a function call there "#LinkClick()" the function needs to return a valid link location, and that is what Set LinkClick = Selection does, it returns the actual selection as link location, so the hyperlink selects what is already selected (means it does nothing at all, but it doesn't complain about an invalid link location).
Option Explicit
Public Function LinkClick() As Range
Set LinkClick = Selection 'make sure a valid link location is returned in the function
ActiveSheet.Range("A1").Value = ActiveSheet.Range("A1").Value + 1
End Function
I am trying to remove duplicate values except the first one.
My solution to this was to conditionally format all duplicates, then, working backwards, clear contents of the formatted cells. This would mean that the first one will stop being formatted once all duplicates are removed.
What I have been trying:
For i = LaC To 5 Step -1
LR = ws.Cells(Rows.Count, LaC).End(xlUp).Row
For j = 2 To LR
cond = (ws.Cells(j, i).DisplayFormat.Interior.ColourIndex.Value)
If cond = 22 Then
ws.Cells(j, i).ClearContents
End If
Next
Next
Basically, if I try ws.Cells(j, i).DisplayFormat.Interior.ColourIndex in the immediate window, it returns 22.
However, if I try this code, I get error:
Object doesn't support this property or method (Error 438)
Any assistance would be greatly appreciated.
If it is not necessary to use VBA you can get rid of duplicates in the GUI via “Data” → “Data Tools” → “Remove Duplicates”. Or, in current versions of Excel O365, you can use the UNIQUE-function like this
UNIQUE(A:A)
assuming your source data is in row A. This will keep up with changing data.
Check this line more carefully:
cond = (ws.Cells(j, i).DisplayFormat.Interior.ColourIndex.Value)
ColorIndex is a Property, not an Object. As such, ColorIndex.Value will return an error. Note that you haven't included the .Value when testing in the Immediate Window, and it works as expected:
?ws.Cells(j, i).DisplayFormat.Interior.ColourIndex 'This will work
?ws.Cells(j, i).DisplayFormat.Interior.ColourIndex.Value 'This will err
Also, there is no need for you to put it in brackets, either:
cond = ws.Cells(j, i).DisplayFormat.Interior.ColourIndex
Ans: I was writing Colour instead of Color.
hei everyone,
am i missing something here?
i'm trying to compare two cells which are both formatted the same.
the way i compare it is such (never mind the "i" variable, the code is inside a for loop):
If SomeSheet.Cells(i, col).Value <> SomeOtherSheet.Cells(i, 4).Value Then
result = False
End If
I have tried converting the values to doubles, but my macro still sees them as different even thought the values are identical.
When I debug the code and try to see the actual values in the immediat window, they are ideed the same:
? SomeSheet.Cells(i, col).Value
310289286,463803
? SomeOtherSheet.Cells(i, 4).Value
310289286,463803
does anyone have any idea?
thanks!
EDIT: As suggested bu FaneDuru in the comments, I tried to round the values and even though i get the same output in the immediate window, the comparrison is correct now. Don't know why. Thanks!
If result has been defined as a boolean variable, the default value will be false. So unless there is another part of the code where result is set to true, It wil always return false.
Maybe try something like this:
If SomeSheet.Cells(i, col).Value = SomeOtherSheet.Cells(i, 4).Value Then
result = True
End If
I know the Headline sounds odd so I will start off with a screenshot:
As you can see, the problem is that the point suddenly changes to a comma when I look up an ID in the UserForm.
Before recalling Infos, I am saving all Information rather straightforward:
with ws
Range("BH" & lastRow).value = Me.payinfoOnTime
Range("BI" & lastRow).value = Me.payinfo30
Range("BJ" & lastRow).value = Me.payinfo60
Range("BK" & lastRow).value = Me.payinfo90
Range("BL" & lastRow).value = Me.payinfo90more
End with
Recalling the respective info for a searched ID is done by:
Set FoundRange = ws.Range("D4:D500").Find(What:=Me.SearchSuppNo, LookIn:=xlValues)
With ws
Me.SEpayinfoontime = FoundRange.Offset(0, 56)
Me.SEpayinfo30 = FoundRange.Offset(0, 57)
Me.SEpayinfo60 = FoundRange.Offset(0, 58)
Me.SEpayinfo90 = FoundRange.Offset(0, 59)
Me.SEpayinfo90more = FoundRange.Offset(0, 60)
end with
The Problem is that later calculations for scores are depending on those textboxes and I constantly get an error, unless I always manually change the commas back to points.
Any ideas how I can fix this?
The line:
Me.SEpayinfoontime = FoundRange.Offset(0, 56)
is in fact:
Me.SEpayinfoontime.Value = FoundRange.Offset(0, 56).Value
When you populate an MSForms.TextBox using the .Value property (typed As Variant), like you implicitly do, and providing a number on the right side, the compiler passes the value to the TextBox as a number, and then the value is automatically converted to string inside the TextBox.
Exactly how that conversion happens does not appear to be documented, and from experiment, it would appear there is a problem with it.
When you freshly start Excel, it would appear assigning .Value will convert the number using the en-us locale, even if your system locale is different. But as soon as you go to the Control Panel and change your current locale to something else, .Value begins to respect the system locale, and changes its result depending on what is currently selected.
It should not be happening and I would see it as an Excel bug.
But if you instead assign the .Text property, the number is converted to string using the current system decimal dot, and that conversion happens outside of the TextBox, because the compiler knows .Text is a string, so it converts the right-hand side number to string beforehand.
So in your situation I would:
Make sure I always use the .Text property explicitly:
Me.SEpayinfoontime.Text = ...
Make sure I explicitly use the correct kind of functions to convert between text and numbers:
Me.SEpayinfoontime.Text = CStr(FoundRange.Offset(0, 56).Value)
MsgBox CInt(Me.SEpayinfoontime.Text) / 10
although this step is optional and represents my personal preference. Given that it's a string on the left side of the assignment, VB will use CStr automatically.
Go to Excel's settings to make sure the "Use system separators" tick is set.
Check what locale is selected in the Control Panel - Language and Regional settings.
If it is not En-Us, I would select En-Us to make sure the decimal separator is a dot there.
Restart Excel.
I am trying to perform some operations over some data, but they are not working and I need to find a way to ignore those cells which don´t meet all the requirements.
Basically, I have a column where some cells have text + numbers in their content and other have only text. I search inside all of them and split TEXT in one column and NUMBER in another one. Then, I run a macro to find the matching text to each one in other column.
But when I try to split TEXT from NUMBERS, I search for the first "(", cause my number format is (10.10.10), but if it is not found, the cell value appears: #VALUE! (ok, it is expected cause the character was not found). Here is the problem: if I run my macro over "#VALUE! it crashs and don´t finish its execution.
I have tried to use
On Error GoTo
in my macro code, but for some reason it doesn´t not handle the "Run time error: Type Mismatch".
For contadorOr = 2 To colO
For contadorDes = 2 To colA
On Error GoTo cont
If InStr(1, Cells(contadorDes, colunaDestino).Value, Cells(contadorOr, colunaOrigem).Value) Then
If InStr(1, Cells(contadorOr, colunaOrigem + 4).Value, Cells(contadorDes, colunaDestino + 1).Value) Then
Cells(contadorOr, colunaOrigem + 5).Value = "Mesma versão"
End If
Exit For
End If
Next contadorDes
cont: Next contadorOr
Any suggestions? I can think of ignoring this error (when it´s happen, my variable contadorOr is incremented and go to no next value) or any way to avoid #VALUE! returned by my functions, but haven´t had success doing that.
Thanks in advance.
Instead of using error handling options you could check if cell which you are going to check/process doesn't return error. This is quite simple as presented below:
If IsError(Cells(contadorDes, colunaDestino).Value) Then
'to do anything if there is error
'usually...do nothing
Else
'do what you want if there is no error
End if