Excel VBA loop through columns

王林
Release: 2024-01-15 23:24:25
forward
663 people have browsed it

excel vba列数循环

excel vba column number loop vlookup

oh. Checked my previous code. It is estimated that it may be caused by errors in your iserror and vlookup results! ! !

Because there is a formula behind the vlookup generation, and the parameters provided to iserror are incorrect, it is caused by this.

I just converted it into a numerical format, and then applied iserror, and then it was ok.

My code is as follows:

For i = 2 To module_number - 1

Range("C" & i).Select: ActiveCell.Value = ActiveCell.Value: If IsError(ActiveCell.Value) Then ActiveCell.Value = 0: Module_wrong = 1: iublink = Cells(ActiveCell.Row, 1) : Exit Sub

Next

So, you just need to split the code and convert it into numerical values.

Welcome to the If Software Changes Life group, let’s work together to improve work efficiency.

VBA vlookup application

Public Sub aaaa()

Dim x As String, w As String

Dim z As Long

x = "Y"

w = "N"

For z = 2 To 10000

If Cells(z, 1) "" Then

Cells(z, 2).FormulaR1C1 = "=IF(ISNUMBER(VLOOKUP(RC[-1],C[1],1,0)),""" & x & """,""" & w & """)"

Cells(z, 2).Value = Cells(z, 2).Value

End If

Next

End Sub

----------------

vlookup will return an error value when the data cannot be found, so it cannot be used directly

If you must use vlookup, please see the following code:

------------------------

Public Sub dsadsa()

Dim x As String, w As String

Dim z As Long, u As Variant

Dim o As Range: Set o = Worksheets("Sheet1").Range("c:c")

x = "Y"

w = "N"

For z = 2 To 6

If Cells(z, 1) "" Then

On Error Resume Next

u = Application.WorksheetFunction.VLookup(Sheets("sheet1").Cells(z, 1), o, 1, False)

If Err.Number = 0 Then

Cells(z, 2) = x

Else

Cells(z, 2) = w

End If

Err.Clear

End If

Next

End Sub

The above is the detailed content of Excel VBA loop through columns. For more information, please follow other related articles on the PHP Chinese website!

source:docexcel.net
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template