The following code is used to store cell values in an array, but when I checked Debug.Printary(j,1), it seems that the values are not displayed (not retrieved).
ary=Array (Cells(j, "E").Value, Cells(j, "F").Value, Cells(j, "G").Value, Cells(j, "H").Value, Cells(j, "I").Value)
What is wrong with the code below?There are no errors.
Dim j As Long
Primary As Variant
For j=1 To LastRow
If Cells(j, "J").Value="JST" Then
aary=Array (Cells(j, "E").Value, Cells(j, "F").Value, Cells(j, "G").Value, Cells(j, "H").Value, Cells(j, "I").Value)
Debug.Printary(j,1)
End If
Next
What is wrong with the code below?There are no errors.
Debug.Printary(j,1) should get an "Index Not in Valid Range" error.
OnError Resume NextIs there a code somewhere that ignores this error?If so, let's comment it out first.
The reason for the error is that the Array function returns a one-dimensional array, so aary is a one-dimensional array.
Nevertheless, we treat it as a two-dimensional array with ary(j,1), which results in the above error.
Debug.Printary(0)
displays the first value of the array (=Cells(j, "E").Value).
If you want to put the first row of columns E through I into an array, you can do it in one row below without looping.
Dim LastRow As Long
LastRow=Cells(Rows.Count, "E").End(xlUp).Row
Primary() As Variant
aary=Range(Cells(1, "E"), Cells(LastRow, "I")).Value
US>'Check the first column (column E) value of the array
Dimi
Fori=1 to LastRow
Debug.Printary(i,1)
Next
By the way, the array substituted with Value as shown above starts with an index of 1.
The Array function starts with 0.
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