Search List of Data

  • Thread starter Thread starter Nigel
  • Start date Start date
N

Nigel

Hi All

Not an area I have explored before so any help appreciated.

I have a range of data in three columns of a worksheet (the data), I have a
userform with a text box control into which the user enters a value
(string).

Another combo box control on the same userform is currently empty. I wish
to use the string entered by the user to search column 1 of the data, if it
is found (and it may find it more than once) the combo box is filled with
the values in the associated second column of the data. If not a message is
displayed.

The user then scans the combobox contents and select the required matching
record. After which a third control on the userfrom shows the value in the
associated third column of the data.

Cheers
Nigel
 
Private Sub Combobox1_Click()
set rng = Range(cells(1,1),cells(rows.count,1).end(xlup))
combobox2.clear
for each cell in rng
if cell.value = combobox1.Value then
combobox2.AddItem cell.offset(0,1).Value
end if
Next
End If

Private Sub Combobox2_Click()
set rng = Range(cells(1,1),cells(rows.count,1).end(xlup))
combobox3.clear
for each cell in rng
if cell.value = combobox1.Value then
if cell.offset(0,1).value = combobox2.Value then
combobox3.AddItem cell.offset(0,2).Value
End if
end if
Next
End If


--
Regards,
Tom Ogilvy


Nigel said:
Hi All

Not an area I have explored before so any help appreciated.

I have a range of data in three columns of a worksheet (the data), I have a
userform with a text box control into which the user enters a value
(string).

Another combo box control on the same userform is currently empty. I wish
to use the string entered by the user to search column 1 of the data, if it
is found (and it may find it more than once) the combo box is filled with
the values in the associated second column of the data. If not a message is
displayed.

The user then scans the combobox contents and select the required matching
record. After which a third control on the userfrom shows the value in the
associated third column of the data.

Cheers
Nigel




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