Excel VBA - vLook-Up if match first 2 characters of one column with another column

[gloriag]

I am a newbie on VBA. I have a table as below (see also the attachment) and need to do the vlookup with "dept" if the left 2 characters of "EE Code" between 91 and 95. Is it possible to have a macro/VBA to complete the task?

		EE Code	dept	How to generate from VBA Macro
		9216000	12345678
vlookup arrary		A123456	23456700
39914123	test	7856370	39914123
38214125	test	9155800	57889056
40964197	test	9326600	38214125
77014155	test	9135800	39914123
		3456999	40964197
		9458000	77014155
		9537903	00123432
		9158000	12345678
		0456999	57889056
		5856370	39914123

[Bob Phillips]

You don't need VBA, just use a formula

=IF(OR(LEFT($C2,2)="91",LEFT($C2,2)="95"),VLOOKUP($D2,$A$3:$B$7,2,FALSE),"" )

[patel]

Sub a()
LR = Cells(Rows.Count, "C").End(xlUp).Row
drow = 2

For r = 2 To LR
  If Val(Left(Cells(r, 3), 2)) >= 91 And Val(Left(Cells(r, 3), 2)) <= 95 Then
    Cells(drow, 6) = Cells(r, 4)
    drow = drow + 1
  End If
Next
End Sub

[gloriag]

Patel, the code doesn't work. The result shall be equal to Col E as shown below (see also the attachment in my first post).

Col A	Col B	Col C	Col D	Col E	Col F
		EE Code	dept	if left 2 characters of EE code= 91 to 95. then vlookup with dept	How to generate from VBA Macro
		9216000	12345678
vlookup arrary		A123456	23456700
39914123	test	7856370	39914123
38214125	test	9155800	57889056
40964197	test	9326600	38214125	test
77014155	test	9135800	39914123	test
		3456999	40964197
		9458000	77014155	test
		9537903	00123432
		9158000	12345678
		0456999	57889056
		5856370	39914123