split來拆解日期

'######用 split來拆解 1992/1/21 這種日期格式, 給y1, m1, d1 的三個下拉選單#######
Private Sub Read_Birthday()
Dim cmA3 As New OleDbCommand("select birthday from resume where resume_no ='" & Session("resume_no") & "'", conn2)
conn2.Open()
Dim drA3 As OleDbDataReader = cmA3.ExecuteReader()
If drA3.Read Then
'Dim birth As String = drA3.Item("birthday").ToString '這種寫法會多跑出 時間, 造成程式錯誤
Dim birth As String = drA3.Item("birthday")
Dim birth_ymd() As String
birth_ymd = Split(birth, "/") '用字串陣列去接1992/2/2的這個字串, 以 / 來拆解成三個數字
y1.SelectedValue = birth_ymd(0)
m1.SelectedValue = birth_ymd(1)
d1.SelectedValue = birth_ymd(2)
End If
drA3.Close()
conn2.Close()
End Sub