正则表达式为7位数后跟可选的3个字母

Question

我是正则表达式的新手,我正在尝试使用正则表达式验证数据库中的收据编号.

我们的收据可以采用以下格式:

• 0123456(Manditory七位数,不多也不少)
• 0126456a(带一个字母az的七位数)
• 0126456ab(两位字母az的七位数)
• 0126456abc(带三个字母az的七位数)

我尝试过使用一堆不同的正则表达式组合,但似乎都没有.现在我有:

(\d)(\d)(\d)(\d)(\d)(\d)(\d)([a-z])?([a-z])?([a-z])?

但这允许超过七位数,并允许超过3个字母.

这是我在Access 2010中的VBA函数,它将验证表达式:

Function ValidateReceiptNumber(ByVal sReceipt As String) As Boolean

    If (Len(sReceipt) = 0) Then
        ValidateReceiptNumber = False
        Exit Function
    End If

    Dim oRegularExpression     As RegExp

'   Sets the regular expression object
    Set oRegularExpression = New RegExp

    With oRegularExpression
'   Sets the regular expression pattern
        .Pattern = "(\d)(\d)(\d)(\d)(\d)(\d)(\d)([a-z])?([a-z])?([a-z])?"

'   Ignores case
        .IgnoreCase = True

'       Test Receipt string
        ValidateReceiptNumber = .Test(sReceipt)

    End With
End Function

Answer 1

您可能需要在末端使用锚点.您的正则表达式可以简化为: -

^\d{7}[a-z]{0,3}$

• \d{7}完全匹配7 digits.你不需要使用\d7次.
• {0,3} 创建一个范围,并匹配前面模式的0到3次重复,
• Caret(^) 匹配行的开头
• Dollar($) 匹配行的结尾.

Answer 2

^(\d){7}[a-z]{0,3}$可能运作良好.在^和$将分别匹配行的开头和结尾.