vba 7 位数字后跟可选的 3 个字母的正则表达式
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Regular expression for 7 digits followed by an optional 3 letters
提问by Lil' Bits
I'm new to regular expressions, and I'm trying to validate receipt numbers in our database with a regular expression.
我是正则表达式的新手，我正在尝试使用正则表达式验证我们数据库中的收据编号。
Our receipts can come in the following formats:
我们的收据可以采用以下格式：
- 0123456 (Manditory seven digits, no more, no less)
- 0126456a (Manditory seven digits with one letter a-z)
- 0126456ab (Manditory seven digits with two letters a-z)
- 0126456abc (Manditory seven digits with three letters a-z)
- 0123456（强制七位数，不多不少）
- 0126456a（强制七位数字加一个字母az）
- 0126456ab（强制七位数字加两个字母az）
- 0126456abc（强制七位数字加三个字母az）
I've tried using a bunch of different regex combinations, but none seem to work. Right now I have:
我试过使用一堆不同的正则表达式组合，但似乎都不起作用。现在我有：
(\d)(\d)(\d)(\d)(\d)(\d)(\d)([a-z])?([a-z])?([a-z])?
But this is allowing for more than seven digits, and is allowing more than 3 letters.
但这允许超过 7 个数字，并且允许超过 3 个字母。
Here is my VBA function within Access 2010 that will validate the expression:
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
回答by Rohit Jain
You probably need to use anchors at the ends. Further your regex can be simplified to: -
您可能需要在末端使用锚点。此外，您的正则表达式可以简化为：-
^\d{7}[a-z]{0,3}$
\d{7}matches exactly7 digits. You don't need to use\d7 times for that.{0,3}creates a range, and matches 0 to 3 repetition of preceding pattern,Caret(^)matches the start of the lineDollar($)matches the end of the line.
\d{7}完全匹配7 digits。您不需要为此使用\d7 次。{0,3}创建一个范围，并匹配前面模式的 0 到 3 个重复，Caret(^)匹配行的开头Dollar($)匹配行尾。
回答by Akshay Joshi
^(\d){7}[a-z]{0,3}$might work well. The ^and $will match the start and end of line respectively.
^(\d){7}[a-z]{0,3}$可能工作得很好。在^和$将分别匹配行的开头和结尾。
回答by Kendall Frey
You may want to make sure you are matching the entire string, by using anchors.
您可能希望通过使用锚点来确保匹配整个字符串。
^(\d)(\d)(\d)(\d)(\d)(\d)(\d)([a-z])?([a-z])?([a-z])?$
You can also simplify the regex. First, you don't need all those parentheses.
您还可以简化正则表达式。首先，您不需要所有这些括号。
^\d\d\d\d\d\d\d[a-z]?[a-z]?[a-z]?$
Also, you can use limited repetition, to prevent repeating yourself.
此外，您可以使用有限的重复，以防止重复自己。
^\d{7}[a-z]{0,3}$
Where {7}means 'exactly 7 times', and {0,3}means '0-3 times'.
其中{7}表示“正好 7 次”，{0,3}表示“0-3 次”。
