Question

请协助适当的Ref对2封信进行匹配,随后将6个整数合并在一起。

These would be valid: 
RJ123456
PY654321
DD321234

These would not
DDD12345
12DDD123

Answer 1

[a-zA-Z]{2}d{6}

[a-zA-Z]{2} means two letters d{6} means 6 digits

If you want only uppercase letters, then:

[A-Z]{2}d{6}

Answer 2

你可以尝试这样的事情:

[a-zA-Z]{2}[0-9]{6}

这里的表述是:

[a-zA-Z]    # Match a single character present in the list below
               # A character in the range between “a” and “z”
               # A character in the range between “A” and “Z”
   {2}         # Exactly 2 times
[0-9]       # Match a single character in the range between “0” and “9”
   {6}         # Exactly 6 times

这将在某个主题中与任何地方相匹配。如果你需要围绕这一主题划定界限,你可以做以下两种工作:

^[a-zA-Z]{2}[0-9]{6}$

以确保整个主题一致。一、导言

or

[a-zA-Z]{2}[0-9]{6}

http://www.google.co.uk/url?sa=t&rct=j&q=word%20boundary%20regex&source=web&cd=1&sqi=2&ved=0CCwQFjA&url=http://www. norm-expressions.info/word约束aries.html&ei=Pv2iT5zYmjiQfyq-Wpw&usg=AFQjprelk=

如“Phrogz”所指出的,你可以通过替换<代码>[0-9]<>>>>而使这一表述变得更可怕,如其他一些答复所示。

[a-zA-Z]{2}d{6}

Answer 3

我取决于你使用什么reg语,但非正式的是:

[:alpha:][:alpha:][:digit:][:digit:][:digit:][:digit:][:digit:][:digit:]

where [:alpha:] = [a-zA-Z] and [:digit:] = [0-9]

如果你使用允许明确重复的常规语言,那么,就象:

[:alpha:]{2}[:digit:]{6}

正确的辛迪加取决于你所使用的特定语言,但这是想法。

Answer 4

Everything you need here can be found in this quickstart guide. A straightforward solution would be [A-Za-z][A-Za-z]dddddd or [A-Za-z]{2}d{6}.

如果你只接受英文大写字母,则将<代码>[A-Za-z]改为[A-Z]。

Answer 5

根据您的reg夫的支持,我可以使用:

[A-Z]{2}d{6}    # Ensure there are "word boundaries" on either side, or

(?<![A-Z])[A-Z]{2}d{6}(?!d) # Ensure there isn t a uppercase letter before
                              # and that there is not a digit after

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