lua:模式匹配和提取电话号码 [英] lua: pattern matching and extracting a phone number

问题描述

在制作Lua中具有以下要求的功能时遇到麻烦:

I am having trouble crafting a function that has the following requirements in Lua:

• 以字符串phone_number和2位数字country_code作为输入.
• phone_number的格式为{1 || "} {country_code} {10或11位手机号码}

• Takes a string phone_number and 2-digit country_code as input.
• phone_number has the form {1 || ""}{country_code}{10 or 11-digit mobile number}

我需要10或11位手机号码作为输出.

I need as output the 10 or 11-digit mobile number.

示例I/O:

phone_number ="552234332344"，country_code ="55" =>"2234332344"

phone_number= "552234332344", country_code= "55" => "2234332344"

phone_number ="15522343323443"，country_code ="55" =>"22343323443"

phone_number= "15522343323443", country_code= "55" => "22343323443"

谢谢！

推荐答案

尝试"(1?)(%d%d)(%d+)".在您的示例中使用它:

Try "(1?)(%d%d)(%d+)". Using this with your examples:

print(("15522343323443"):match("(1?)(%d%d)(%d+)"))
print(("5522343323443"):match("(1?)(%d%d)(%d+)"))

将打印:

1   55  22343323443
55  22343323443

如果电话号码中恰好需要10或11位数字，则指定%d 10次，然后添加%d?. %d是与任何数字匹配的字符类，问号修饰符与前一个字符或字符类0或1次匹配.

If you need exactly 10 or 11 digits in the phone number, then specify %d 10 times and then add %d?. %d is a character class that matches any number and question mark modifier matches the previous character or a character class 0 or 1 time.

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