java - 如何在不拆分字符串的情况下检查字符串是否具有某种模式？

Question

所以这就是问题所在。有人可以告诉我正则表达式模式是什么样子吗？

Write a program that determines if the input string matches the following format:

Format: PRODUCT_ID.PRODUCT_CATEGORY-LOCATOR_TYPE[LOCATOR_LOT]

• PRODUCT_ID = always starts with # followed by 3 zeros, followed by a numeric value that can be 1-7 digits, in the range 1-9999999
• PRODUCT_CATEGORY = 1-4 uppercase alphabetic characters
• LOCATOR_TYPE = a single uppercase X, Y or Z character
• LOCATOR_LOT = 1-2 numeric digits, in the range 1-99

All other format characters are the literal characters

Return true if it matches and false otherwise.

这是函数声明:

public boolean checkPattern(String s){    
}

我尝试拆分字符串，然后检查每个字符，但它变得非常复杂。

这是到目前为止我所得到的正则表达式:

String regex = "#000^([1-9]|[1-9][0-9]|[1-9][0-9][0-9]|[1-9][0-9][0-9][0-9]|[1-9][0-9][0-9][0-9][0-9]|[1-9][0-9][0-9]|[1-9][0-9][0-9])$";

这是我开始做的事情，但它又长又复杂，甚至不完整(只检查产品 ID)，我认为我没有走在正确的轨道上

Answer 1

这是一个有效的正则表达式:

#000[1-9]\d{0,6}\.[A-Z]{1,4}\-[XYZ]\[[1-9]\d?\]

这是一个分割:

#000    # match the literal characters, #000
[1-9]   # any digit 1 to 9 (to ensure there are no preceding zeroes)
\d      # any digit character; equivalent to [0-9]
{0,6}   # perform the preceding match (\d) anywhere from 0 to 6 times
        #    (0,6 instead of 1,7 because we already matched the first digit above)

\.      # match a dot character. Must be escaped with a backslash \ as
        #    the unescaped dot will match *anything* in regex otherwise.

[A-Z]   # any uppercase alphabetic character.
{1,4}   # will repeat the preceding match anywhere from 1 to 4 times.

\-      # match a hyphen character. escaping is optional here.

[XYZ]   # any of X, Y, or Z.

\[      # a literal [ character. Must be escaped.

[1-9]   # matches 1 to 9
\d      # any digit; equivalent to [0-9]
?       # Makes the preceding match optional. Equivalent to {0,1}

\]      # a literal ] character. Must be escaped.

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其他说明:

网站 RegExr.com 是一个非常好的工具，可以帮助您更好地理解正则表达式。