我需要一个Java正则表达式

Question

I am currently using the following regular expression:

^[a-zA-Z]{0,}(\\*?)?[a-zA-Z0-9]{0,}

to check a string to start with an alpha character and end with alphanumeric characters and have an asterisk(*) anywhere in the string but only a maximum of one time. The problem here is that if the given string still passes if it starts with a number but doesn't have an *, which should fail. How can I rework the regex to fail this case?

ex.

TE - pass

*TE - pass

TE* - pass

T*E - pass

*9TE - pass

*TE* - fail (multiple asterisk)

9E - fail (starts with number)

EDIT: Sorry to introduce a late edit but I also need to ensure that the string is 8 characters or less, can I include that in the regex as well? Or should I just check the string length after the regex validation?

Answer 1

This passes your example:

"^([a-zA-Z]+\\*?|\\*)[a-zA-Z0-9]*$"

It says:
  start with: [a-zA-Z]+\\*? (a letter and maybe a star)
              | (or)
              \\* a single star
  and end with [a-zA-Z0-9]* (an alphanumeric character)

Code to test it:

public static void main(final String[] args) {
    final Pattern p = Pattern.compile("^([a-zA-Z]+\\*?|\\*)\\w*$");

    System.out.println(p.matcher("TE").matches());
    System.out.println(p.matcher("*TE").matches());
    System.out.println(p.matcher("TE*").matches());
    System.out.println(p.matcher("T*E").matches());
    System.out.println(p.matcher("*9TE").matches());
    System.out.println(p.matcher("*TE*").matches());
    System.out.println(p.matcher("9E").matches());
}

Per Stargazer, if you allow alphanumeric before the star, then use this:

^([a-zA-Z][a-zA-Z0-9]*\\*?|\\*)\\w*$

Answer 2

One possible way is to separate into 2 conditions:

^(?=[^*]*\*?[^*]*$)[a-zA-Z*][a-zA-Z0-9*]*$

• The (?=[^*]*\\*?[^*]*$) part ensures there is at most one * in the string.
• The [a-zA-Z*][a-zA-Z0-9*]* part ensures it starts with an alphabet or a * , and followed by only alphanumerals or * .

Answer 3

It might be easier to develop and maintain later if you just break your regular expressions into a few pieces, eg, one for the start and end, and one for the asterisk. I am not sure what the overall performance effect would be, you would have simpler expressions but have to run a few of them.

Answer 4

寻找两种可能的模式，一种以*开头，另一种以字母char开头：

^[a-zA-Z][a-zA-Z0-9]*(\\*?)?[a-zA-Z0-9]*|\*[a-zA-Z0-9]*

Answer 5

How about this, it's easier to read:

boolean pass = input.replaceFirst("\\*", "").matches("^[a-zA-Z].*\\w$");

Assuming I read right, you want to:

1. Start with an alpha character
2. End with an alphanumeric character
3. Allow up to one * anywhere

Answer 6

^([a-zA-Z][a-zA-Z0-9]*\\*|\\*|[a-zA-Z])([a-zA-Z0-9])*$

下半部的括号是为了清楚起见，可以安全地排除在外。

Answer 7

This was a tough one (liked the challenge), but here it is:

^(\*[a-zA-Z0-9]+|[a-zA-Z]+[\*]{1}[a-zA-Z]*)$

In order to comply with T9*Z, as pointed out on another post with StarGazer712, I had to change it to:

^(\*[a-zA-Z0-9]+|[a-zA-Z]{1}[a-zA-Z0-9]*[\*]{1}[a-zA-Z0-9]*)$

Answer 8

This is Python, it'll need some massaging for Java:

>>> import re
>>> p = re.compile('^([a-z][^*]*[*]?[^*]*[a-z0-9]|[*][^*]*[a-z0-9]|[a-z][^*]*[*])$', re.I)
>>> for test in ['TE', '*TE', 'TE*', 'T*E', '*9TE', '*TE*', '9E']:
...  if p.match(test):
...   print test, 'pass'
...  else:
...   print test, 'fail'
... 
TE pass
*TE pass
TE* pass
T*E pass
*9TE pass
*TE* fail
9E fail

Hope I didn't miss anything.

Answer 9

At most one asterisk, alphabetic characters anywhere and numbers anywhere but at start.

    String alpha = "[a-zA-Z]";
    String alnum = "[a-zA-Z0-9]";

    String asteriskNone = "^" + alpha + "+" + alnum + "*";
    String asteriskStart = "^\\*" + alnum + "*";
    String asteriskInside = "^" + alpha + "+" + alnum + "+\\*" + alnum + "*";
    String yourRegex = asteriskNone + "|" + asteriskStart + "|"
            + asteriskInside;
    String[] tests = {"TE","*TE","TE*","T*E","*9TE","*TE*", "9E"};
    for (String test : tests)
        System.out.println(test + " " + (test.matches(yourRegex)?"PASS":"FAIL"));