RegEx-匹配<a>Java</a>中的整个<a>标签</a>

Question

I'm trying to match this <a href="**something**"> using regex in java using this code:

Pattern regex = Pattern.compile("<([a-z]+) *[^/]*?>");
                Matcher matcher = regex.matcher(string);
                string= matcher.replaceAll("");

I'm not really familiar with regex. What am I doing wrong? Thanks

Answer 1

If you just want to find the start tag you could use:

"<a(?=[>\\s])[^>]*>"

If you are trying to get the href attribute it would be better to use:

"<a\\s+[^>]*href=(['\"])(.*?)\\1[^>]*>"

This would capture the link into capturing group 2.

Answer 2

To give you an idea of why people always say "don't try to parse HTML with a regular expression", here'e a simplified regex for matching an <a> tag:

<\s*a(?:\s+[a-z]+(?:\s*=\s*(?:[a-z0-9]+|"[^"]*"|'[^']*'))?)*\s*>

It actually is possible to match a tag with a regular expression. It just isn't as easy as most people expect.

All of HTML, on the other hand, is not "regular" and so you can't do it with a regular expression. (The "regex" support in many/most languages is actually more powerful than "regular", but few are powerful enough to deal with balanced structures like those in HTML.)

Here's a breakdown of what the above expression does:

<\s*             < and possibly some spaces
a                "a"
(?:              0 or more...
  \s+              some spaces
  [a-z]+           attribute name (simplified)
  (?:              and maybe...
    \s*=\s*          an equal sign, possibly with surrounding spaces
    (?:              and one of:
      [a-z0-9]+        - a simple attribute value (simplified)
      |"[^"]*"         - a double-quoted attr value
      |'[^']*'         - a single quoted atttr value
    )
  )?
)*
\s*>             possibly more spaces and then >

(The comments at the start of each group also talk about the operator at the end of the group, or even in the group.)

There are possibly other simplifications here -- I wrote this from memory, not from the spec. Even if you follow the spec to the letter, browsers are even more fault tolerant and will accept all sorts of invalid input.

Answer 3

you can just match against:

"<a[^>]*>"

If the * is "greedy" in java (what I think it is, this is correct) But you cannot match < a whatever="foo" > with that, because of the whitespaces.

Although the following is better, but more complicated to understand:

"<\\s*a\\s+[^>]*>"

(The double \\\\ is needed because \\ is a special char in a java strings)

This handles optional whitespaces before a and at minimum one whitespace after a . So you don't match <abcdef> which is not a correct a tag. (I assume your a tag stands isolated in one line and you are not working with multiline mode enabled. Else it gets far far more complicated.) your last *[^/]*?> seems a little bit strange, maybe it doesn't work cause of that.

Ok lets check what you are doing:

<([a-z]+) *[^/]*?>

<([a-z]+)

match something that contains an < followed by a [az] at least one time. This is grouped by the brackets.

Now you use a * which means the defined group ([az])* may appear multiple time, or not.

[^/]*

This means now match everything, but a / or nothing (because of the * )

The question mark is just wrong, not sure how this is interpreted.

Last char > matched as last element, which must appear.

To sum up, your expression is just wrong and cannot work :)

Take a look at: http://www.regular-expressions.info/

This is a good starting point.