如何捕获前两位或三或四个字符的11位数字，使用Regex字符可以是点

Question

I need to search in a HTML page a sequence of digits that could be like that:

p.fg 67389109321 or pfg 67389109321 or pf 67389109321

After parsing the HTML page I transform it in a string:

 String  Pagestring  = Page.toString().toLowerCase().replaceAll("  <[^>]+>","");

and using this Regex to capture the 11 digits:

final Matcher m = Pattern.compile(("(?<!\\d)\\d{11}(?!\\d)")).matcher(Page );

But it captures the first instance of 11 digits. I need to include the above options.

Answer 1

Straight forward: define the possible beginnings and separate them by "or" ( | ), then go for the 11 digits:

(p\.fg|pfg|p\.f) \d{11}

This means:

• ( : delimiter for the or operation
• p\\.fg : literal pf.g
• | : or
• pfg : literal pfg
• | : or
• p\\.f : literal pf
• ) : delimiter for the or operation
• : literal space
• \\d{11} : eleven digits

Try it online

That said, removing HTML tags the way you do ( replaceAll(" <[^>]+>",""); ) is not reliable. Use a HTML specific tool like HtmlAgilityPack . That regex might fail on HTML like

<tag attribute=">"/>

Answer 2

Regex : p(?:\\.?fg|\\.f)\\s\\d{11}

Details:

• (?:) Non-capturing group
• | Or
• \\s Matches any whitespace character

Java code :

String string = "p.fg 67389109321 or  pfg 67389109321 or  p.f 67389109321";
Matcher matches = Pattern.compile("p(?:\\.?fg|\\.f)\\s\\d{11}").matcher(string);
while (matches.find()) {
    System.out.println(matches.group(0));
}

Output:

p.fg 67389109321
pfg 67389109321
p.f 67389109321

Code demo