PHP中的Regex：将字符串中第一个单词之后的所有单词截断，并将所有单词截断为第一个字符

Question

I'm quite terrible at regexes.

I have a string that may have 1 or more words in it (generally 2 or 3), usually a person name, for example:

$str1 = 'John Smith';
$str2 = 'John Doe';
$str3 = 'David X. Cohen';
$str4 = 'Kim Jong Un';
$str5 = 'Bob';

I'd like to convert each as follows:

$str1 = 'John S.';
$str2 = 'John D.';
$str3 = 'David X. C.';
$str4 = 'Kim J. U.';
$str5 = 'Bob';

My guess is that I should first match the first word, like so:

preg_match( "^([\w\-]+)", $str1, $first_word )

then all the words after the first one... but how do I match those? should I use again preg_match and use offset = 1 in the arguments? but that offset is in characters or bytes right?

Anyway after I matched the words following the first, if the exist, should I do for each of them something like:

$second_word = substr( $following_word, 1 ) . '. ';

Or my approach is completely wrong?

Thanks

ps - it would be a boon if the regex could maintain the whole first two words when the string contain three or more words... (eg 'Kim Jong U.').

Answer 1

It can be done in single preg_replace using a regex.

You can search using this regex:

^\w+(?:$| +)(*SKIP)(*F)|(\w)\w+

And replace by:

$1.

RegEx Demo

Code:

$name = preg_replace('/^\w+(?:$| +)(*SKIP)(*F)|(\w)\w+/', '$1.', $name);

Explanation:

• (*FAIL) behaves like a failing negative assertion and is a synonym for (?!)
• (*SKIP) defines a point beyond which the regex engine is not allowed to backtrack when the subpattern fails later
• (*SKIP)(*FAIL) together provide a nice alternative of restriction that you cannot have a variable length lookbehind in above regex.
• ^\\w+(?:$| +)(*SKIP)(*F) matches first word in a name and skips it (does nothing)
• (\\w)\\w+ matches all other words and replaces it with first letter and a dot.

Answer 2

You could use a positive lookbehind assertion.

(?<=\h)([A-Z])\w+

OR

Use this regex if you want to turn Bob F to Bob F.

(?<=\h)([A-Z])\w*(?!\.)

Then replace the matched characters with \\1.

DEMO

Code would be like,

preg_replace('~(?<=\h)([A-Z])\w+~', '\1.', $string);

DEMO

• (?<=\\h)([AZ]) Captures all the uppercase letters which are preceeded by a horizontal space character.

• \\w+ matches one or more word characters.

• Replace the matched chars with the chars inside the group index 1 \\1 plus a dot will give you the desired output.

Answer 3

A simple solution with only look-ahead and word boundary check:

preg_replace('~(?!^)\b(\w)\w+~', '$1.', $string);

• (\\w)\\w+ is a word in the name, with the first character captured
• (?!^)\\b performs a word boundary check \\b , and makes sure the match is not at the start of the string (?!^) .

Demo