如何在不丢失角色的情况下preg_split？

I have a string like this

$string = "Hello; how are you;Hey, I am fine";
$new = preg_split("/;\w/", $string);
print_r($new);

I am trying to split the string only when there is no white-space between the words and ";". But when I do this, I lose the H from Hey. It's probably because the split happens through the recognition of ;H. Could someone tell me how to prevent this?

My output:

$array  =   [
  0 => [
    0 => 'Hello; how are you ',
    1 => 0,
  ],
  1 => [
    0 => 'ey, I am fine',
    1 => 21,
  ],
]

You might use a word boundary \b:

\b;\b

$string = "Hello; how are you;Hey, I am fine";
$new = preg_split("/\b;\b/", $string);
print_r($new);

Demo

Or a negative lookahead and negative lookbehind

(?<! );(?! )

Demo

You are capturingthe \w in your regex.You dont want that. Therefore, do this:

$new = preg_split("/;(?=\w)/", $string);

A capture group is defined in brackets, but the ?= means match but don't capture.

Check it out here https://3v4l.org/Q77LZ

Use split with this regex ;(?=\w) then you will not lose the H

Lookarounds cost more steps. In terms of pattern efficiency, a word boundary is better and maintains the intended "no-length" character consumption.

In well-formed English, you won't ever have to check for a space before a semi-colon, so only 1 word boundary seems sufficient (I don't know if malformed English is possible because it is not represented in your sample string).

If you want to acquire the offset value, preg_split() has a flag for that.

Code: (Demo)

$string = "Hello; how are you;Hey, I am fine";
$new = preg_split("/;\b/", $string, -1, PREG_SPLIT_OFFSET_CAPTURE);
var_export($new);

Output:

array (
  0 => 
  array (
    0 => 'Hello; how are you',
    1 => 0,
  ),
  1 => 
  array (
    0 => 'Hey, I am fine',
    1 => 19,
  ),
)