PHP preg_replace find a word in string and remove other characters

Question

I am trying to remove from a string all characters that do not match to a list of words.

my list of words could be:

a string can look like this:

my result should look like this:

my approach:

preg_replace('/(person|animal)/i', '', '123-ea-person.jpg')

output:
123-ea-.jpg

expected output:
person

How can I reverse the pattern to get the result?

Solution proposed by @Syscall works preg_replace('/(.*)(person|animal)(.*)/i', '$2$4', '123-ea-person.png'); if you delete extension condition — Juan
– Juan, Commented Jan 25, 2023 at 20:11
@Juan Yay, this works fine strtolower(preg_replace('/(.*)(person|animal)(.*)/i', '$2', '123-ea-person.jpg 456456-on-Person.jpg a-animal-dog.png')); — pino
– pino, Commented Jan 25, 2023 at 20:28
How about e.g. preg_match('~\b(person|animal)\b~i', $str, $out) — bobble bubble
– bobble bubble, Commented Jan 25, 2023 at 21:02
What is the desired result if person or animal occur more than once in the string? — mickmackusa
– mickmackusa ♦, Commented Jan 25, 2023 at 23:10

user3783243 · Accepted Answer · 2023-01-25 20:25:53Z

3

strtolower with PCRE verbs could achieve your goal:

strtolower(preg_replace('/(?:person|animal)(*SKIP)(*FAIL)|./i', '', '123-ea-person.jpg 456456-on-Person.jpg a-animal-dog.png'));

with this approach a person or animal match would be skipped, all other matches would be replaced with nothing.

answered Jan 25, 2023 at 20:25

user3783243

5,2125 gold badges27 silver badges45 bronze badges

Sign up to request clarification or add additional context in comments.

1 Answer 1