Let's say I have the following code:
$string = "Hello! This is a test. Hello this is a test!"
echo str_replace("Hello", "Bye", $string);
This will replace all Hello in $string with Bye. How can I e.g. exclude all where there's ! after Hello.
Means, I want this output: Hello! This is a test. Bye this is a test!
Is there a way in php to do that?
The solution using preg_repalce function with specific regex pattern:
$string = "Hello! This is a test. Hello this is a test!";
$result = preg_replace("/Hello(?!\!)/", "Bye", $string);
print_r($result);
The output:
Hello! This is a test. Bye this is a test!
(?!\!) - lookahead negative assertion, matches Hello word only if it's NOT followed by '!'
You'll need a regular expression:
echo preg_replace("/Hello([^!])/", "Bye$1", $string);
[] is a character class and the ^ means NOT. So Hello NOT followed by !. Capture in () the NOT ! that is after Hello so you can use it in the replacement as $1 (first capture group).
