Question

Consider:

$a = 'How are you?';

if ($a contains 'are')
    echo 'true';

Suppose I have the code above, what is the correct way to write the statement if ($a contains 'are')?

Answer 1

Now with PHP 8 you can do this using str_contains:

if (str_contains('How are you', 'are')) { 
    echo 'true';
}

Please note: The str_contains function will always return true if the $needle (the substring to search for in your string) is empty.

$haystack = 'Hello';
$needle   = '';

if (str_contains($haystack, $needle)) {
    echo "This returned true!";
}

You should first make sure the $needle (your substring) is not empty.

$haystack = 'How are you?';
$needle   = '';

if ($needle !== '' && str_contains($haystack, $needle)) {
    echo "This returned true!";
} else {
    echo "This returned false!";
}

Output: This returned false!

It’s also worth noting that the new str_contains function is case-sensitive.

$haystack = 'How are you?';
$needle   = 'how';

if ($needle !== '' && str_contains($haystack, $needle)) {
    echo "This returned true!";
} else {
    echo "This returned false!";
}

Output: This returned false!

RFC

Before PHP 8

You can use the strpos() function which is used to find the occurrence of one string inside another one:

$haystack = 'How are you?';
$needle   = 'are';

if (strpos($haystack, $needle) !== false) {
    echo 'true';
}

Note that the use of !== false is deliberate (neither != false nor === true will return the desired result); strpos() returns either the offset at which the needle string begins in the haystack string, or the boolean false if the needle isn’t found. Since 0 is a valid offset and 0 is "falsey", we can’t use simpler constructs like !strpos($a, 'are').