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Answer:

The problem here is that strpos() returns the starting position index of $str1 in $str2 (if found), otherwise it returns false. So in this example, strpos() returns 0 (which is then coerced to false when referenced in the if statement). That’s why the code doesn’t work properly.
The correct solution would be to explicitly compare the value returned by strpos() to false as follows:

$str1 = 'yabadabadoo';
$str2 = 'yaba';
if (strpos($str1,$str2) !== false) {
 echo """ . $str1 . "" contains "" . $str2 . """;
} else {
 echo """ . $str1 . "" does not contain "" . $str2 . """;
}

Note that we used the !== operator, not just the != operator. If we use !=, we’ll be back to the problem that 0 is coerced to false when referenced in a boolean expression, so 0 != false will evaluate to false.

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