How to iterate loop inside a string searching for any word aftera fixed keyword?

Question

Suppose I have a sting as "PHP Paddy,PHP Pranav,PHP Parth", now i have a count as 3,now how should I iterate loop in the string aiming on string after "PHP " to display the all the names?

Alright This is the string "BEGIN IF (NEW.name != OLD.name) THEN INSERT INTO jos_menuaudit set menuid=OLD.id, oldvalue = OLD.name, newvalue = NEW.name, field = "name"; END IF; IF (NEW.alias != OLD.alias) THEN INSERT INTO jos_menuaudit set menuid=OLD.id, oldvalue = OLD.alias, newvalue = NEW.alias, field = "alias"; END IF; END" in which i am searching the particular word after " IF (NEW.", and after that particualar others strings should not b displayed, hence whenever in a loop it finds " IF (NEW." I musr get a word just next to it. and in this way an array should b ready for to use.

Answer 1

Use explode and substr:

$str =  "PHP Paddy,PHP Pranav,PHP Parth";    
$ary = explode(',', $str);
foreach($ary as $str){
   echo substr($str, 4);
}

Answer 2

you can do:

$str = "PHP Paddy,PHP Pranav,PHP Parth";
foreach(explode(',',$str) as $p) {
    echo substr($p,4)."\n";
}

• First we split the string on comma to get an array with 3 elements
• Next we iterate through the array and print each element skipping the first 4 char.

Answer 3

Or you can (more or less) "combine" the explode() and substr() thingy into one call of preg_match_all()

$pattern = '/(?<=PHP )[^,]+/';
$input = "PHP Paddy,PHP Pranav,PHP Parth";
preg_match_all($pattern, $input, $captures);
foreach( $captures[0] as $e ) {
  echo $e, "\n";
}

prints

Paddy
Pranav
Parth

Answer 4

Also can use strstr

$str = "PHP Paddy,PHP Pranav,PHP Parth";
foreach(explode(',',$str) as $st) {
echo strstr($st,' ')."\n";
}

Answer 5

$str =  "PHP Paddy Paddy,PHP Pranav,PHP Parth";
$s = array_filter(preg_split("/,|PHP/",$str));
print_r($s);