如何在PHP脚本中正确使用“ INCLUDE”

Question

I have a PHP script that goes through a list of movie data and parses that data to return the movie title and rating for several businesses. I'd like to have the data parsing script as a single file to be included in several different files, so if I need to edit the movie data parser script, it will be the same for every file it's included in.

This is the code of the data parsing script that works perfectly. for the filename, I need to replace that with a variable, so when included in another file, I can change that filename variable to the name of the correct movie list. Originally, I had each business their own php data parsing script, but it's not very maintainable.

    foreach (glob('mov/FILENAME.*mov') as $filename){ 
    $theData = file_get_contents($filename) or die("Unable to retrieve file data");
}

$string = $theData;
$titles = explode("\n", $string);

function getInfo($string){
    $Ratings = ['G', 'PG', 'PG-13', 'R', 'NR', 'XXX'];
    $split = preg_split("/\"(.+)\"/", $string, 0, PREG_SPLIT_DELIM_CAPTURE);
    if(count($split) == 3){ 
        preg_match("/(".implode("|", $Ratings).")\s/", $split[0], $matches);
        $rating = $matches[0];
        return ["title" => $split[1], "rating" => $rating];
    }
    return false;
}

FINALLY TO THE QUESTION:

In the data parsing script. How would I implement a variable where it says 'FILENAME', and when I include this script in another file. How would I assign the variable to the correct filename and output/run the script?

Answer 1

You should implement the code in the include file as a function, and FILENAME should be a parameter to the function. Then all the other scripts can include the file and call the function with the appropriate filename.

Similarly, $theData should be the return value of the function, not a global variable.