Question

Basically i created a some code that creates a file….
I can use that code and works perfectly in the IF statements in my code… now… because i have to use it twice… I decided to make it a function… but as soon as i make it a function it doesnt work…

anyone can tell me why??
I am just calling the function inside others IF statements as
if (createFile()) {
echo "it worked";
}

function createFile () {
        //Creates File and populates it.
        $fOpen = fopen($dbFile, 'w');

            $fString .= "<?php\n";
            $fString .= "// Database Constants\n";
            $fString .= "define(\"DB_SERVER\", \"$server\");\n";
            $fString .= "define(\"DB_USER\", \"$username\");\n";
            $fString .= "define(\"DB_PASS\", \"$password\");\n";
            $fString .= "define(\"DB_NAME\", \"$dbname\");\n";
            $fString .= "?>";

        fwrite($fOpen, $fString);
        fclose($fOpen);
}

Answer 1

You need to pass in the value for $dbFile to the function. Like

function createFile($dbFile) {

This is typically considered a better approach than declaring the $dbFile variable as global inside the function.