Answers for "PHP $argv echo with number of words"

PHP $argv echo

for ($i=0; $i < $argc; $i++) {
    echo $argv[$i] . "\n";
}

Posted by: Guest on July-02-2021

PHP $argv echo with number of words

<?php 

echo "There are $argc arguments\n";

for ($i=0; $i < $argc; $i++) {
    echo $argv[$i] . "\n";
}

Posted by: Guest on July-02-2021

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