Question

I wrote the code as follows:

<?php
array_shift($argv);
$taskid=$argv[0];
$file1=file_get_contents($argv[1]);
$file2=fopen($argv[2],"w");
echo $taskid."\n".$file1."\n".$file2."\n";
?>

I execute this program as php myfile.php 1 1.txt 2.txt
Where 1 is the taskid, 1.txt is the input file and 2.txt is the output file.
I want to modify this program in such a way that it runs even though I don’t pass any arguments that is like php myfile.php should run. I want to place an if condition for that. I worked with the conditions like if(*argv<=1) and if(argc==NULL) etc., but none of them are working.

I need a perfect condition that checks if no arguments are passed then my program should show a user friendly message.

Can someone help me??

Answer 1

if ($argc !== 4) {
  // show message