php/bash script adds a parameter to command line - php

Question

<?php var_dump($argv); ?>
ok - run it directly - get what's expected.
$ php /tmp/check-arg.php -s test yellow bus
array(5) {
[0]=>
string(18) "/tmp/check-arg.php"
[1]=>
string(2) "-s"
[2]=>
string(4) "test"
[3]=>
string(6) "yellow"
[4]=>
string(3) "bus"
}
ok - so I want to run this script as if it were a command (it's a php script that replaces an existing command), so I created this script
$ vi /tmp/testingcommand
php /tmp/check-arg.php $1 $2 $3 $4 $5 $6 $7 $8 $9 $10
(edit - change the $10 to {10} is solution - or use "$#" instead of $1 $2...)
so I should be able to just
$ /tmp/testingcommand -s test yellow bus
array(6) {
[0]=>
string(18) "/tmp/check-arg.php"
[1]=>
string(2) "-s"
[2]=>
string(4) "test"
[3]=>
string(6) "yellow"
[4]=>
string(3) "bus"
[5]=>
string(3) "-s0"
}
OK - so where did that "-s0" come from? I've done some fiddling and it's what is in $argv[1] (-s) and a "0" (so in this case -s0)
Any ideas? happened on our RHEL7 as well as a Fedora 30 setup
since the script can be run either directly (php program.php) or via a script I can't just ignore the last index of argv[]
I guess I could check argv[0] for '...php' and keep all indexs and if no ...php then ignore last index

Answer 1

-s0 came from $10. That's $1 followed by 0.
Use ${10} to access parameter 10. You need curly braces whenever the parameter number is more than one digit.
Note that your code won't work properly if any of the arguments have spaces, because you're not quoting the variables. The variable value will undergo word splitting and wildcard expansion.
But if you quote all the variables, you'll get explicit '' values for the arguments that weren't supplied, which is probably not desired, either.
The correct way to reference all the arguments is with "$#".
php /tmp/check-arg.php "$#"