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Bash: Get The Last Argument Passed to a Shell Script

I'm writing a bash wrapper script that will pass arguments to the command. I need to find out the last argument if I call the wrapper as follows:

./wrapper -a -b --longarg=foo thisfilename.txt
./wrapper -a -b thisfilename.txt
./wrapper -a --next=true thisfilename.txt

=> $@ is all of them.
=> $0 is script name.
=> $1 is first arg.

I want thisfilename.txt stored in a shell variable called $last. How do I find the last argument passed to a shell script written in bash or ksh under Unix like operating systems?
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HowTo Run a Script In Linux

How do I run a Linux shell script? How can I run a script in Linux operating system using command line options?
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Linux / UNIX: Command For Executing a Shell Script

How do I execute a script under UNIX or Linux like operating system using command prompt?
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Bourne Shell Exit Status Examples

Can you explains and provide us "Bourne Shell Exit Status Code" examples?
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