Bash Get All Command Line Arguments Before Last Parameter In $@

by on June 19, 2007 · 0 comments· LAST UPDATED June 19, 2012

in

I'm writing a wrapper bash shell script that will get the last argument (a domain name) from the command line into a shell variable called $_domain. I need to find all other parameters before last parameter in $@ and stored in a shell variable called $allargs. So that I can pass them as follows:

/path/to/real/binary "$allargs" "$_domain"

How do I do this using bash shell under Unix like operating systems?

You can store all command line arguments or parameter in a bash array as follows:

 
array=( $@ )
 

First, you need to find out length of an array:

 
len=${#array[@]}
 

Next, get the last command line argument from an array (i.e. $@ stored in an array):

 
_domain=${array[$len-1]}
 

Finally, extract and store all command line parameters before last parameter in $@:

 
args=${array[@]:0:$len-1}
 

Putting it all together:

 
#!/bin/bash
array=( $@ )
len=${#array[@]}
_domain=${array[$len-1]}
_args=${array[@]:0:$len-1}
 
echo "Domain: $_domain"
echo "All Args before $_domain are: $_args"
 

Run it as follows:
./script -p -y --zzz cyberciti.biz
Sample outputs:

Domain: cyberciti.biz
All Args before cyberciti.biz are: -p -y --zzz

Another sample run:
$ ./script -p -y --zzz --delete cyberciti.biz
Sample outputs:

Domain: cyberciti.biz
All Args before cyberciti.biz are: -p -y --zzz --delete
TwitterFacebookGoogle+PDF versionFound an error/typo on this page? Help us!

{ 0 comments… add one now }

Leave a Comment

Tagged as: , , , , , , , , ,

Previous Faq:

Next Faq: