Incorrect string replacement

Question

I know this is stupid, but I am new to the shell and I really have no idea what is going on here. The following is intended to get the directory name after the zip file has been extracted. So, if that works, the result will be like that, / p>

$test.sh helloworld.zip helloworld 

Take a look at test.sh:

 #! /bin/sh length=echo `expr index "$1" .zip` a=$1 echo $(a:0:length} 

However, I received an unsuccessful replacement, and this is just a mess.
And when I mention the "shell". I'm just talking about the shell because I don't know the difference between bash or others. I just use ubuntu 10.04 and use the terminal. And I assume that I am using bash.

Answer 1

If your shell is the latest version of bash enough, then the parameter extension should work.

In many other shells this will not work, and the bad substitution error is how the shell says: "You asked to replace the parameter, but that does not make sense to me."

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Also, given the script:

 #! /bin/sh length=echo `expr index "$1" .zip` a=$1 echo $(a:0:length} 

The second line exports the length variable with the echo value for the command that is created when expr index "$1" .zip is run. It does not assign length . This should be fair:

 length=$(expr index "${1:?}" .zip) 

where the notation ${1:?} generates an error if $1 not specified (if the script is called without arguments).

The last line should be:

 echo ${a:0:$length} 

Note that if $1 contains filename.zip , the output of expr index $1 .zip is 2, since the letter i appears at index 2 in filename.zip . If the goal is to get the base file name without the .zip extension, then the classic way to do this is:

 base=$(basename $1 .zip) 

and more modern way:

 base=${1%.zip} 

There is a difference; if the name is /path/to/filename.zip , the classic output is filename , and modern is /path/to/filename . You can get classic output with:

 base=${1%.zip} base=${base##*/} 

Or, in the classic version, you can get the path with:

 base=$(dirname $1)/$(basename $1 .zip)`.) 

If file names may contain spaces, you need to consider using double quotes, especially in calls to basename and dirname .

Answer 2

Try it in bash :

 echo $1 len=$(wc -c <<< "$1") a="${1}.zip" echo ${a:0:$len} 

Adapt it to your needs.