CODEWITHSUNDEEP
bashshell
connor.roche
connor.roche

Reputation: 55

Can't pass /bin/*

I am trying to pass a to fill an array with a list of directories using a simple shell script. However when I pass /bin/* into the program as the first parameter $1 turn to /bin/bash (the first directory in /bin). My code is below for reference. Can I not pass * in a parameter?

#!/bin/bash

declare -a d
placeholder=0

for filename in ${1}; do
     d[${placeholder}]=${filename:5}
     ((placeholder++))
 done

echo "$1"

The result of the echo is /bin/bash

Thanks!

Upvotes: 0

Views: 37

Answers (2)

Charles Duffy
Charles Duffy

Reputation: 295687

A glob is replaced by the shell with a list of names it matches.

Take advantage of this, and read the list of names passed on your argument list:

#!/bin/bash

declare -a d
placeholder=0

for filename in "$@"; do               # or just ''for filename; do''
     d[${placeholder}]=${filename:5}
     ((placeholder++))
done

# print your array
declare -p d

# or, to print your array's contents one entry per line:
printf '%s\n' "${d[@]}"

...or, far easier:

#!/bin/bash
d=( "${@##*/}" )           # assuming the :5 is intended to leave only filenames
declare -p d

Upvotes: 1

John3136
John3136

Reputation: 29266

* is being expanded by the shell before it calls your script. Try calling as yourscript '/bin/*' (note the quotes).

Upvotes: 0

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