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Expand Variables and Wildcards into another variable.

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# 1  
Expand Variables and Wildcards into another variable.

Dear Forum members,

I am having trouble getting the complete filename (and directory path) in a variable.

Output directory mentioned in the code have three files:


Code is as shown below:
I am tryin to get get the full path of the above files (including directory path in variable file_path).



while IFS= read -r city_name
 ## None of the below command  worked.    

    # eval file_path=`${output_dir}/*MIGRATE*${city_name}*.out`
    # file_path=${eval echo `ls ${output_dir} | grep "MIGRATE"`}
    # eval file_path=`ls ${output_dir} | grep "MIGRATE"`
    echo $file_path
done < ${city_list}    

##  city_list is the list of cities for which I want to display details

Can anybody help with the command?

# 2  
files_list=`ls ${output_dir}/*MIGRATE*${city_name}*.out`

This User Gave Thanks to anbu23 For This Post:
# 3  
Thanks Anbu23 for quick reply. The command sent by you is working.

But what is the difference between below two?


files_list=`ls ${output_dir}/*MIGRATE*${city_name}*.out`



files_list=`echo ${output_dir}/*MIGRATE*${city_name}*.out`

both seems to be working..
# 4  
A very, very simple example:
Last login: Wed May 15 18:04:12 on ttys000
AMIGA:amiga~> ls Au*
AudioScope.Manual         AudioScope_Quick_Start.Notes
AMIGA:amiga~> echo Au*
AudioScope.Circuits AudioScope.Config AudioScope.Manual AudioScope_Quick_Start.Notes
AMIGA:amiga~> _

# 5  
You see the difference with
echo "$file_path"

The quotes prevent the shell to do word-splitting on command arguments.

ls is an external command. The efficient per-line formatter is printf "%s\n" that 1. is a shell-builtin and 2. does no extra I/O.

What do you want to do with the $file_path?
It can make sense to not expand it too early. For example
for f in $file_path

is smarter than
file_path=`ls ${output_dir}/*MIGRATE*${city_name}*.out`
for f in $file_path

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