16

I have a folder named 'sample' and it has 3 files in it. I want to write a shell script which will read these files inside the sample folder and post it to an HTTP site using curl.

I have written the following for listing files inside the folder:

for dir in sample/*; do
        echo $dir;
        done

But it gives me the following output:

sample/log

sample/clk

sample/demo

It is attaching the parent folder in it. I want the output as follows (without the parent folder name)

log

clk

demo

How do I do this?

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1
  • 1
    use 'cd' first. If you need 'curl' it, you still need enter the directory. Also you can use echo ${dir##*/} to take off the path. Commented Apr 6, 2017 at 15:41

4 Answers 4

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41

Use basename to strip the leading path off of the files:

for file in sample/*; do
    echo "$(basename "$file")"
done

Though why not:

( cd sample; ls )
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5

Assuming your shell supports it you could use parameter expansion

for path in sample/*; do
    printf -- '%s\n' "${path##*/}"
done

or you could just change to that directory and do the listing there

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2
  • 1
    The -- is not needed here. The first argument is a fixed string with no dashes. Commented May 29, 2018 at 10:21
  • Also, ${parameter##pattern} is supported by all POSIX shells. Commented Jun 19, 2018 at 13:22
5

It depends what you want to do with the directories.

To simply print the name, without a check whether it is a directory you could use ls:

ls -1 sample

Better would be find, because you can use filters:

find sample -type d -maxdepth 1 -printf '%f\n'

If you want to run commands on the files, you should use find and not a for loop:

find sample -type d -maxdepth 1 -exec basename {} \;
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1
  • Yes. Just to clarify. ls sample/* will show sample/log, etc. ls sample will show only log. Commented Sep 10, 2020 at 15:37
4

Because *nix systems allow nearly any character to be part of a filename (including including whitespace, newlines, commas, pipe symbols, etc.), you should never parse the output of the "ls" command in a shell script. It's not reliable. See Why you shouldn't parse the output of ls.

Use "find" to create a list of files. If you are using Bash, you can insert the output of "find" into an array. Example below, with the caveat that I used a non-working "curl" command!

searchDir="sample/"
oldFiles=()
while IFS= read -r -d $'\0' foundFile; do
    oldFiles+=("$foundFile")
done < <(find "$searchDir" -maxdepth 1 -type f -print0 2> /dev/null)

if [[ ${#oldFiles[@]} -ne 0 ]]; then
    for file in "${oldFiles[@]}"; do
        curl -F ‘data=@"$file"’ UPLOAD_ADDRESS
    done
fi
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3
  • What is the directory here? $'\0' ? I want to pass a directory that's in a variable... Commented Nov 19, 2018 at 17:30
  • searchDir is the directory variable. "-d $'\0'" sets the delimiter of the read command to null. Commented Nov 27, 2018 at 2:59
  • How do I achieve the same thing in a sh shell? I am using #!/bin/sh Commented Jul 3, 2019 at 3:10

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