Extract filename and path from URL in bash script

Question

In my bash script I need to extract just the path from the given URL. For example, from the variable containing string:

http://login:password@example.com/one/more/dir/fi

Answer 1

I wrote a function to that will extract any part or the URL. I've only tested it in bash. Usage:

url_parse <url> [url-part]

example:

$ url_parse "http://example.com:8080/home/index.html" path
home/index.html

code:

url_parse() {
  local -r url=$1 url_part=$2
  #define url tokens and url regular expression
  local -r protocol='^[^:]+' user='[^:@]+' password='[^@]+' host='[^:/?#]+' \
    port='[0-9]+' path='\/([^?#]*)' query='\?([^#]+)' fragment='#(.*)'
  local -r auth="($user)(:($password))?@"
  local -r connection="($auth)?($host)(:($port))?"
  local -r url_regex="($protocol):\/\/($connection)?($path)?($query)?($fragment)?$"
  #parse url and create an array
  IFS=',' read -r -a url_arr <<< $(echo $url | awk -v OFS=, \
    "{match(\$0,/$url_regex/,a);print a[1],a[4],a[6],a[7],a[9],a[11],a[13],a[15]}")

  [[ ${url_arr[0]} ]] || { echo "Invalid URL: $url" >&2 ; return 1 ; }

  case $url_part in
    protocol) echo ${url_arr[0]} ;;
    auth)     echo ${url_arr[1]}:${url_arr[2]} ;; # ex: john.doe:1234
    user)     echo ${url_arr[1]} ;;
    password) echo ${url_arr[2]} ;;
    host-port)echo ${url_arr[3]}:${url_arr[4]} ;; #ex: example.com:8080
    host)     echo ${url_arr[3]} ;;
    port)     echo ${url_arr[4]} ;;
    path)     echo ${url_arr[5]} ;;
    query)    echo ${url_arr[6]} ;;
    fragment) echo ${url_arr[7]} ;;
    info)     echo -e "protocol:${url_arr[0]}\nuser:${url_arr[1]}\npassword:${url_arr[2]}\nhost:${url_arr[3]}\nport:${url_arr[4]}\npath:${url_arr[5]}\nquery:${url_arr[6]}\nfragment:${url_arr[7]}";;
    "")       ;; # used to validate url
    *)        echo "Invalid URL part: $url_part" >&2 ; return 1 ;;
  esac
}

Answer 2

This perl one-liner works for me on the command line, so could be added to your script.

echo 'http://login:password@example.com/one/more/dir/file.exe?a=sth&b=sth' | perl -n -e 'm{http://[^/]+(/[^?]+)};print $1'

Note that this assumes there will always be a '?' character at the end of the string you want to extract.

Answer 3

There are built-in functions in bash to handle this, e.g., the string pattern-matching operators:

1. '#' remove minimal matching prefixes
2. '##' remove maximal matching prefixes
3. '%' remove minimal matching suffixes
4. '%%' remove maximal matching suffixes

For example:

FILE=/home/user/src/prog.c
echo ${FILE#/*/}  # ==> user/src/prog.c
echo ${FILE##/*/} # ==> prog.c
echo ${FILE%/*}   # ==> /home/user/src
echo ${FILE%%/*}  # ==> nil
echo ${FILE%.c}   # ==> /home/user/src/prog

All this from the excellent book: "A Practical Guide to Linux Commands, Editors, and Shell Programming by Mark G. Sobell (http://www.sobell.com/)

Answer 4

gawk

echo "http://login:password@example.com/one/more/dir/file.exe?a=sth&b=sth" | awk -F"/" '
{
 $1=$2=$3=""
 gsub(/\?.*/,"",$NF)
 print substr($0,3)
}' OFS="/"

output

# ./test.sh
/one/more/dir/file.exe

Answer 5

url="http://login:password@example.com/one/more/dir/file.exe?a=sth&b=sth"

GNU grep

$ grep -Po '\w\K/\w+[^?]+' <<<$url
/one/more/dir/file.exe

BSD grep

$ grep -o '\w/\w\+[^?]\+' <<<$url | tail -c+2
/one/more/dir/file.exe

ripgrep

$ rg -o '\w(/\w+[^?]+)' -r '$1' <<<$url
/one/more/dir/file.exe

---

To get other parts of URL, check: Getting parts of a URL (Regex).

Answer 6

I agree that "cut" is a wonderful tool on the command line. However, a more purely bash solution is to use a powerful feature of variable expansion in bash. For example:

pass_first_last='password,firstname,lastname'

pass=${pass_first_last%%,*}

first_last=${pass_first_last#*,}

first=${first_last%,*}

last=${first_last#*,}

or, alternatively,

last=${pass_first_last##*,}