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#1 2024-05-02 18:04:12

tethys
Member
Registered: 2019-08-13
Posts: 105

[SOLVED] Accessing bash script parameters

I have a simple bash script where I am trying to pass the bash script parameters to a grep in the script:

#!/bin/bash

grepargs=""
for var in "$@"; do grepargs+="-e "; grepargs+="$var "; done
echo "${grepargs% }|" # eliminate trailing space
grep ${grepargs% } file.txt

If I run the script with two words as argument everything is fine:

$ myscript word1 word2
-e word1 -e word2|

However, I want to be able to run the script also with a sentence as an argument, like this:

$ myscript word1 "word2 word3"
-e word1 -e word2 word3|
grep: word3: File or directory not found

$ myscript word1 '"word2 word3"'
-e word1 -e "word2 word3"|
grep: word3": File or directory not found

The first syntax won't work obviously but why does not the second syntax work?
I get the same error message in both cases, although the "${grepargs% }" variable seems to be correct in the second case!

Last edited by tethys (2024-05-02 20:34:42)

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#2 2024-05-02 18:42:54

Trilby
Inspector Parrot
Registered: 2011-11-29
Posts: 29,857
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Re: [SOLVED] Accessing bash script parameters

Minimal fix:

- grep ${grepargs% } file.txt
+ grep "${grepargs% }" file.txt

Better script:

#!/bin/sh

grepargs=$(printf -- '-e "%s" ' "$@")
eval grep $grepargs file.txt

Best script:

printf '%s\n' "$@" | grep -f - file.txt

Last edited by Trilby (2024-05-02 18:56:21)


"UNIX is simple and coherent" - Dennis Ritchie; "GNU's Not Unix" - Richard Stallman

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#3 2024-05-02 20:34:19

tethys
Member
Registered: 2019-08-13
Posts: 105

Re: [SOLVED] Accessing bash script parameters

Wow, thank you Trilby! Such a simple solution for my original script!

I am not familiar with "printf" but I tested it, it works, and it is better than my original script so I will adopt it: with "printf" I do not have to double-quote the arguments of the script (i.e. I can use "word2 word3" instead of '"word2 word3"')!

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#4 2024-05-03 00:00:39

sukolyn
Member
Registered: 2024-02-14
Posts: 16

Re: [SOLVED] Accessing bash script parameters

I might be missing something :

#!/bin/bash

args=( "$@" )
echo "$(IFS='|'; echo "${args[*]}")"
$ myScript a 'b c' 'd e f'
a|b c|d e f
$

though, args array is not necessary :

#!/bin/bash

echo "$(IFS='|'; echo "$*")"
$ myScript.new a 'b c' 'd e f'
a|b c|d e f
$

Last edited by sukolyn (2024-05-03 00:04:05)

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#5 2024-05-03 00:27:57

Trilby
Inspector Parrot
Registered: 2011-11-29
Posts: 29,857
Website

Re: [SOLVED] Accessing bash script parameters

sukolyn, what are you trying to show there?  How does that relate to this thread?


"UNIX is simple and coherent" - Dennis Ritchie; "GNU's Not Unix" - Richard Stallman

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#6 2024-05-03 00:31:30

sukolyn
Member
Registered: 2024-02-14
Posts: 16

Re: [SOLVED] Accessing bash script parameters

tethys wants to build a regex expression from script's parameters, doesn't he ?
this is what I show.

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#7 2024-05-03 05:33:09

mountaintrek
Member
Registered: 2024-02-01
Posts: 19

Re: [SOLVED] Accessing bash script parameters

It looks like you are building grep's parameter list dynamically. You sort of have to think like the bash parser. There is a snippet below that builds the parameters in an array, which bash will parse into words correctly. You might want to look at using a case statement in a loop so you could expand what your script can do. Have fun.

#!/bin/bash
# Usage: ./cmd p1 p2 -- file_to_search

declare -a gp=()
declare -- i=1 file=''

while (( i <= ${#@} )); do
  if [[ "${1}" == '--' ]]; then
    file="${2}"
    break
  fi
  # add to array
  gp+=('-e')
  gp+=("$1")
  shift 1
done

# Print array contents
declare -p gp
echo

# Execute the command
if (( "${#gp}" >= 2 )); then
  echo grep "${gp[@]}" "${file:?Missing file}"
  grep "${gp[@]}" "${file:?Missing file}"
else
  printf -- 'Less then 2 parameters\n'
fi

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#8 2024-05-03 06:30:12

tethys
Member
Registered: 2019-08-13
Posts: 105

Re: [SOLVED] Accessing bash script parameters

Thank you all for your inputs.
What I want is to build a grep of the kind

grep [OPTION...] -e PATTERNS ... [FILE...] 

inside a bash script. The bash script itself filters the output of a log file like this:

tail -f file.log | grep -e PATTERN1  -e PATTERN2 ...

And I want to be give the PATTERNS as parameters to the bash script. Actually I do not need regex, but I want to be able to filter certain sentences (containing spaces) in the "file.log".

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#9 2024-05-03 11:50:38

sukolyn
Member
Registered: 2024-02-14
Posts: 16

Re: [SOLVED] Accessing bash script parameters

grep -e pat1 -e pat2 is equivalent to grep -E 'pat1|pat2'

Actually I do not need regex

each expression given to grep is a regex.

Last edited by sukolyn (2024-05-03 11:58:35)

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