Bash: How to process all given arguments and store them into the array

Question

I'm beginner in bash scripting and I'm having trouble with following script.

I want to process all given arguments from stdin. Then I check, if these arguments are ordinary text files. If yes, I want to store them into the array and later I want to go through the whole array. But I am getting an error : word unexpexted on the line with files+=("$@") I've tried to write it like this files=("$@") but then I get following error on the line : "(" unexpected (expecting "fi")

I would be really grateful for any advice. Thank you in advance.

for file in "${argv[@]}"; do
    if [ -d "$file" ] 
    then
        echo "Error: '"$file"' is directory!" > /dev/stderr
    continue
    fi  


    if [[! -f "$file"] || [! -r "$file"]] 
    then
        echo "Error: '"$file"'!" > /dev/stderr
    continue
    fi  

    file "$file" | grep text >& /dev/null

    if [ ! $status ]
    then
    files+=("$@") 
    else
    echo "Error: '"$file"' not a text file!" > /dev/stderr
    fi
done

for file in "${files[@]}"; do
# .....
done

What is the $status variable? And don't you mean files+=("$file") ? — cdleonard
– cdleonard, Commented Nov 16, 2012 at 10:51
Why are you constantly UNquoting the part of your echo that MUST be quoted, ie. the $file? — lhunath
– lhunath, Commented Jun 5, 2013 at 1:04
status is never assigned; should be quoted in the [. Make sure you're running this with bash, and not sh. You don't read stdin anywhere and there's no indication of what argv is. You should write to &2, not /dev/stderr. To test a command, do if command, not command; if [ .... — lhunath
– lhunath, Commented Jun 5, 2013 at 1:07

Gilles Quénot · Accepted Answer · 2012-11-16 13:45:04Z

1

Try doing this :

#!/bin/bash

files=( )

for file; do
    if ([[ -f "$file && -r "$file" ]] && file "$file" | grep -q -i "text"); then
        files+=( "$file" )
    fi
done

for f in ${files[@]}; do
    # something with "$f"
done

Another version, with error handling :

#!/bin/bash

files=( )

for file; do
    if [[ ! -f "$file ]]; then
        echo >&2 "$file is not a regular file"
        continue
    fi

    if [[ ! -r "$file ]]; then
        echo >&2 "$file is not readable for $USER"
        continue
    fi

    if ! file "$file" | grep -q -i "text"; then
        echo >&2 "$file is not a text file"
        continue
    fi

    files+=( "$file" )
done

for f in ${files[@]}; do
    # something with "$f"
done

NOTE

argv doesn't exists literally in bash, for file is sufficient
instead of using non existent variable $status, use predefined variable $?
no need to test last status, you can do something shorter like grep -q pattern file && do_something
echo >&2 mean to redirect to STDERR

edited Nov 16, 2012 at 13:45

answered Nov 16, 2012 at 10:56

Gilles Quénot

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4 Comments

quida Over a year ago

I tried your versions, but at the line with files=( ) I get an error : "(" unexpected

quida Over a year ago

I tried to write it like: declare -a file or declare -a file='()' (I found it in some tutorial)but then I get error: word unexpected (expecting ")") or : declare not found

Gilles Quénot Over a year ago

Seems youre using the wrong shell. Try this : chmod +x script.sh; ./script.sh with the new version (added a shebang)

Anders Johansson Over a year ago

or more portably change the first line to #! /usr/bin/env bash (allows bash to be anywhere on your $PATH), alternatively try explicitly with bash script.sh to see if it runs then.

gniourf_gniourf · Accepted Answer · 2012-12-28 11:51:25Z

Here is a script I just coded and seems to do what you ask... Just replace the .. in the third test with whatever you need. Unfortunately, I have never used arrays in the manner you do above, so I just wrote it my way. I hope it helps. Just run it as bash {scriptname}.sh. Anything entered into standard in will process.

#!/bin/bash

checkfile()
{

for i in $token
    do  

        if [ -f "${i}" ]; then
            {
            echo "It's a file"
            }   
        elif [ -d "${i}" ]; then
            {
            echo "It's a directory"
            }       
        elif [ -z "${i}" ]; then
            {
                :
            }
        fi

    done


}

while ( : )
do
    read token
    checkfile
    sleep 2
done

Here is the debug output in bash:

+ read token
a
+ checkfile
+ for i in '$token'
+ '[' -f a ']'
+ '[' -d a ']'
+ echo 'It'\''s a directory'
It's a directory
+ sleep 2
+ :
+ read token
b
+ checkfile
+ for i in '$token'
+ '[' -f b ']'
+ echo 'It'\''s a file'
It's a file
+ sleep 2
+ :
+ read token
a
+ checkfile
+ for i in '$token'
+ '[' -f a ']'
+ '[' -d a ']'
+ echo 'It'\''s a directory'
It's a directory
+ sleep 2
+ :
+ read token

Community · Accepted Answer · 2017-05-23 11:43:06Z

Sputnick's answer is a very good solution. Just in case you want to stick with your own implementation, make sure you correct the following line:

if [[! -f "$file"] || [! -r "$file"]]

You should always leave spaces between the brackets and the test expression: [[ ! -f "$file" ]]. Also, since you are using the || operator, use double brackets rather than single ones:

if [[ ! -f "$file" ]] || [[ ! -r "$file" ]]

Change the line files+=("$@") to either files+=( "$file" ) or files[${#files[@]}]="$file".

Not sure what you wanted to do with your status variable. The test [ ! $status ] will return True for an unassigned status, a variable assigned like status= or any status=$(command >& /dev/null). If you want to test against the exit status of the file ... command, use integer tests rather than string comparison: if [ $? -eq 0 ] or if (($? == 0)).

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