[SOLVED] how can I create a string variable in bash in a specific format in bash

mia_tech · 05-29-2014, 02:44 AM

I want to create a variable containing a string in the following format.

Code:

var='"pic1.jpg", "pic2.jpg", "pic3.jpg", "picN.jpg"'

so far I'm able to create pic1.jpg,pic2.jpg,pic3.jpg, but I would like to add quotation marks around each filename eg "pic1.jpg"
I'm using the following command

Code:

file=*.jpg
allPics=$(echo $file | tr ' ' ', ')

but this only takes care of the command between files, but I also need to add quotation. I also tried this approach

Code:

ls *.jpg | \
while read line; do
allPics="${allPics}, \"$line\""
done

but that doesn't seem to work.... if anyone has a better suggestion?
Thanks

grail · 05-29-2014, 02:54 AM

I am a little confused at what you need

Are you looking to assign a single line with all the file names quoted and separated by commas to a variable or are you looking to create an array of the file names quoted?

Option 1:

Code:

var='"pic1.jpg", "pic2.jpg", "pic3.jpg", "picN.jpg"'

Option 2:

Code:

var=('"pic1.jpg"', '"pic2.jpg"', '"pic3.jpg"', '"picN.jpg"')

There could be other alternatives depending on how you wish to retrieve the file names, ie. both of these are currently manual.

mia_tech · 05-29-2014, 02:57 AM

Quote:

Originally Posted by grail

I am a little confused at what you need

Are you looking to assign a single line with all the file names quoted and separated by commas to a variable or are you looking to create an array of the file names quoted?

Option 1:

Code:

var='"pic1.jpg", "pic2.jpg", "pic3.jpg", "picN.jpg"'

Option 2:

Code:

var=('"pic1.jpg"', '"pic2.jpg"', '"pic3.jpg"', '"picN.jpg"')

There could be other alternatives depending on how you wish to retrieve the file names, ie. both of these are currently manual.

good question... the answer is "Option 1"
I will re-edit the original post to reflect the changes

colucix · 05-29-2014, 03:00 AM

One method among many others:

Code:

var=$(echo -n *.jpg | sed -r 's/^|$/"/g; s/ /", "/g')

or

Code:

var=$(echo *.jpg | awk '$1=$1{printf "\"%s\"", $0}'  OFS='", "')

mia_tech · 05-29-2014, 03:21 AM

Quote:

Originally Posted by colucix

One method among many others:

Code:

var=$(echo -n *.jpg | sed -r 's/^|$/"/g; s/ /", "/g')

or

Code:

var=$(echo *.jpg | awk '$1=$1{printf "\"%s\"", $0}'  OFS='", "')

that worked great....
but I'm just wondering if it would be possible to do it with a loop like:

Code:

all=""
ls *.jpg | \
while read line; do
all="${all}, \"$line\""
done

why is the avobe example not working... is the appropiate way to concatenate string is

Code:

all="${all}$line"

colucix · 05-29-2014, 03:43 AM

It is because you pass input to the loop trough a pipe: this opens a subshell where all the variables are local, so that every change made to the variable is lost when the loop terminates. Try this to see what happens:

Code:

all=""
ls *.jpg | \
while read line; do
  all="${all}, \"$line\""
  echo "Inside loop: $all"
done
echo "Outside loop: $all"

You can try to avoid the subshell by means of something like this:

Code:

for file in *.jpg
do
  all="${all}, \"$file\""
done

but you have to adjust something to get the desired output, yet.

colucix · 05-29-2014, 03:53 AM

This one should work as desired (it uses a conditional expression to initialize the variable properly):

Code:

for file in *.jpg
do
  [[ -z $all ]] && all="\"$file\"" || all="$all, \"$file\""
done

grail · 05-29-2014, 11:34 AM

Here is one a little different:

Code:

files=( *.jpg )

printf -v all "\"%s\", " ${files[@]}

echo "${files%,}"

Or maybe:

Code:

files=( *.jpg )
files=( ${files[@]/#/\"} )
files=( ${files[@]/%/\"} )

IFS=","
all="${files[*]}"
IFS=

echo "$all"