Shell script pass arguments with spaces

Question

I want to pass arguments from one shell script ( say script1 ) to another. Some of the arguments contain spaces. So I included quotes in the arguments and before passing to script2, I echoed it. Here is how it is,

echo $FL gives
-filelist "/Users/armv7/My build/normal/My build.LinkFilelist" -filelist "/Users/arm64/My build/normal/My build.LinkFilelist"

But when I do

script2  -arch armv7 -arch arm64 -isysroot /Applications/blahblah/iPhoneOS8.1.sdk $FL

and in the script2 if I do,

 for var in "$@"
  do
      echo "$var"
  done

I still get

"-arch"
"armv7"
"-arch"
"arm64"
"isysroot"
"/Applications/blahblah/iPhoneOS8.1.sdk"
"-filelist"
""/Users/armv7/My"
"build/normal/My"            // I want all these 3 lines together
build.LinkFilelist"" 
"-filelist"
""/Users/arm64/My"
"build/normal/My"
build.LinkFilelist""

Can someone please correct my error ? What should I do to get the mentioned argument as a whole.

mywiki.wooledge.org/BashFAQ/050 Don't stick quoted strings inside variables and expect them to work, they won't. Just pass the arguments directly. — Etan Reisner
– Etan Reisner, Commented Jan 29, 2015 at 16:10
How do I set up the FL variable ? Like this -filelist /Users/armv7/My build/normal/My build.LinkFilelist -filelist /Users/arm64/My build/normal/My build.LinkFilelist ? — Theguy
– Theguy, Commented Jan 29, 2015 at 16:23

Gordon Davisson · Accepted Answer · 2015-01-30 04:00:23Z

3

Embedding quotes in a variable's value doesn't do anything useful. As @Etan Reisner said, refer to http://mywiki.wooledge.org/BashFAQ/050. In this case, the best answer is probably to store FL as an array, rather than a plain variable:

FL=(-filelist "/Users/armv7/My build/normal/My build.LinkFilelist" -filelist "/Users/arm64/My build/normal/My build.LinkFilelist")

Note that the quotes aren't stored as part of the array elements; instead, they're used to force the paths to be treated single array elements, rather than broken up by the spaces. Then refer to it with "${FL[@]}", which makes bash treat each element as an argument:

script2 -arch armv7 -arch arm64 -isysroot /Applications/blahblah/iPhoneOS8.1.sdk "${FL[@]}"

answered Jan 30, 2015 at 4:00

Gordon Davisson

128k16 gold badges141 silver badges168 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Aman · Accepted Answer · 2015-01-29 16:33:45Z

0

1- Use the following (put "" around FL):

script2  -arch armv7 -arch arm64 -isysroot /Applications/blahblah/iPhoneOS8.1.sdk "$FL"

2- Then inside your script2 use (to extract the variable based on the format that you are aware of):

for arg; do # default for a for loop is to iterate over "$@"
   case $arg in
    '-filelist'*) input=${arg} ;;
      esac
done

3- Now you can break the input parameter to whatever format you want using awk.

answered Jan 29, 2015 at 16:33

Aman

1,2078 silver badges13 bronze badges

Collectives™ on Stack Overflow

Shell script pass arguments with spaces

2 Answers 2

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related