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

Answer 1

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[@]}"

Answer 2

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.