[SOLVED] Combining variables in Bash to form a command sent to AppleScript using the osascript command

Issue

I'm trying to make this script work. It's a Bash script that is meant to take some variables, put them together and use the result to send an AppleScript command. Manually pasting the string echoed from the variable to_osa behind osascript -e to the terminal works as I want and expect it to. But when I try to combine the command osascript -e and the string to_osa, it does not work. How can I make this work?

the_url="\"http://stackoverflow.com/questions/1521462/looping-through-the-content-of-a-file-in-bash\""
the_script='tell application "Safari" to set the URL of the front document to '
delimiter="'"
to_osa=${delimiter}${the_script}${the_url}${delimiter}
echo ${to_osa}
osascript -e ${to_osa}

In addition to working manually the script also works when I write the desired command to a script and then execute it:

echo "osascript -e" $to_osa > ~/Desktop/outputfile.sh
sh  ~/Desktop/outputfile.sh

Solution

As a general rule, don't put double-quotes in the variable, put them around the variable. In this case it's more complicated, since you have some double-quotes for bash-level quoting, and some for AppleScript-level quoting; in this case, the AppleScript-level quotes go in the variable, the bash-level quotes go around the variable:

the_url="\"http://stackoverflow.com/questions/1521462/looping-through-the-content-of-a-file-in-bash\""
the_script='tell application "Safari" to set the URL of the front document to '
osascript -e "${the_script}${the_url}"

BTW, using echo to check things like this is highly misleading. echo is telling you what's in the variable, not what'll be executed when you reference the variable on a command line. The biggest difference is that echo prints its arguments after they've been through bash parsing (quote and escape removal, etc), but when you say "Manually pasting the string ... works" you're saying it's what you want before parsing. If the quotes are there in the echoed string, that means bash didn't recognize them as quotes and remove them. Compare:

string='"quoted string"'
echo $string          # prints the string with double-quotes around it because bash doesnt't recognize them in a variable
echo "quoted string"  # prints *without* quotes because bash recognizes and removes them

Answered By - Gordon Davisson

Answer Checked By - Robin (WPSolving Admin)