CODEWITHSUNDEEP
bashvariablessyntaxapplescriptosascript
Bernd
Bernd

Reputation: 695

osascript using bash variable with a space

I am using osascript in Bash to display a message in Notification Center (Mac OS X) via Apple Script. I am trying to pass a text variable from Bash to the script. For a variable without spaces, this works just fine, but not for one with spaces:

Defining

var1="Hello"
var2="Hello World"

and using

osascript -e 'display notification "'$var1'"'

works, but using

osascript -e 'display notification "'$var2'"'

yields

syntax error: Expected string but found end of script.

What do I need to change (I am new to this)? Thanks!

Upvotes: 23

Views: 24543

Answers (2)

Idriss Neumann
Idriss Neumann

Reputation: 3838

You could try to use instead :

osascript -e "display notification \"$var2\""

Or :

osascript -e 'display notification "'"$var2"'"'

This fixes the problem of manipulation of variables that contains spaces in bash. However, this solution doesn't protect against injections of osascript code. So it would be better to choose one of Charles Duffy's solutions or to use bash parameter expansion :

# if you prefer escape the doubles quotes
osascript -e "display notification \"${var2//\"/\\\"}\""
# or
osascript -e 'display notification "'"${var2//\"/\\\"}"'"'

# if you prefer to remove the doubles quotes
osascript -e "display notification \"${var2//\"/}\""
# or
osascript -e 'display notification "'"${var2//\"/}"'"'

Thank to mklement0 for this very useful suggestion !

Upvotes: 38

Charles Duffy
Charles Duffy

Reputation: 295288

This version is completely safe against injection attacks, unlike variants trying to use string concatenation.

osascript \
  -e "on run(argv)" \
  -e "return display notification item 1 of argv" \
  -e "end" \
  -- "$var2"

...or, if one preferred to pass code in on stdin rather than argv:

osascript -- - "$var2" <<'EOF'
  on run(argv)
    return display notification item 1 of argv
  end
EOF

Upvotes: 20

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