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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!
250
There are no data types. A variable in bash can contain a number, a character, a string of characters. You have no need to declare a variable, just assigning a value to its reference will create it.
To initialize a string, you directly start with the name of the variable followed by the assignment operator(=) and the actual value of the string enclosed in single or double quotes. Output: This simple example initializes a string and prints the value using the “$” operator.
Variable Scope of Bash Functions Scope refers to the visibility of variables. By default, every variable has a global scope that means it is visible everywhere in the script. You can also create a variable as a local variable.
You could try to use instead :
osascript -e "display notification \"$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 !
51
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