JavaScript has a function for this:
'world'.startsWith('w')
true
How can I test this with shell? I have this code:
if [ world = w ]
then
echo true
else
echo false
fi
but it fails because it is testing for equality. I would prefer using a builtin, but any utilities from this page would be acceptable:
http://pubs.opengroup.org/onlinepubs/9699919799/idx/utilities.html
If your shell is bash: within double brackets, the right-hand side of the == operator is a pattern unless fully quoted:
if [[ world == w* ]]; then
echo true
else
echo false
fi
Or more tersely: [[ world == w* ]] && echo true || echo false [*]
If you are not targetting bash specifically: use the case statement for pattern matching
case "world" in
w*) echo true ;;
*) echo false ;;
esac
[*] but you need to be careful with the A && B || C form because C will be executed if either A fails or B fails. The if A; then B; else C; fi form will only execute C if A fails.
Set a variable to that value (here we set $str to world):
str=world
Then:
if [ -n "${str%%w*}" ]
then
echo false
else
echo true
fi
w* from the start of $str. w* matches w followed by any number of characters, so it'll match the whole string if possible.$str does not start with w.Or:
if [ "$str" = "${str#w}" ]
then
echo false
else
echo true
fi
w from $str, if possible.$str.$str does not start with w.The standard sh syntax for that is with a case construct:
string=world
start=w
case $string in
("$start"*) printf '%s\n' "$string starts with $start";;
(*) printf '%s\n' "$string does not start with $start";;
esac
Note the quotes around $start which are necessary in the cases where $start contains wildcard characters. For instance, with start='*', without them, $start* would always match. Same goes for the [[ $string = "$start"* ]] of ksh/bash.
