问题
I'm writing a script in UNIX where I have to check whether the first character in a string is "/" and if it is, branch.
For example I have a string:
/some/directory/file
I want this to return 1, and:
server@10.200.200.20:/some/directory/file
to return 0.
回答1:
Many ways to do this. You could use wildcards in double brackets:
str="/some/directory/file"
if [[ $str == /* ]]; then echo 1; else echo 0; fi
You can use substring expansion:
if [[ ${str:0:1} == "/" ]] ; then echo 1; else echo 0; fi
Or a regex:
if [[ $str =~ ^/ ]]; then echo 1; else echo 0; fi
回答2:
Consider the case statement as well which is compatible with most sh-based shells:
case "$STRING" in
/*)
echo 1
;;
*)
echo 0
;;
esac
回答3:
$ foo="/some/directory/file"
$ [ ${foo:0:1} == "/" ] && echo 1 || echo 0
1
$ foo="server@10.200.200.20:/some/directory/file"
$ [ ${foo:0:1} == "/" ] && echo 1 || echo 0
0
回答4:
cut -c1
This is POSIX, and unlike case actually extracts the first char if you need it for later:
myvar=abc
first_char="$(printf '%s' "$myvar" | cut -c1)"
if [ "$first_char" = a ]; then
echo 'starts with a'
else
echo 'does not start with a'
fi
awk substr
is another POSIX but less efficient alternative:
printf '%s' "$myvar" | awk '{print substr ($0, 0, 1)}'
printf '%s'
is to avoid problems with escape characters: https://stackoverflow.com/a/40423558/895245 e.g.:
myvar='\n'
printf '%s' "$myvar" | cut -c1
outputs \
as expected.
${::}
does not seem to be POSIX.
See also: How to extract the first two characters of a string in shell scripting?
来源:https://stackoverflow.com/questions/18488270/how-to-check-the-first-character-in-a-string-in-bash-or-unix-shell