code:
read location
accepted inputs are
"/home/(10 chars)" or "/(10 chars)" or "(10 chars)"
how to check for valid input? and how to cut the 10 chars from $location variable?
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3 Answers
You need something like this:
case "$location" in
/home/??????????)
echo $location
;;
/??????????)
echo $location
;;
??????????)
echo $location
;;
*)
d=$(dirname "$location")
b=$(basename "$location")
echo $d/${b:0:10}
esac
answered Nov 1, 2011 at 21:05
Diego Torres Milano
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Comments
I would use a grep expression like so:
echo $location | grep -xq "\w\{10\}\|/\w\{10\}\|/home/\w\{10\}"
This matches lines which are exactly one of the following cases( caused by the -x ) and doesn't print the matching line (caused by the -q)
- 10 characters
- 10 characters with a leading /
- 10 character preceded by '/home/'
To use this in a script, just drop it in an if statement like so:
if echo "$location" | grep -xq "\w\{10\}\|/\w\{10\}\|/home/\w\{10\}"; then
# location valid
else
# location not valid
fi
1 Comment
milahu
please fix the grep solution to
printf '%s' "$input" | tr $'\n' ' ' | grep -q -E '^...$'. i use tr $'\n' ' ' to replace every newline with an illegal character. here i assume that space is an illegal character. see also Check if a string matches a regex in Bash scriptYou want substitution operators. ${VAR#/home/} evaluates to the value of $VAR with a leading /home/ stripped, if it exists. By comparing the result of these expressions to $VAR itself you can determine whether your value matches, and what the stripped version is. And ${#VAR} gives you the length of a variable value.
