I'm trying to return a function if the NAMESPACE variable is blank or if the VERSION variable doesn't match the correct pattern.
# return usage if namespace is blank or version doesn't match the version format.
if [[ "$NAMESPACE" == "" || "$VERSION" =~ ^([0-9]\.([1-9]|[1-9][0-9])\.[0-9])$ ]];
then
usage
fi
Currently I'm using =~ which returns true if the pattern is x.xx.x or x.x.x. But I'm having trouble finding what the operator would be for not equal (something similar to !=~)
You can negate after the || for a (A OR NOT B):
if [[ "$NAMESPACE" == "" || ! "$VERSION" =~ ^([0-9]\.([1-9]|[1-9][0-9])\.[0-9])$ ]];
then
usage
fi
Note that you need to have spaces around the !.
Alternatively you can reverse the (A OR NOT B) to NOT (NOT A AND B):
if ! [[ "$NAMESPACE" != "" && "$VERSION" =~ ^([0-9]\.([1-9]|[1-9][0-9])\.[0-9])$ ]];
then
usage
fi
Related
I was trying to write myself some handy scripts in order to legitimately slacking off work more efficiently, and this question suddenly popped up:
Given a very long string $LONGEST_EVER_STRING and several keywords strings like $A='foo bar' , $B='omg bbq' and $C='stack overflow'
How should I use exact keyword matching as a condition in the case statement?
for word in $LONGEST_EVER_STRING; do
case $word in
any exact match in $A) do something ;;
any exact match in $B) do something ;;
any exact match in $C) do something ;;
*) do something else;;
esac
done
I know I can write in this way but it looks really ugly:
for word in $LONGEST_EVER_STRING; do
if [[ -n $(echo $A | fgrep -w $word) ]]; then
do something;
elif [[ -n $(echo $B | fgrep -w $word) ]]; then
do something;
elif [[ -n $(echo $C | fgrep -w $word) ]]; then
do something;
else
do something else;
fi
done
Does anyone have an elegant solution? Many thanks!
You could use a function to do a little transform in your A, B, C variables and then:
shopt -s extglob
Ax="+(foo|bar)"
Bx="+(omg|bbq)"
Cx="+(stack|overflow)"
for word in $LONGEST_EVER_STRING; do
case $word in
$Ax) do something ;;
$Bx) do something ;;
$Cx) do something ;;
*) do something else;;
esac
done
I would just define a function for this. It'll be slower than grep for large wordlists, but faster than starting up grep many times.
##
# Success if the first arg is one of the later args.
has() {
[[ $1 = $2 ]] || {
[[ $3 ]] && has "$1" "${#:3}"
}
}
$ has a b c && echo t || echo f
f
$ has a b c a d e f && echo t || echo f
t
A variation on /etc/bashrc's "pathmunge"
for word in $LONGEST_EVER_STRING; do
found_word=false
for list in " $A " " $B " " $C "; do
if [[ $list == *" $word "* ]]; then
found_word=true
stuff with $list and $word
break
fi
done
$found_word || stuff when not found
done
if [[ " " =~ ^[0-9]*$ ]]; then echo "si"; else echo "no"; fi; //Echoes No
if [[ "" =~ ^[0-9]*$ ]]; then echo "si"; else echo "no"; fi; //Echoes Yes
Is this a bug or am I missing something?
This is as expected. You specified 0 or more times (*) a digit ([0-9]). An empty string is 0 times that.
Use a + (which means "1 or more times") instead of a *:
if [[ " " =~ ^[0-9]+$ ]]; then echo "si"; else echo "no"; fi; // Should echo No
if [[ "" =~ ^[0-9]+$ ]]; then echo "si"; else echo "no"; fi; // Should echo No
The first one is a space, which does not match the [0-9]* regex.
The second is empty, which is [0-9]* because * also implies 0 ocurrencies. If you make it match at least one ocurrency with +, then it is false:
$ if [[ " " =~ ^[0-9]+$ ]]; then echo "si"; else echo "no"; fi;
no
* in a regex means "0 or more", so with nothing in the target string, the regex trivially matches.
[0-9]* matches zero or more digits, so yes, it matches the empty string. If you don't want to match the empty string, use [0-9]+, which matches one or more digits.
I need to ensure that the last character in a string is a /
x="test.com/"
if [[ $x =~ //$/ ]] ; then
x=$x"extention"
else
x=$x"/extention"
fi
at the moment, false always fires.
Like this, for example:
$ x="test.com/"
$ [[ "$x" == */ ]] && echo "yes"
yes
$ x="test.com"
$ [[ "$x" == */ ]] && echo "yes"
$
$ x="test.c/om"
$ [[ "$x" == */ ]] && echo "yes"
$
$ x="test.c/om/"
$ [[ "$x" == */ ]] && echo "yes"
yes
$ x="test.c//om/"
$ [[ "$x" == */ ]] && echo "yes"
yes
You can index strings in Bash using ${var:index} and ${#var} to get the length of the string. Negative indices means the moving from the end to the start of the string so that -1 is index of the last character:
if [[ "${x:${#x}-1}" == "/" ]]; then
# last character of x is /
fi
Your condition was slightly incorrect. When using =~, the rhs is considered a pattern, so you'd say pattern and not /pattern/.
You'd have got expected results if you said
if [[ $x =~ /$ ]] ; then
instead of
if [[ $x =~ //$/ ]] ; then
You can do this generically using bash substrings $(string:offset:length} - length is optional
#x is the length of x
Therefore
$n = 1 # 1 character
last_char = ${x:${#x} - $n}
For future references,
$ man bash
has all the magic
${parameter:offset:length}
Substring Expansion. Expands to up to length characters of parameter
starting at the character specified by offset. If length is
omitted, expands to the substring of parameter starting at the
character specified by offset. length and offset are arithmetic
expressions ...
I have some text that looks like this:
(something1)something2
However something1 and something2 might also have some parentheses inside them such as
(some(thing)1)something(2)
I want to extract something1 (including internal parentheses if there are any) to a variable. Since I can count on the text always starting with an opening parentheses, I'm hoping that I can do something where I match the first parenthesis to the correct closing parentheses, and extract the middle.
Everything I have tried so far has the potential to match the wrong ending parentheses.
If you have perl, the:
perl -MText::Balanced -nlE 'say [Text::Balanced::extract_bracketed( $_, "()" )]->[0]' <<EOF
(something1)something2
(some(thing)1)something(2)
(some(t()()hing)()1)()something(2)
EOF
will prints
(something1)
(some(thing)1)
(some(t()()hing)()1)
Since this is apparently something that is impossible with regular expressions, I have resorted to pickup the the characters 1 by 1:
first=""
count=0
while test -n "$string"
do
char=${string:0:1} # Get the first character
if [[ "$char" == ")" ]]
then
count=$(( $count - 1 ))
fi
if [[ $count > 0 ]]
then
first="$first$char"
fi
if [[ "$char" == "(" ]]
then
count=$(( $count + 1 ))
fi
string=${string:1} # Trim the first character
if [[ $count == 0 ]]
then
second="$string"
string=""
fi
done
You can do it with perl:
echo "(some(thing)1)something(2)" | perl -ne '$_ =~ /(\((?:\(.*\)|[^(])*\))|\w+/s; print $1;'
awk can do it:
#!/bin/awk -f
{
for (i=1; i<=length; ++i) {
if (numLeft == 0 && substr($0, i, 1) == "(") {
leftPos = i
numLeft = 1
} else if (substr($0, i, 1) == "(") {
++numLeft
} else if (substr($0, i, 1) == ")") {
++numRight
}
if (numLeft && numLeft == numRight) {
print substr($0, leftPos, i-leftPos+1)
next
}
}
}
Input:
(something1)something2
(some(thing)1)something(2)
Output:
(something1)
(some(thing)1)
Lets say I do this:
my ($a,$b,$let) = $version =~ m/^(\d+)\.(\d+)\.?([A-Za-z])?$/;
so this will match for instance: 1.3a, 1.3,...
I want to have a default value for $let if let is not available, lets say, default 0.
so for 1.3 I will get:
$a = 1
$b = 3
$let = 0
is it possible? (from the regex it self, without using additional statements)
Thanks,
This will work - updated to use bitwise or instead of ternary operator.
my ($a,$b,$let) = ($version =~ m/^(\d+)\.(\d+)\.?([A-Za-z])?$/)
&& ($1,$2,$3 || 0 );
Here is a test script
&t("1.3");
&t("1.3a");
&t("1.3.a");
sub t {
$version = shift;
my ($a,$b,$let) = ($version =~ m/^(\d+)\.(\d+)\.?([A-Za-z])?$/)
&& ($1,$2,$3 || 0 );
print "\n result $a.$b.$let";
}
Output is
result 1.3.0
result 1.3.a
result 1.3.a
original solution using ternary operator also works
my ($a,$b,$let) = ($version =~ m/^(\d+)\.(\d+)\.?([A-Za-z])?$/)
&& (defined $3 ? ($1,$2,$3) : ($1,$2,0));
$let should have a default value of undef. You can test on that if you need to.