In Xcode 8 I have a Run Script that is comparing a string to a regex:
if [ "$MOBILE_BUNDLE_IDENTIFIER" =~ ".+(Debug)" ]
then
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Debug
elif [ "$MOBILE_BUNDLE_IDENTIFIER" =~ ".+(Test)" ]
then
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Production
elif [ "$MOBILE_BUNDLE_IDENTIFIER" =~ ".+(ProductionTest)" ]
then
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Test
else
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Store
fi
The variable $MOBILE_BUNDLE_IDENTIFIER will be one of the following:
com.cnt.Filer
com.cnt.Filer.ProductionTest
com.cnt.Filer.Debug
com.cnt.Filer.Test
When I run this I get the following error:
line 4: [: =~: binary operator expected
line 7: [: =~: binary operator expected
line 10: [: =~: binary operator expected
In the full script lines 4, 7 and 10 are the then statements.
Does anyone know how I can successfully compare my variable to the regex?
You should be using [[ string =~ regex ]] for regex in BASH
Don't quote regex
Looks like you don't even need regex, you can just do string comparison using ==
Your script can be this:
if [[ $MOBILE_BUNDLE_IDENTIFIER == *Debug* ]]; then
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Debug
elif [[ $MOBILE_BUNDLE_IDENTIFIER == *Test* ]]; then
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Production
elif [[ $MOBILE_BUNDLE_IDENTIFIER == *ProductionTest* ]]; then
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Test
else
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Store
fi
PS: You may also consider using case
case "$MOBILE_BUNDLE_IDENTIFIER" in
*Debug*)
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Debug
;;
*Test*)
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Production
;;
*ProductionTest*)
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Test
;;
*)
RESOURCE_PATH=${SRCROOT}/Resources/Clients/Cnt/Files/Store
;;
esac
Related
I am trying to understand the logic of Bash algorithm.
When I tried this, it printed "a":
a=a;[ $a == "a" ] && echo $a
So far so good. Then I tried the following and it printed "a" again:
a=a;[[ $a == "a" ]] && echo $a
Now I introduced an error by using arithmetic comparison:
a=abc;[ $a -eq "abc" ] && echo $a
I got an error message that makes sense:
-bash: [: abc: integer expression expected
Then I tried to do this with double bracket and got no error, but "abc":
a=abc;[[ $a -eq "abc" ]] && echo $a
I can sort of explain it (bash is trying to be accomodating), but then I got something that puzzles me. If I do that, I get an error message about recursion:
a=a;[[ $a -eq "a" ]] && echo $a
-bash: [[: a: expression recursion level exceeded (error token is "a")
If I use single brackets, there is no recursion but a reasonable error "integer expression expected":
a=a;[ $a -eq "a" ] && echo $a
-bash: [: abc: integer expression expected
This is weird. What Bash is trying to do in that "recursion" case with double brackets? I am talking about:
a=a;[[ $a -eq "a" ]] && echo $a
-bash: [[: a: expression recursion level exceeded (error token is "a")
It's trying to coerce a into a numeric form. It does this by treating its contents as a variable name to dereference; when the result of this is a loop, you get a "recursion level exceeded" error.
Thus:
a=b
b=10
[[ $a -eq 10 ]]
...is true.
This question already has answers here:
Shell equality operators (=, ==, -eq)
(4 answers)
Closed 4 years ago.
i am studying shell script right now. I started to learn how to work with more complex if statements. What's wrong with this code bellow? I read other similar questions here in stackoverflow, but i couldnt resolve my problem. Now im verifying if the first, second or third argument is null. In the future i pretend to verify based in some regex or something like that.
Thanks!!
The code (line 9):
if [ "$1" -eq "" ] || [ "$2" -eq "" ] || [ "$3" -eq "" ] then ...
line 9: [: : integer expression expected line 9: [: : integer
expression expected line 9: [: : integer expression expected
-eq performs an arithmetic comparison between two numbers. Use = for string comparisons. Or better yet, use [[ and ==.
[[ $1 == "" ]]
[ "$1" = "" ]
You can also use -z and -n to directly test whether a value is empty/non-empty.
[[ -n $value ]] # [[ $value != "" ]]
[[ -z $value ]] # [[ $value == "" ]]
use [[ and ]] for the more modern / complex operators. This is a bashism, so beware.
I found this question for how to do conditional assignment in bash, but what I'm trying to do is a little more complex, and I can't seem to get the syntax right. The condition in my case is to test a variable to see if it exists, and the output is concatenated to a string.
Here's what I have so far:
fwversion="${BASH_REMATCH[1]}.$(( [[ BASH_REMATCH[2] ]] ? BASH_REMATCH[2] : 0 ))"
Which produces this error message:
bash: line 41: [[ BASH_REMATCH[2] ]] ? BASH_REMATCH[2] : 0 : syntax error:
operand expected (error token is "[[ BASH_REMATCH[2] ]] ? BASH_REMATCH[2] : 0 ")
Here's what I'm trying to achieve as C++ code:
std::string fwversion = BASH_REMATCH[1] + "." + ((BASH_REMATCH[2]) ? : BASH_REMATCH[2] : 0);
What's the correct syntax to do this in bash? Thanks.
Looks like [[ ... ]] are not understood in an arithmetic expression.
I'd do this:
fwversion=${BASH_REMATCH[1]}
[[ ${BASH_REMATCH[2]} ]] && fwversion+=${BASH_REMATCH[2]} || fwversion+=0
or
[[ ${BASH_REMATCH[2]} ]] && ext=${BASH_REMATCH[2]} || ext=0
fwversion="${BASH_REMATCH[1]}.$ext"
On second thought, I wouldn't do that at all, I'd use the power of the shell's parameter expansion
str="foo:bar"
if [[ $str =~ ^([a-z]+):([a-z]*)$ ]]; then
echo "${BASH_REMATCH[1]}.${BASH_REMATCH[2]:-0}"
fi
foo.bar
str="foo:"
if [[ $str =~ ^([a-z]+):([a-z]*)$ ]]; then
echo "${BASH_REMATCH[1]}.${BASH_REMATCH[2]:-0}"
fi
foo.0
./build_binaries.sh: line 43: [: ==: unary operator expected
I have this line (line 43) in my bash script which looks correct to me, but it keeps throwing error.
if [ ${platform} == "macosx" ]; then
Error:
./foo.sh: line 43: [: ==: unary operator expected
This is on OSX.
The problem is that $platform is an empty string. The usual workaround is to put it in quotes:
if [ "${platform}" == "macosx" ]
Example:
$ unset x
$ [ $x == 3 ]
-bash: [: ==: unary operator expected
$ [ "$x" == "3" ]
$
One possibility is to use a single =. That's the classic notation. Some shells allow ==, but others do not.
Also, you should enclose the ${platform} in double quotes; I think that it is an empty string, and this is confusing things.
platform=
if [ $platform == mac ]; then echo hi; else echo lo; fi
if [ "$platform" == mac ]; then echo hi; else echo lo; fi
This produces the error you're seeing on the second line.
The following bash script is giving me problems:
#!/bin/bash
if [[ $VAR -eq "<EMPTY>" ]]; then echo "Hello World!"; fi
Bash fails, complaining:
line 3: [[: <EMPTY>: syntax error: operand expected (error token is "<EMPTY>")
How can I test if the string contained in VAR is equivalent to the string "<EMPTY>"?
You are using the wrong operator. == is for strings, -eq is for numbers.
#!/bin/bash
if [[ $VAR == "<EMPTY>" ]]; then echo "Hello World!"; fi
Inside [[ ... ]], -eq has a different meaning: it is used to compare integers. You can use (( ... )) to compare integeres with normal operators, though. Use the following for strings:
[[ $VAR == "<EMPTY>" ]]