I am trying to understand a test script, which includes the following segment:
SCRIPT_PATH=${0%/*}
if [ "$0" != "$SCRIPT_PATH" ] && [ "$SCRIPT_PATH" != "" ]; then
cd $SCRIPT_PATH
fi
What does the ${0%/*} stand for? Thanks
It is called Parameter Expansion. Take a look at this page and the rest of the site.
What ${0%/*} does is, it expands the value contained within the argument 0 (which is the path that called the script) after removing the string /* suffix from the end of it.
So, $0 is the same as ${0} which is like any other argument, eg. $1 which you can write as ${1}. As I said $0 is special, as it's not a real argument, it's always there and represents name of script. Parameter Expansion works within the { } -- curly braces, and % is one type of Parameter Expansion.
%/* matches the last occurrence of / and removes anything (* means anything) after that character. Take a look at this simple example:
$ var="foo/bar/baz"
$ echo "$var"
foo/bar/baz
$ echo "${var}"
foo/bar/baz
$ echo "${var%/*}"
foo/bar
Related
I've got a few Unix shell scripts where I need to check that certain environment variables are set before I start doing stuff, so I do this sort of thing:
if [ -z "$STATE" ]; then
echo "Need to set STATE"
exit 1
fi
if [ -z "$DEST" ]; then
echo "Need to set DEST"
exit 1
fi
which is a lot of typing. Is there a more elegant idiom for checking that a set of environment variables is set?
EDIT: I should mention that these variables have no meaningful default value - the script should error out if any are unset.
Parameter Expansion
The obvious answer is to use one of the special forms of parameter expansion:
: ${STATE?"Need to set STATE"}
: ${DEST:?"Need to set DEST non-empty"}
Or, better (see section on 'Position of double quotes' below):
: "${STATE?Need to set STATE}"
: "${DEST:?Need to set DEST non-empty}"
The first variant (using just ?) requires STATE to be set, but STATE="" (an empty string) is OK — not exactly what you want, but the alternative and older notation.
The second variant (using :?) requires DEST to be set and non-empty.
If you supply no message, the shell provides a default message.
The ${var?} construct is portable back to Version 7 UNIX and the Bourne Shell (1978 or thereabouts). The ${var:?} construct is slightly more recent: I think it was in System III UNIX circa 1981, but it may have been in PWB UNIX before that. It is therefore in the Korn Shell, and in the POSIX shells, including specifically Bash.
It is usually documented in the shell's man page in a section called Parameter Expansion. For example, the bash manual says:
${parameter:?word}
Display Error if Null or Unset. If parameter is null or unset, the expansion of word (or a message to that effect if word is not present) is written to the standard error and the shell, if it is not interactive, exits. Otherwise, the value of parameter is substituted.
The Colon Command
I should probably add that the colon command simply has its arguments evaluated and then succeeds. It is the original shell comment notation (before '#' to end of line). For a long time, Bourne shell scripts had a colon as the first character. The C Shell would read a script and use the first character to determine whether it was for the C Shell (a '#' hash) or the Bourne shell (a ':' colon). Then the kernel got in on the act and added support for '#!/path/to/program' and the Bourne shell got '#' comments, and the colon convention went by the wayside. But if you come across a script that starts with a colon, now you will know why.
Position of double quotes
blong asked in a comment:
Any thoughts on this discussion? https://github.com/koalaman/shellcheck/issues/380#issuecomment-145872749
The gist of the discussion is:
… However, when I shellcheck it (with version 0.4.1), I get this message:
In script.sh line 13:
: ${FOO:?"The environment variable 'FOO' must be set and non-empty"}
^-- SC2086: Double quote to prevent globbing and word splitting.
Any advice on what I should do in this case?
The short answer is "do as shellcheck suggests":
: "${STATE?Need to set STATE}"
: "${DEST:?Need to set DEST non-empty}"
To illustrate why, study the following. Note that the : command doesn't echo its arguments (but the shell does evaluate the arguments). We want to see the arguments, so the code below uses printf "%s\n" in place of :.
$ mkdir junk
$ cd junk
$ > abc
$ > def
$ > ghi
$
$ x="*"
$ printf "%s\n" ${x:?You must set x} # Careless; not recommended
abc
def
ghi
$ unset x
$ printf "%s\n" ${x:?You must set x} # Careless; not recommended
bash: x: You must set x
$ printf "%s\n" "${x:?You must set x}" # Careful: should be used
bash: x: You must set x
$ x="*"
$ printf "%s\n" "${x:?You must set x}" # Careful: should be used
*
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
abc
def
ghi
$ x=
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
bash: x: You must set x
$ unset x
$ printf "%s\n" ${x:?"You must set x"} # Not quite careful enough
bash: x: You must set x
$
Note how the value in $x is expanded to first * and then a list of file names when the overall expression is not in double quotes. This is what shellcheck is recommending should be fixed. I have not verified that it doesn't object to the form where the expression is enclosed in double quotes, but it is a reasonable assumption that it would be OK.
Try this:
[ -z "$STATE" ] && echo "Need to set STATE" && exit 1;
Your question is dependent on the shell that you are using.
Bourne shell leaves very little in the way of what you're after.
BUT...
It does work, just about everywhere.
Just try and stay away from csh. It was good for the bells and whistles it added, compared the Bourne shell, but it is really creaking now. If you don't believe me, just try and separate out STDERR in csh! (-:
There are two possibilities here. The example above, namely using:
${MyVariable:=SomeDefault}
for the first time you need to refer to $MyVariable. This takes the env. var MyVariable and, if it is currently not set, assigns the value of SomeDefault to the variable for later use.
You also have the possibility of:
${MyVariable:-SomeDefault}
which just substitutes SomeDefault for the variable where you are using this construct. It doesn't assign the value SomeDefault to the variable, and the value of MyVariable will still be null after this statement is encountered.
Surely the simplest approach is to add the -u switch to the shebang (the line at the top of your script), assuming you’re using bash:
#!/bin/sh -u
This will cause the script to exit if any unbound variables lurk within.
${MyVariable:=SomeDefault}
If MyVariable is set and not null, it will reset the variable value (= nothing happens).
Else, MyVariable is set to SomeDefault.
The above will attempt to execute ${MyVariable}, so if you just want to set the variable do:
MyVariable=${MyVariable:=SomeDefault}
In my opinion the simplest and most compatible check for #!/bin/sh is:
if [ "$MYVAR" = "" ]
then
echo "Does not exist"
else
echo "Exists"
fi
Again, this is for /bin/sh and is compatible also on old Solaris systems.
bash 4.2 introduced the -v operator which tests if a name is set to any value, even the empty string.
$ unset a
$ b=
$ c=
$ [[ -v a ]] && echo "a is set"
$ [[ -v b ]] && echo "b is set"
b is set
$ [[ -v c ]] && echo "c is set"
c is set
I always used:
if [ "x$STATE" == "x" ]; then echo "Need to set State"; exit 1; fi
Not that much more concise, I'm afraid.
Under CSH you have $?STATE.
For future people like me, I wanted to go a step forward and parameterize the var name, so I can loop over a variable sized list of variable names:
#!/bin/bash
declare -a vars=(NAME GITLAB_URL GITLAB_TOKEN)
for var_name in "${vars[#]}"
do
if [ -z "$(eval "echo \$$var_name")" ]; then
echo "Missing environment variable $var_name"
exit 1
fi
done
We can write a nice assertion to check a bunch of variables all at once:
#
# assert if variables are set (to a non-empty string)
# if any variable is not set, exit 1 (when -f option is set) or return 1 otherwise
#
# Usage: assert_var_not_null [-f] variable ...
#
function assert_var_not_null() {
local fatal var num_null=0
[[ "$1" = "-f" ]] && { shift; fatal=1; }
for var in "$#"; do
[[ -z "${!var}" ]] &&
printf '%s\n' "Variable '$var' not set" >&2 &&
((num_null++))
done
if ((num_null > 0)); then
[[ "$fatal" ]] && exit 1
return 1
fi
return 0
}
Sample invocation:
one=1 two=2
assert_var_not_null one two
echo test 1: return_code=$?
assert_var_not_null one two three
echo test 2: return_code=$?
assert_var_not_null -f one two three
echo test 3: return_code=$? # this code shouldn't execute
Output:
test 1: return_code=0
Variable 'three' not set
test 2: return_code=1
Variable 'three' not set
More such assertions here: https://github.com/codeforester/base/blob/master/lib/assertions.sh
This can be a way too:
if (set -u; : $HOME) 2> /dev/null
...
...
http://unstableme.blogspot.com/2007/02/checks-whether-envvar-is-set-or-not.html
None of the above solutions worked for my purposes, in part because I checking the environment for an open-ended list of variables that need to be set before starting a lengthy process. I ended up with this:
mapfile -t arr < variables.txt
EXITCODE=0
for i in "${arr[#]}"
do
ISSET=$(env | grep ^${i}= | wc -l)
if [ "${ISSET}" = "0" ];
then
EXITCODE=-1
echo "ENV variable $i is required."
fi
done
exit ${EXITCODE}
Rather than using external shell scripts I tend to load in functions in my login shell. I use something like this as a helper function to check for environment variables rather than any set variable:
is_this_an_env_variable ()
local var="$1"
if env |grep -q "^$var"; then
return 0
else
return 1
fi
}
The $? syntax is pretty neat:
if [ $?BLAH == 1 ]; then
echo "Exists";
else
echo "Does not exist";
fi
The command is:
[ -d $x ] && echo $x | grep "${1:-.*}"
I have run it separately, and [ -d $x ] && echo $x just outputs the directory name. What does the ${1:-.*} mean?
In the script you refer to, grep is called. Its first argument, what
it will search for, is either the first script parameter $1, if one
was given, or .*, which matches anything, if no parameters were given.
"$1" or "${1}" in a bash script is replaced with the first argument
the script was called with. Sometimes it's necessary to process that
string a little, which can be done with shell parameter expansion, as
Etan Reisner helpfully pointed out. :- is one such tool; there are
several others.
"${1:-.*}" means "if parameter 1 is unset or null (i.e., no such
parameter was given), then substitute the part after :; in this case,
.*.
Example script pe:
#!/bin/bash
printf 'parameter count = %d\n' $#
printf 'parameter 1 is "%s"\n' "$1"
printf 'parameter 1 is "%s"\n' "${1:-(not given)}"
Output:
$ ./pe 'foo bar'
parameter count = 1
parameter 1 is "foo bar"
parameter 1 is "foo bar"
$ ./pe
parameter count = 0
parameter 1 is ""
parameter 1 is "(not given)"
I want to implement a bash function which test is the 1st argument is actually a variable, defined somewhere.
For instance, in my .bashrc :
customPrompt='yes';
syntaxOn='no';
[...]
function my_func {
[...]
# I want to test if the string $1 is the name of a variable defined up above
# so something like:
if [[ $$1 == 'yes' ]];then
echo "$1 is set to yes";
else
echo "$1 is not set or != to yes";
fi
# but of course $$1 doesn't work
}
output needed :
$ my_func customPrompt
> customPrompt is set to yes
$ my_func syntaxOn
> syntaxOn is set but != to yes
$ my_func foobar
> foobar is not set
I tried a lot of test, like -v "$1", -z "$1", -n "$1", but all of them test $1 as a string not as a variable.
(please correct me if I make not myself clear enought)
In the bash you can use the indirect variable subtituion.
t1=some
t2=yes
fufu() {
case "${!1}" in
yes) echo "$1: set to yes. Value: ${!1}";;
'') echo "$1: not set. Value: ${!1:-UNDEF}";;
*) echo "$1: set to something other than yes. Value: ${!1}";;
esac
}
fufu t1
fufu t2
fufu t3
prints
t1: set to something other than yes. Value: some
t2: set to yes. Value: yes
t3: not set. Value: UNDEF
The ${!variablename} in bash mean indirect variable expansion. Described in the e.g. https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
Whrere:
The basic form of parameter expansion is ${parameter}. The value of
parameter is substituted. The braces are required when parameter is a
positional parameter with more than one digit, or when parameter is
followed by a character that is not to be interpreted as part of its
name.
If the first character of parameter is an exclamation point (!), a
level of variable indirection is introduced. Bash uses the value of
the variable formed from the rest of parameter as the name of the
variable; this variable is then expanded and that value is used in the
rest of the substitution, rather than the value of parameter itself.
This is known as indirect expansion. The exceptions to this are the
expansions of ${!prefix } and ${!name[#]} described below. The
exclamation point must immediately follow the left brace in order to
introduce indirection.
Also, check this: https://stackoverflow.com/a/16131829/632407 how to modify in a function a value of the variable passed indirectly.
You can check variable set or not by simply like
if [[ $var ]]
then
echo "Sorry First set variable"
else
echo $var
fi
You can do something like this for your script
customPrompt='yes';
syntaxOn='no';
function my_func
{
if [[ ${!1} ]];then
echo "$1 is set to ${!1}";
else
echo "$1 is not set";
fi
}
my_func customPrompt
my_func syntaxOn
my_func foobar
Output:
customPrompt is set to yes
syntaxOn is set to no
foobar is not set
You can customize the function as per you requirement by simply making some comparison conditions.
For more details you can check this answer
If you really want to check if your variable is set or unset (not just empty), use this format:
function my_func {
if [[ -z ${!1+.} ]]; then
echo "$1 is not set."
elif [[ ${!1} == yes ]]; then
echo "$1 is set to yes"
else
echo "$1 is set to \"${!1}\"."
fi
}
You're going to have problems...
The Bash shell is a very wily creature. Before you execute anything, Bash comes in and interpolates your command. Your command or shell script never sees whether or not you have a variable as a parameter.
$ set -x
set -x
$ foo=bar
+ foo=bar
$ echo "$foo"
+ echo bar
bar
$ set +x
The set -x turns on debugging mode in the shell. It shows you what a command actually executes. For example, I set foo=bar and then do echo $foo. My echo command doesn't see $foo. Instead, before echo executes, it interpolates $foo with bar. All echo sees at this point is that it's suppose to take bar as its argument (not $foo).
This is awesomely powerful. It means that your program doesn't have to sit there and interpret the command line. If you typed echo *.txt, echo doesn't have to expand *.txt because the shell has already done the dirty work.
For example, here's a test shell script:
#! /bin/sh
if [[ $1 = "*" ]]
then
echo "The first argument was '*'"
else
"I was passed in $# parameters"
fi
Now, I'll run my shell script:
$ test.sh *
I was passed in 24 parameters
What? Wasn't the first parameter of my script a *? No. The shell grabbed * and expanded it to be all of the files and directories in my directory. My shell script never saw the *. However, I can do this:
$ test.sh '*'
The first argument was '*'
The single quotes tell the shell not to interpolate anything. (Double quotes prevent globbing, but still allow for environment variable expansion).
This if I wanted to see if my first parameter is a variable, I have to pass it in single quotes:
$ test.sh '$foo'
And, I can do this as a test:
if [[ $1 != ${1#$} ]]
then
echo "The first parameter is the variable '$1'"
fi
The ${1#$} looks a bit strange, but it's just ${var#pattern}. This removes pattern from the left most side of $var. I am taking $1 and removing the $ if it exists. This gets expanded in the shell as:
if [[ $foo != foo ]]
which is true.
So, several things:
First, you've got to stop the shell from interpolating your variable. That means you have to use single quotes around the name.
You have to use pattern matching to verify that the first parameter starts with a $.
Once you do that, you should be able to use your variable with ${$1} in your script.
It is pretty clear that with shell scripting this sort of thing can be accomplished in a huge number of ways (more than most programming languages) because of all the different variable expansion methods and programs like test and [ and [[, etc.
Right now I'm just looking for
DIR=$1 or .
Meaning, my DIR variable should contain either what is specified in the first arg or the current directory.
What is the difference between this and DIR=${1-.}?
I find the hyphen syntax confusing, and seek more readable syntax.
Why can't I do this?
DIR="$1" || '.'
I'm guessing this means "if $1 is empty, the assignment still works (DIR becomes empty), so the invalid command '.' never gets executed."
I see several questions here.
“Can I write something that actually reflects this logic”
Yes. There are a few ways you can do it. Here's one:
if [[ "$1" != "" ]]; then
DIR="$1"
else
DIR=.
fi
“What is the difference between this and DIR=${1-.}?”
The syntax ${1-.} expands to . if $1 is unset, but expands like $1 if $1 is set—even if $1 is set to the empty string.
The syntax ${1:-.} expands to . if $1 is unset or is set to the empty string. It expands like $1 only if $1 is set to something other than the empty string.
“Why can't I do this? DIR="$1" || '.'”
Because this is bash, not perl or ruby or some other language. (Pardon my snideness.)
In bash, || separates entire commands (technically it separates pipelines). It doesn't separate expressions.
So DIR="$1" || '.' means “execute DIR="$1", and if that exits with a non-zero exit code, execute '.'”.
How about this:
DIR=.
if [ $# -gt 0 ]; then
DIR=$1
fi
$# is the number of arguments given to the script, and -gt means "greater than", so you basically set DIR to the default value, and if the user has specified an argument, then you set DIR to that instead.
I use a simple helper function to make such assignments look cleaner. The function below accepts any number of arguments, but returns the first one that's not the empty string.
default_value() {
# Return the first non-empty argument
while [[ "$1" == "" ]] && [[ "$#" -gt "0" ]]; do
shift
done
echo $1
}
x=$(default_value "$1" 0)
I am little unfamiliar with the 'if...then...fi' and the 'for' statements syntax.
Could anyone explain what the "$2/$fn" and "/etc/*release" in the code snippets below mean?...specifically on the use of the forward slash....and the asterisk...
if [ -f "$filename" ]; then
if [ ! -f "$2/$fn" ]; then
echo "$fn is missing from $2"
missing=$((missing + 1))
fi
fi
and
function system_info
{
if ls /etc/*release 1>/dev/null 2>&1; then
echo "<h2>System release info</h2>"
echo "<pre>"
for i in /etc/*release; do
# Since we can't be sure of the
# length of the file, only
# display the first line.
head -n 1 $i
done
uname -orp
echo "</pre>"
fi
} # end of system_info
...thx for the help...
/etc/*release : here the * will match any number of any characters, so any thing /etc/0release , /etc/asdfasdfr_release etc will be matched. Simply stated, it defined all the files in the /etc/ directory which ends with the string release.
The $2 is the 2nd commandline argument to the shell script, and $fn is some other shell variable. The "$2/$fn" after the variable substitutions will make a string, and the [ -f "$2/$fn" ] will test if the string formed after the substitution forms a path to a regular file which is specified by the -f switch. If it is a regular file then the body of if is executed.
In the for loop the loop will loop for all the files ending with the string release in the directory /etc (the path). At each iteration i will contain the next such file name, and for each iteration the first 1 line of the file is displayed with the head command by getting the file name from variable i within the body.
It is better to check the manual man bash and for if condition check man test . Here is a good resource: http://tldp.org/LDP/Bash-Beginners-Guide/html/
The forward slash is the path separator, and the * is a file glob character. $2/$fn is a path where $2 specifies the directory and $fn is the filename. /etc/*release expands to the space separated list of all the files in /etc whose name ends in "release"
Dollar sign marks variable. The "-f" operator means "file exsists".
So,
[ -f "$filename" ]
checks if there is file named the same as value contained in $filename variable.
Simmilar, if we assume that $2 = "some_folder", and $fn = "some_file", expression
[ ! -f "$2/$fn" ]
returns true if file some_folder/some_file doesn't exsist.
Now, about asterisk - it marks "zero or more of any character(s)". So, expression:
for i in /etc/*release; do
will iterate trough all folders named by that pattern, for example:
/etc/release, /etc/666release, /etc/wtf_release...
I hope this helps.