Understand this .bashrc script (curly braces, eval, ...) - bash

I have some difficulty understanding what is written in my ubuntu's .bashrc which is shown in part below.
Here is what I don't understand :
What is the purpose of curly braces and the -/+ symbols used after :? (ex. : ${debian_chroot:-} and ${debian_chroot:+($debian_chroot)})
The eval command.
How the following snippet of code works.
[ -x /usr/bin/lesspipe ] && eval "$(SHELL=/bin/sh lesspipe)"
if [ -z "${debian_chroot:-}" ] && [ -r /etc/debian_chroot ]; then
debian_chroot=$(cat /etc/debian_chroot)
fi
if [ "$color_prompt" = yes ]; then
PS1='${debian_chroot:+($debian_chroot)}\[\033[01;32m\]\u#\h\[\033[00m\]:\[\033[01;34m\]\w\[\033[00m\]\$ '
else
PS1='${debian_chroot:+($debian_chroot)}\u#\h:\w\$ '
fi

${var:-default} means $var if $var is defined and otherwise "default"
${var:+value} means if $var is defined use "value"; otherwise nothing
The second one might seem a little wierd, but your code snippet shows a typical use:
${debian_chroot:+($debian_chroot)}
This means "if $debian_chroot is defined, then insert it inside parentheses."
Above, "defined" means "set to some non-null value". Unix shells typically don't distinguish between unset variables and variables set to an empty string, but bash can be told to raise an error condition if an unset variable is used. (You do this with set -u.) In this case, if debian_chroot has never been set, $debian_chroot will cause an error, while ${debian_chroot:-} will use $debian_chroot if it has been set, and otherwise an empty string.

Related

How can I override a bash variable at the time of executing it in console? [duplicate]

I have a whole bunch of tests on variables in a bash (3.00) shell script where if the variable is not set, then it assigns a default, e.g.:
if [ -z "${VARIABLE}" ]; then
FOO='default'
else
FOO=${VARIABLE}
fi
I seem to recall there's some syntax to doing this in one line, something resembling a ternary operator, e.g.:
FOO=${ ${VARIABLE} : 'default' }
(though I know that won't work...)
Am I crazy, or does something like that exist?
Very close to what you posted, actually. You can use something called Bash parameter expansion to accomplish this.
To get the assigned value, or default if it's missing:
FOO="${VARIABLE:-default}" # If variable not set or null, use default.
# If VARIABLE was unset or null, it still is after this (no assignment done).
Or to assign default to VARIABLE at the same time:
FOO="${VARIABLE:=default}" # If variable not set or null, set it to default.
For command line arguments:
VARIABLE="${1:-$DEFAULTVALUE}"
which assigns to VARIABLE the value of the 1st argument passed to the script or the value of DEFAULTVALUE if no such argument was passed. Quoting prevents globbing and word splitting.
If the variable is same, then
: "${VARIABLE:=DEFAULT_VALUE}"
assigns DEFAULT_VALUE to VARIABLE if not defined.
The colon builtin (:) ensures the variable result is not executed
The double quotes (") prevent globbing and word splitting.
Also see Section 3.5.3, Shell Parameter Expansion, in the Bash manual.
To answer your question and on all variable substitutions
echo "${var}"
echo "Substitute the value of var."
echo "${var:-word}"
echo "If var is null or unset, word is substituted for var. The value of var does not change."
echo "${var:=word}"
echo "If var is null or unset, var is set to the value of word."
echo "${var:?message}"
echo "If var is null or unset, message is printed to standard error. This checks that variables are set correctly."
echo "${var:+word}"
echo "If var is set, word is substituted for var. The value of var does not change."
You can escape the whole expression by putting a \ between the dollar sign and the rest of the expression.
echo "$\{var}"
Even you can use like default value the value of another variable
having a file defvalue.sh
#!/bin/bash
variable1=$1
variable2=${2:-$variable1}
echo $variable1
echo $variable2
run ./defvalue.sh first-value second-value output
first-value
second-value
and run ./defvalue.sh first-value output
first-value
first-value
see here under 3.5.3(shell parameter expansion)
so in your case
${VARIABLE:-default}
FWIW, you can provide an error message like so:
USERNAME=${1:?"Specify a username"}
This displays a message like this and exits with code 1:
./myscript.sh
./myscript.sh: line 2: 1: Specify a username
A more complete example of everything:
#!/bin/bash
ACTION=${1:?"Specify 'action' as argv[1]"}
DIRNAME=${2:-$PWD}
OUTPUT_DIR=${3:-${HOMEDIR:-"/tmp"}}
echo "$ACTION"
echo "$DIRNAME"
echo "$OUTPUT_DIR"
Output:
$ ./script.sh foo
foo
/path/to/pwd
/tmp
$ export HOMEDIR=/home/myuser
$ ./script.sh foo
foo
/path/to/pwd
/home/myuser
$ACTION takes the value of the first argument, and exits if empty
$DIRNAME is the 2nd argument, and defaults to the current directory
$OUTPUT_DIR is the 3rd argument, or $HOMEDIR (if defined), else, /tmp. This works on OS X, but I'm not positive that it's portable.
Then there's the way of expressing your 'if' construct more tersely:
FOO='default'
[ -n "${VARIABLE}" ] && FOO=${VARIABLE}
It is possible to chain default values like so:
DOCKER_LABEL=${GIT_TAG:-${GIT_COMMIT_AND_DATE:-latest}}
i.e. if $GIT_TAG doesnt exist, take $GIT_COMMIT_AND_DATE - if this doesnt exist, take "latest"
Here is an example
#!/bin/bash
default='default_value'
value=${1:-$default}
echo "value: [$value]"
save this as script.sh and make it executable.
run it without params
./script.sh
> value: [default_value]
run it with param
./script.sh my_value
> value: [my_value]
If you want 1 liner for your if-then-else, then we can consider rewriting:
if [ -z "${VARIABLE}" ]; then
FOO='default'
else
FOO=${VARIABLE}
fi
with semicolons:
if [ -z ${VARIABLE} ]; then FOO=`default`; else FOO=${VARIABLE}; fi
Alternatively, you can drop the if-then-else-fi keywords if you use boolean operators such as:
[ -z "${VARIABLE}" ] && FOO='default' || FOO=${VARIABLE}
Generalizing, the boolean operator pattern is:
conditional && then_command || else_command

Bash If statement null

Looking for the correct syntax that looks at a bash variable and determines if its null, if it is then do ... otherwise continue on.
Perhaps something like if [ $lastUpdated = null?; then... else...
Just test if the variable is empty:
if [ -z "$lastUpdated" ]; then
# not set
fi
Expanding on #chepner's comments, here's how you could test for an unset (as opposed to set to a possibly empty value) variable:
if [ -z "${lastUpdated+set}" ]; then
The ${variable+word} syntax gives an empty string if $variable is unset, and the string "word" if it's set:
$ fullvar=somestring
$ emptyvar=
$ echo "<${fullvar+set}>"
<set>
$ echo "<${emptyvar+set}>"
<set>
$ echo "<${unsetvar+set}>"
<>
To sum it all up: There is no real null value in bash. Chepner's comment is on point:
The bash documentation uses null as a synonym for the empty string.
Therefore, checking for null would mean checking for an empty string:
if [ "${lastUpdated}" = "" ]; then
# $lastUpdated is an empty string
fi
If what you really want to do is check for an unset or empty (i.e. "", i.e. 'null') variable, use trojanfoe's approach:
if [ -z "$lastUpdated" ]; then
# $lastUpdated could be "" or not set at all
fi
If you want to check weather the variable is unset, but are fine with empty strings, Gordon Davisson's answer is the way to go:
if [ -z ${lastUpdated+set} ]; then
# $lastUpdated is not set
fi
(Parameter Expansion is what's going here)

What does the following line of a bash script do?

Usually work in Windows, but trying to setup RabbitMQ on my Mac. Can someone let me know what the line below does?
[ "x" = "x$RABBITMQ_NODE_IP_ADDRESS" ] && [ "x" != "x$NODE_IP_ADDRESS" ] && RABBITMQ_NODE_IP_ADDRESS=${NODE_IP_ADDRESS}
Specifically, I'm curious about the [ "x" = "x$RAB..."] syntax.
If the RABBITMQ_NODE_IP_ADDRESS variable is empty/doesn't exist, it'll evaluate as "x" = "x" , which is true.
So it basically says, if RABBITMQ_NODE_IP_ADDRESS isn't set and NODE_IP_ADDRESS is set, set RABBITMQ_NODE_IP_ADDRESS=NODE_IP_ADDRESS
The "x" is used (somewhat superstitiously*) to prevent errors if the variable is null or unset. Most of the time the quotes take care of that for you. By putting the literal first and the variable second you eliminate errors in cases where the variable contains a string that starts with a dash, since test (aka [) would think it is an operator. In the case of your example, it would be preferable to use the -z and -n operators that test whether a variable is empty (null or unset) or not empty, respectively.
POSIX shells, such as Bourne (works in Bash, too):
[ -z $RABBITMQ_NODE_IP_ADDRESS ] && [ -n $NODE_IP_ADDRESS" ] && RABBITMQ_NODE_IP_ADDRESS=${NODE_IP_ADDRESS}
Bash (and ksh and zsh):
[[ -z $RABBITMQ_NODE_IP_ADDRESS && -n $NODE_IP_ADDRESS" ]] && RABBITMQ_NODE_IP_ADDRESS=${NODE_IP_ADDRESS}
* There may be some shells that need the "x", but some people do that "because it's always been done that way".
The "x" is not always superstitious, even in my relatively new bash (4.0.33).
Let's put the operation between parens. Empty variables are fine:
$ a=""
$ b=""
$ if [ '(' "$a" = "$b" ')' ]; then echo both_equal; fi
both_equal
But the ! operator for instance is not:
$ a='!'
$ if [ '(' "$a" = "$b" ')' ]; then echo both_equal; fi
bash: [: `)' expected, found
This is not a problem if we write "x$a" = "x$b" instead of "$a" = "$b".
The bracket [ is a test operator, which you can think of as an if statement. This is checking to see if the shell variable RABBITMQ_NODE_IP_ADDRESS is empty. Unfortunately, if you try to compare to an empty string "", the shell eliminates it before it does the test and your binary comparison operator only gets one (or maybe zero) operands. To prevent that error, it is a common practice to concatenate an "x" on each side of the =. Thus, instead of
[ "" = "<variable>" ]
becoming
[ = value ]
and yielding an error,
[ "X" = "X<variable>" ]
becomes
[ X = Xvalue ]
and the comparison may continue

Assigning default values to shell variables with a single command in bash

I have a whole bunch of tests on variables in a bash (3.00) shell script where if the variable is not set, then it assigns a default, e.g.:
if [ -z "${VARIABLE}" ]; then
FOO='default'
else
FOO=${VARIABLE}
fi
I seem to recall there's some syntax to doing this in one line, something resembling a ternary operator, e.g.:
FOO=${ ${VARIABLE} : 'default' }
(though I know that won't work...)
Am I crazy, or does something like that exist?
Very close to what you posted, actually. You can use something called Bash parameter expansion to accomplish this.
To get the assigned value, or default if it's missing:
FOO="${VARIABLE:-default}" # If variable not set or null, use default.
# If VARIABLE was unset or null, it still is after this (no assignment done).
Or to assign default to VARIABLE at the same time:
FOO="${VARIABLE:=default}" # If variable not set or null, set it to default.
For command line arguments:
VARIABLE="${1:-$DEFAULTVALUE}"
which assigns to VARIABLE the value of the 1st argument passed to the script or the value of DEFAULTVALUE if no such argument was passed. Quoting prevents globbing and word splitting.
If the variable is same, then
: "${VARIABLE:=DEFAULT_VALUE}"
assigns DEFAULT_VALUE to VARIABLE if not defined.
The colon builtin (:) ensures the variable result is not executed
The double quotes (") prevent globbing and word splitting.
Also see Section 3.5.3, Shell Parameter Expansion, in the Bash manual.
To answer your question and on all variable substitutions
echo "${var}"
echo "Substitute the value of var."
echo "${var:-word}"
echo "If var is null or unset, word is substituted for var. The value of var does not change."
echo "${var:=word}"
echo "If var is null or unset, var is set to the value of word."
echo "${var:?message}"
echo "If var is null or unset, message is printed to standard error. This checks that variables are set correctly."
echo "${var:+word}"
echo "If var is set, word is substituted for var. The value of var does not change."
You can escape the whole expression by putting a \ between the dollar sign and the rest of the expression.
echo "$\{var}"
Even you can use like default value the value of another variable
having a file defvalue.sh
#!/bin/bash
variable1=$1
variable2=${2:-$variable1}
echo $variable1
echo $variable2
run ./defvalue.sh first-value second-value output
first-value
second-value
and run ./defvalue.sh first-value output
first-value
first-value
see here under 3.5.3(shell parameter expansion)
so in your case
${VARIABLE:-default}
FWIW, you can provide an error message like so:
USERNAME=${1:?"Specify a username"}
This displays a message like this and exits with code 1:
./myscript.sh
./myscript.sh: line 2: 1: Specify a username
A more complete example of everything:
#!/bin/bash
ACTION=${1:?"Specify 'action' as argv[1]"}
DIRNAME=${2:-$PWD}
OUTPUT_DIR=${3:-${HOMEDIR:-"/tmp"}}
echo "$ACTION"
echo "$DIRNAME"
echo "$OUTPUT_DIR"
Output:
$ ./script.sh foo
foo
/path/to/pwd
/tmp
$ export HOMEDIR=/home/myuser
$ ./script.sh foo
foo
/path/to/pwd
/home/myuser
$ACTION takes the value of the first argument, and exits if empty
$DIRNAME is the 2nd argument, and defaults to the current directory
$OUTPUT_DIR is the 3rd argument, or $HOMEDIR (if defined), else, /tmp. This works on OS X, but I'm not positive that it's portable.
Then there's the way of expressing your 'if' construct more tersely:
FOO='default'
[ -n "${VARIABLE}" ] && FOO=${VARIABLE}
It is possible to chain default values like so:
DOCKER_LABEL=${GIT_TAG:-${GIT_COMMIT_AND_DATE:-latest}}
i.e. if $GIT_TAG doesnt exist, take $GIT_COMMIT_AND_DATE - if this doesnt exist, take "latest"
Here is an example
#!/bin/bash
default='default_value'
value=${1:-$default}
echo "value: [$value]"
save this as script.sh and make it executable.
run it without params
./script.sh
> value: [default_value]
run it with param
./script.sh my_value
> value: [my_value]
If you want 1 liner for your if-then-else, then we can consider rewriting:
if [ -z "${VARIABLE}" ]; then
FOO='default'
else
FOO=${VARIABLE}
fi
with semicolons:
if [ -z ${VARIABLE} ]; then FOO=`default`; else FOO=${VARIABLE}; fi
Alternatively, you can drop the if-then-else-fi keywords if you use boolean operators such as:
[ -z "${VARIABLE}" ] && FOO='default' || FOO=${VARIABLE}
Generalizing, the boolean operator pattern is:
conditional && then_command || else_command

Test for a Bash variable being unset, using a function

A simple Bash variable test goes:
${varName:? "${varName} is not defined"}
I'd like to reuse this, by putting it in a function. How can I do it?
The following fails
#
# Test a variable exists
tvar(){
val=${1:? "${1} must be defined, preferably in $basedir"}
if [ -z ${val} ]
then
echo Zero length value
else
echo ${1} exists, value ${1}
fi
}
I.e., I need to exit if the test fails.
Thanks to lhunath's answer, I was led to a part of the Bash man page that I've overlooked hundreds of times:
When not performing substring expansion, bash tests for a parameter that is unset or null; omitting the colon results in a test only for a parameter that is unset.
This prompted me to create the following truth table:
Unset
Set, but null
Set and not null
Meaning
${var-_}
T
F
T
Not null or not set
${var:-_}
T
T
T
Always true, use for subst.
$var
F
F
T
'var' is set and not null
${!var[#]}
F
T
T
'var' is set
This table introduces the specification in the last row. The Bash man page says "If name is not an array, expands to 0 if name is set and null otherwise." For purposes of this truth table, it behaves the same even if it's an array.
You're looking for indirection.
assertNotEmpty() {
: "${!1:? "$1 is empty, aborting."}"
}
That causes the script to abort with an error message if you do something like this:
$ foo=""
$ assertNotEmpty foo
bash: !1: foo is empty, aborting.
If you just want to test whether foo is empty, instead of aborting the script, use this instead of a function:
[[ $foo ]]
For example:
until read -p "What is your name? " name && [[ $name ]]; do
echo "You didn't enter your name. Please, try again." >&2
done
Also, note that there is a very important difference between an empty and an unset parameter. You should take care not to confuse these terms! An empty parameter is one that is set, but just set to an empty string. An unset parameter is one that doesn't exist at all.
The previous examples all test for empty parameters. If you want to test for unset parameters and consider all set parameters OK, whether they're empty or not, use this:
[[ ! $foo && ${foo-_} ]]
Use it in a function like this:
assertIsSet() {
[[ ! ${!1} && ${!1-_} ]] && {
echo "$1 is not set, aborting." >&2
exit 1
}
}
Which only aborts the script when the parameter name you pass denotes a parameter that isn't set:
$ ( foo="blah"; assertIsSet foo; echo "Still running." )
Still running.
$ ( foo=""; assertIsSet foo; echo "Still running." )
Still running.
$ ( unset foo; assertIsSet foo; echo "Still running." )
foo is not set, aborting.
You want to use [ -z ${parameter+word} ]
Some part of man bash:
Parameter Expansion
...
In each of the cases below, word is subject to tilde expansion, parameter expansion, command substitution, and
arithmetic expansion. When not performing substring expansion, bash tests for a parameter that is unset or null;
omitting the colon results in a test only for a parameter that is unset.
...
${parameter:+word}
Use Alternate Value. If parameter is null or unset, nothing is substituted, otherwise the expansion of
word is substituted.
...
in other words:
${parameter+word}
Use Alternate Value. If parameter is unset, nothing is substituted, otherwise the expansion of
word is substituted.
some examples:
$ set | grep FOOBAR
$ if [ -z "${FOOBAR+something}" ]; then echo "it is unset"; fi
it is unset
$ declare FOOBAR
$ if [ -z "${FOOBAR+something}" ]; then echo "it is unset"; fi
$ FOOBAR=
$ if [ -z "${FOOBAR+something}" ]; then echo "it is unset"; fi
$ FOOBAR=1
$ if [ -z "${FOOBAR+something}" ]; then echo "it is unset"; fi
$ unset FOOBAR
$ if [ -z "${FOOBAR+something}" ]; then echo "it is unset"; fi
it is unset
$
This function tests for variables that are currently set. The variable may even be an array. Note that in Bash: 0 == TRUE, 1 == FALSE.
function var.defined {
eval '[[ ${!'$1'[#]} ]]'
}
# Typical usage of var.defined {}
declare you="Your Name Here" ref='you';
read -p "What's your name: " you;
if var.defined you; then # Simple demo using literal text
echo "BASH recognizes $you";
echo "BASH also knows a reference to $ref as ${!ref}, by indirection.";
fi
unset you # Have just been killed by a master :D
if ! var.defined $ref; then # Standard demo using an expanded literal value
echo "BASH doesn't know $ref any longer";
fi
read -s -N 1 -p "Press any key to continue...";
echo "";
So to be clear here, the function tests literal text. Every time a command is called in Bash, variables are generally 'swapped-out' or 'substituted' with the underlying value unless:
$varRef ($) is escaped: $varRef
$varRef is single quoted '$varRef'
I.e., I need to exit if the test fails.
The code:
${varName:? "${varName} is not defined"}
will return a nonzero exit code when there is not a variable named "varName". The exit code of the last command is saved in $?.
About your code:
val=${1:? "${1} must be defined, preferably in $basedir"}
Maybe it is not doing what you need. In the case that $1 is not defined, the "${1}" will be substituted with nothing. Probably you want use the single quotes that literally writes ${1} without substitution.
val=${1:? '${1} must be defined, preferably in $basedir'
I am unsure if this is exactly what you want, but a handy trick I use when writing a new and complex script is to use "set -o":
set -o # Will make the script bomb out when it finds an unset variable
For example,
$ grep '$1' chex.sh
case "$1" in
$ ./chex.sh
./chex.sh: line 111: $1: unbound variable
$ ./chex.sh foo
incorrect/no options passed.. exiting
if set | grep -q '^VARIABLE='
then
echo VARIABLE is set
fi

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