vars="a,b"
a="True"
b="False"
IFS=","
for var in $vars; do
if [[ "$var" = "True" ]]; then
echo "True found"
fi
done
I would expect the above bash script to print out "True found". But it does not print anything. Any ideas as to why ?
Make this:
if [[ "${!var}" = "True" ]]; then
${!varname} expands the variable named in $varname. Otherwise, you get the name itself, not the contents of the variable with that name.
See BashFAQ #6 for far more details.
Related
Here is my code
vmname="$1"
EXCEPTLIST="desktop-01|desktop-02|desktop-03|desktop-04"
if [[ $vmname != #(${EXCEPTLIST}) ]]; then
echo "${vmname}"
else
echo "Its in the exceptlist"
fi
The above code works perfectly but my question is , the EXCEPTLIST can be a long line, say 100 server names. In that case its hard to put all that names in one line. In that situation is there any way to make the variable EXCEPTLIST to be a multiline variable ? something like as follows:
EXCEPTLIST="desktop-01|desktop-02|desktop-03| \n
desktop-04|desktop-05|desktop-06| \n
desktop-07|desktop-08"
I am not sure but was thinking of possibilities.
Apparently I would like to know the terminology of using #(${})- Is this called variable expansion or what ? Does anyone know the documentation/explain to me about how this works in bash. ?
One can declare an array if the data/string is long/large. Use IFS and printf for the format string, something like:
#!/usr/bin/env bash
exceptlist=(
desktop-01
desktop-02
desktop-03
desktop-04
desktop-05
desktop-06
)
pattern=$(IFS='|'; printf '#(%s)' "${exceptlist[*]}")
[[ "$vmname" != $pattern ]] && echo good
In that situation is there any way to make the variable EXCEPTLIST to be a multiline variable ?
With your given input/data an array is also a best option, something like:
exceptlist=(
'desktop-01|desktop-02|desktop-03'
'desktop-04|desktop-05|desktop-06'
'desktop-07|desktop-08'
)
Check what is the value of $pattern variable one way is:
declare -p pattern
Output:
declare -- pattern="#(desktop-01|desktop-02|desktop-03|desktop-04|desktop-05|desktop-06)"
Need to test/check if $vmname is an empty string too, since it will always be true.
On a side note, don't use all upper case variables for purely internal purposes.
The $(...) is called Command Substitution.
See LESS=+'/\ *Command Substitution' man bash
In addition to what was mentioned in the comments about pattern matching
See LESS=+/'(pattern-list)' man bash
See LESS=+/' *\[\[ expression' man bash
s there any way to make the variable EXCEPTLIST to be a multiline variable ?
I see no reason to use matching. Use a bash array and just compare.
exceptlist=(
desktop-01
desktop-02
desktop-03
desktop-04
desktop-05
desktop-06
)
is_in_list() {
local i
for i in "${#:2}"; do
if [[ "$1" = "$i" ]]; then
return 0
fi
done
return 1
}
if is_in_list "$vmname" "${EXCEPTLIST[#]}"; then
echo "is in exception list ${vmname}"
fi
#(${})- Is this called variable expansion or what ? Does anyone know the documentation/explain to me about how this works in bash. ?
${var} is a variable expansion.
#(...) are just characters # ( ).
From man bash in Compund commands:
[[ expression ]]
When the == and != operators are used, the string to the right of the operator is considered a pattern and matched according to the rules
described below under Pattern Matching, as if the extglob shell option were enabled. ...
From Pattern Matching in man bash:
#(pattern-list)
Matches one of the given patterns
[[ command receives the #(a|b|c) string and then matches the arguments.
There is absolutely no need to use Bash specific regex or arrays and loop for a match, if using grep for raw string on word boundary.
The exception list can be multi-line, it will work as well:
#!/usr/bin/sh
exceptlist='
desktop-01|desktop-02|desktop-03|
deskop-04|desktop-05|desktop-06|
desktop-07|deskop-08'
if printf %s "$exceptlist" | grep -qwF "$1"; then
printf '%s is in the exceptlist\n' "$1"
fi
I wouldn't bother with multiple lines of text. This is would be just fine:
EXCEPTLIST='desktop-01|desktop-02|desktop-03|'
EXCEPTLIST+='desktop-04|desktop-05|desktop-06|'
EXCEPTLIST+='desktop-07|desktop-08'
The #(...) construct is called extended globbing pattern and what it does is an extension of what you probably already know -- wildcards:
VAR='foobar'
if [[ "$VAR" == fo?b* ]]; then
echo "Yes!"
else
echo "No!"
fi
A quick walkthrough on extended globbing examples: https://www.linuxjournal.com/content/bash-extended-globbing
#!/bin/bash
set +o posix
shopt -s extglob
vmname=$1
EXCEPTLIST=(
desktop-01 desktop-02 desktop-03
...
)
if IFS='|' eval '[[ ${vmname} == #(${EXCEPTLIST[*]}) ]]'; then
...
Here's one way to load a multiline string into a variable:
fn() {
cat <<EOF
desktop-01|desktop-02|desktop-03|
desktop-04|desktop-05|desktop-06|
desktop-07|desktop-08
EOF
}
exceptlist="$(fn)"
echo $exceptlist
As to solving your specific problem, I can think of a variety of approaches.
Solution 1, since all the desktop has the same desktop-0 prefix and only differ in the last letter, we can make use of {,} or {..} expansion as follows:
vmname="$1"
found=0
for d in desktop-{01..08}
do
if [[ "$vmname" == $d ]]; then
echo "It's in the exceptlist"
found=1
break
fi
done
if (( !found )); then
echo "Not found"
fi
Solution 2, sometimes, it is good to provide a list in a maintainable clear text list. We can use a while loop and iterate through the list
vmname="$1"
found=0
while IFS= read -r d
do
if [[ "$vmname" == $d ]]; then
echo "It's in the exceptlist"
found=1
break
fi
done <<EOF
desktop-01
desktop-02
desktop-03
desktop-04
desktop-05
desktop-06
desktop-07
desktop-08
EOF
if (( !found )); then
echo "Not found"
fi
Solution 3, we can desktop the servers using regular expressions:
vmname="$1"
if [[ "$vmname" =~ ^desktop-0[1-8]$ ]]; then
echo "It's in the exceptlist"
else
echo "Not found"
fi
Solution 4, we populate an array, then iterate through an array:
vmname="$1"
exceptlist=()
exceptlist+=(desktop-01 desktop-02 desktop-03 deskop-04)
exceptlist+=(desktop-05 desktop-06 desktop-07 deskop-08)
found=0
for d in ${exceptlist[#]}
do
if [[ "$vmname" == "$d" ]]; then
echo "It's in the exceptlist"
found=1
break;
fi
done
if (( !found )); then
echo "Not found"
fi
Please make me understand this code.
Is there assignment happening or comparison in bash shell scripting?
if test x${property} = x; then
some condition
fi
It's never coming inside of this if block
Here actually you are checking whether the variable is empty or not. If that variable is emplty then if executes that block.
You can see the following explained.
> echo ${property}
> if test x${property} = x; then echo variable is empty;fi
variable is empty
> property='some data'
> if test "x${property}" = x; then echo variable is empty;fi
You can also check it with -z option.
if [ -z "${property}" ]; then
echo "variable is empty"
else
echo "variable is not empty"
fi
It's just a way of comparison that's been done in the old shells, which didn't have many options. It tests whether property is empty. In modern shells, such as ksh or bash, you would use -z and the equivalent condition would be:
if [[ -z ${property} ]]; then
some condition
fi
I have a large number of variables in my script, and I want the script to error out if any one of the variables are empty.
I know I can:
if [[ -z "$var_1" ]] || [[ -z "$var_2" ]] || ... [[ -z "$var_n" ]]; then
# failure message
fi
However, I cannot inform the user which variable was empty if I do it in this way. Is there an alternative approach to the above so that I can inform the user about the empty variable?
#!/bin/sh
foo=(var_1 var_2 var_n)
for bar in ${foo[*]}
do
if [[ ! ${!bar} ]]
then
echo $bar is empty
fi
done
Just use ${var:?var is empty or unset} the first time you reference the variable. If empty strings are acceptable and you only care if the variables are set, do ${var?var is unset}. Using ? in the parameter expansion causes the shell to terminate and if the variable is (empty or) unset.
I have written a function which checks whether one or more variables passed as arguments are empty (got the idea from here)
Here is what the script looks like:
#!/bin/bash
is_variable_defined?() {
for VAR in "$#"; do
: "${!VAR? "ERROR: $VAR is undefined."}"
done
}
TEST='a'
is_variable_defined? TEST TEST2
And here is the output:
/path/to/script.sh: line 4: !VAR: ERROR: TEST2 is undefined.
However what I would like to output is:
TEST2 is undefined
I have tried tweaking the : "${!VAR? "ERROR: $VAR is undefined."}" line but whatever I do it breaks.
Does anybody know what to modify in the script to get the output I want?
Are you seeing if the variable is undefined or simply a null string?
Null Test
is_variable_defined() {
for var in $#
do
if [ -z ${!var} ] # Or if [[ ${!var} ]]
then
echo "Var $var has no value"
fi
done
}
Undefined Test
is_variable_defined() {
for var in $#
do
if ! set | grep -q "^$var="
then
echo "Var $var is not set"
fi
done
The last is looking at the output of the set command to see if the variable is listed there. The -q will make grep quiet. And you look at the exit status to see if it found the string. If your grep doesn't support the -q parameter, you can try grep "^$var=" 2> /dev/null.
Using the [[ $foo ]] syntax for testing won't necessarily work. The below will print $foo is not set even though it was set to an empty string:
foo=""
if [[ $foo ]]
then
echo "\$foo = '$foo'"
else
echo "\$foo is not set"
fi
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