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)"
Related
I got a bash script , for example
[root#test ~]# cat 1.sh
#!/bin/bash
var1=""
var2=""
touch "$var1"
touch $var2
it output as:
[root#test ~]# bash -x 1.sh
+ var1=
+ var2=
+ touch ''
touch: cannot touch ‘’: No such file or directory
+ touch
touch: missing file operand
so if I use "$var" over command , when it is empty , it will become '' and breaks my command
while if I use $var without double quotes , it works but shell check keep complaint to double quote to prevent word split
how can I workaround this ?
-------------edit -------
the touch here is example , please don't mind the command here , the goal is to make it output nothing
e.g.
my_command $var1 $var2 $var3 $var4
these var can be empty , so if I double quoted it
my_command "$var1" "$var2" "$var3" "$var4"
let's say if var1 and 2 is empty , then it will run as
var1=""
var2=""
var3="some"
var4="thing"
my_command "$var1" "$var2" "$var3" "$var4"
it output as
my_command '' '' some thing , and broken due to '' in argument.
if I do
var1=""
var2=""
var3="some"
var4="thing"
my_command $var1 $var2 $var3 $var4
it goes as my_command some thing which just works , but shell check keep complaints :(
You can omit empty variables from an argument list like this:
my_command ${var1:+"$var1"} ${var2:+"$var2"} ${var3:+"$var3"} ${var4:+"$var4"}
Explanation: ${var:+something} expands to something if var is defined and non-empty. So ${var:+"$var"} expands to a properly double-quoted reference to $var if the variable isn't empty. If it's empty or undefined, the whole thing expands to nothing, and since the outer reference isn't double-quoted it gets eliminated by word splitting.
When not quoted, empty string var=""; touch $var expands to nothing (as if you did not add it).
But when quoted var=""; touch "$var" then it expands to the empty string from var and this is an invalid filename argument to touch.
how can I workaround this ?
There is nothing to workaround as it works as expected. The presented command touch "$var" works as expected. When var is empty, one empty argument is being passed to touch command.
As creating a file with empty name is invalid, happily touch complains with a message.
the goal is to make it output nothing
Then check if the argument to touch is non-empty before running it.
if [ -n "$var" ]; then
somecommand "$var"
fi
If you want to omit an argument from a list if it's empty, use bash array to accumulate existing arguments and pass them properly quoted to the command:
args=()
if [[ -n "$var1" ]]; then
args+=("$var1")
fi
if [[ -n "$var2" ]]; then
args+=("$var2")
fi
somecommand "${args[#]}"
The POSIX compatible alternatives to bash arrays is to use set -- "$#" "$var" posititional arguments to accumulate the arguments or str+="$(printf " %q" "$var") properly quoted string to be evalulated.
If you want to "make it output nothing" literally, then silence stdout and stderr of a command, typically by redirecting to /dev/null:
{ somecommand ...; } >/dev/null 2>&1
You should check the validity of your variables before trying to use them. For example, you can use if [[ -z $VAR ]]:
if [[ -z $var1 || -z $var2 || -z $var3 || -z $var4 ]]; then
echo "Error: argument cannot be empty"
exit 1
fi
my_command "$var1" "$var2" "$var3" "$var4"
You can also split it into various ifs in case you want different messages for each one, for example.
I am trying to understand a test script, which includes the following segment:
SCRIPT_PATH=${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 "${var%/*}"
foo/bar
I'm trying to create a Bash script that will extract the last parameter given from the command line into a variable to be used elsewhere. Here's the script I'm working on:
#!/bin/bash
# compact - archive and compact file/folder(s)
eval LAST=\$$#
FILES="$#"
NAME=$LAST
# Usage - display usage if no parameters are given
if [[ -z $NAME ]]; then
echo "compact <file> <folder>... <compressed-name>.tar.gz"
exit
fi
# Check if an archive name has been given
if [[ -f $NAME ]]; then
echo "File exists or you forgot to enter a filename. Exiting."
exit
fi
tar -czvpf "$NAME".tar.gz $FILES
Since the first parameters could be of any number, I have to find a way to extract the last parameter, (e.g. compact file.a file.b file.d files-a-b-d.tar.gz). As it is now the archive name will be included in the files to compact. Is there a way to do this?
To remove the last item from the array you could use something like this:
#!/bin/bash
length=$(($#-1))
array=${#:1:$length}
echo $array
Even shorter way:
array=${#:1:$#-1}
But arays are a Bashism, try avoid using them :(.
Portable and compact solutions
This is how I do in my scripts
last=${#:$#} # last parameter
other=${*%${!#}} # all parameters except the last
EDIT
According to some comments (see below), this solution is more portable than others.
Please read Michael Dimmitt's commentary for an explanation of how it works.
last_arg="${!#}"
Several solutions have already been posted; however I would advise restructuring your script so that the archive name is the first parameter rather than the last. Then it's really simple, since you can use the shift builtin to remove the first parameter:
ARCHIVENAME="$1"
shift
# Now "$#" contains all of the arguments except for the first
Thanks guys, got it done, heres the final bash script:
#!/bin/bash
# compact - archive and compress file/folder(s)
# Extract archive filename for variable
ARCHIVENAME="${!#}"
# Remove archive filename for file/folder list to backup
length=$(($#-1))
FILES=${#:1:$length}
# Usage - display usage if no parameters are given
if [[ -z $# ]]; then
echo "compact <file> <folder>... <compressed-name>.tar.gz"
exit
fi
# Tar the files, name archive after last file/folder if no name given
if [[ ! -f $ARCHIVENAME ]]; then
tar -czvpf "$ARCHIVENAME".tar.gz $FILES; else
tar -czvpf "$ARCHIVENAME".tar.gz "$#"
fi
Just dropping the length variable used in Krzysztof Klimonda's solution:
(
set -- 1 2 3 4 5
echo "${#:1:($#-1)}" # 1 2 3 4
echo "${#:(-$#):($#-1)}" # 1 2 3 4
)
I would add this as a comment, but don't have enough reputation and the answer got a bit longer anyway. Hope it doesn't mind.
As #func stated:
last_arg="${!#}"
How it works:
${!PARAM} indicates level of indirection. You are not referencing PARAM itself, but the value stored in PARAM ( think of PARAM as pointer to value ).
${#} expands to the number of parameters (Note: $0 - the script name - is not counted here).
Consider following execution:
$./myscript.sh p1 p2 p3
And in the myscript.sh
#!/bin/bash
echo "Number of params: ${#}" # 3
echo "Last parameter using '\${!#}': ${!#}" # p3
echo "Last parameter by evaluating positional value: $(eval LASTP='$'${#} ; echo $LASTP)" # p3
Hence you can think of ${!#} as a shortcut for the above eval usage, which does exactly the approach described above - evaluates the value stored in the given parameter, here the parameter is 3 and holds the positional argument $3
Now if you want all the params except the last one, you can use substring removal ${PARAM%PATTERN} where % sign means 'remove the shortest matching pattern from the end of the string'.
Hence in our script:
echo "Every parameter except the last one: ${*%${!#}}"
You can read something in here: Parameter expansion
Are you sure this fancy script is any better than a simple alias to tar?
alias compact="tar -czvpf"
Usage is:
compact ARCHIVENAME FILES...
Where FILES can be file1 file2 or globs like *.html
Try:
if [ "$#" -gt '0' ]; then
/bin/echo "${!#}" "${#:1:$(($# - 1))}
fi
Array without last parameter:
array=${#:1:$#-1}
But it's a bashism :(. Proper solutions would involve shift and adding into variable as others use.
#!/bin/bash
lastidx=$#
lastidx=`expr $lastidx - 1`
eval last='$'{$lastidx}
echo $last
Alternative way to pull the last parameter out of the argument list:
eval last="\$$#"
eval set -- `awk 'BEGIN{for(i=1;i<'$#';i++) printf " \"$%d\"",i;}'`
#!/bin/sh
eval last='$'$#
while test $# -gt 1; do
list="$list $1"
shift
done
echo $list $last
I can't find a way to use array-subscript notation on $#, so this is the best I can do:
#!/bin/bash
args=("$#")
echo "${args[$(($#-1))]}"
This script may work for you - it returns a subrange of the arguments, and can be called from another script.
Examples of it running:
$ args_get_range 2 -2 y a b "c 1" d e f g
'b' 'c 1' 'd' 'e'
$ args_get_range 1 2 n arg1 arg2
arg1 arg2
$ args_get_range 2 -2 y arg1 arg2 arg3 "arg 4" arg5
'arg2' 'arg3'
$ args_get_range 2 -1 y arg1 arg2 arg3 "arg 4" arg5
'arg2' 'arg3' 'arg 4'
# You could use this in another script of course
# by calling it like so, which puts all
# args except the last one into a new variable
# called NEW_ARGS
NEW_ARGS=$(args_get_range 1 -1 y "$#")
args_get_range.sh
#!/usr/bin/env bash
function show_help()
{
IT="
Extracts a range of arguments from passed in args
and returns them quoted or not quoted.
usage: START END QUOTED ARG1 {ARG2} ...
e.g.
# extract args 2-3
$ args_get_range.sh 2 3 n arg1 arg2 arg3
arg2 arg3
# extract all args from 2 to one before the last argument
$ args_get_range.sh 2 -1 n arg1 arg2 arg3 arg4 arg5
arg2 arg3 arg4
# extract all args from 2 to 3, quoting them in the response
$ args_get_range.sh 2 3 y arg1 arg2 arg3 arg4 arg5
'arg2' 'arg3'
# You could use this in another script of course
# by calling it like so, which puts all
# args except the last one into a new variable
# called NEW_ARGS
NEW_ARGS=\$(args_get_range.sh 1 -1 \"\$#\")
"
echo "$IT"
exit
}
if [ "$1" == "help" ]
then
show_help
fi
if [ $# -lt 3 ]
then
show_help
fi
START=$1
END=$2
QUOTED=$3
shift;
shift;
shift;
if [ $# -eq 0 ]
then
echo "Please supply a folder name"
exit;
fi
# If end is a negative, it means relative
# to the last argument.
if [ $END -lt 0 ]
then
END=$(($#+$END))
fi
ARGS=""
COUNT=$(($START-1))
for i in "${#:$START}"
do
COUNT=$((COUNT+1))
if [ "$QUOTED" == "y" ]
then
ARGS="$ARGS '$i'"
else
ARGS="$ARGS $i"
fi
if [ $COUNT -eq $END ]
then
echo $ARGS
exit;
fi
done
echo $ARGS
This works for me, with sh and bash:
last=${*##* }
others=${*%${*##* }}
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
how can I know the number of tokens in a bash variable (whitespace-separated tokens) - or at least, wether it is one or there are more.
The $# expansion will tell you the number of elements in a variable / array. If you're working with a bash version greater than 2.05 or so you can:
VAR='some string with words'
VAR=( $VAR )
echo ${#VAR[#]}
This effectively splits the string into an array along whitespace (which is the default delimiter), and then counts the members of the array.
EDIT:
Of course, this recasts the variable as an array. If you don't want that, use a different variable name or recast the variable back into a string:
VAR="${VAR[*]}"
I can't understand why people are using those overcomplicated bashisms all the time. There's almost always a straight-forward, no-bashism solution.
howmany() { echo $#; }
myvar="I am your var"
howmany $myvar
This uses the tokenizer built-in to the shell, so there's no discrepancy.
Here's one related gotcha:
myvar='*'
echo $myvar
echo "$myvar"
set -f
echo $myvar
echo "$myvar"
Note that the solution from #guns using bash array has the same gotcha.
The following is a (supposedly) super-robust version to work around the gotcha:
howmany() ( set -f; set -- $1; echo $# )
If we want to avoid the subshell, things start to get ugly
howmany() {
case $- in *f*) set -- $1;; *) set -f; set -- $1; set +f;; esac
echo $#
}
These two must be used WITH quotes, e.g. howmany "one two three" returns 3
set VAR='hello world'
echo $VAR | wc -w
here is how you can check.
if [ `echo $VAR | wc -w` -gt 1 ]
then
echo "Hello"
fi
Simple method:
$ VAR="a b c d"
$ set $VAR
$ echo $#
4
To count:
sentence="This is a sentence, please count the words in me."
words="${sentence//[^\ ]} "
echo ${#words}
To check:
sentence1="Two words"
sentence2="One"
[[ "$sentence1" =~ [\ ] ]] && echo "sentence1 has more than one word"
[[ "$sentence2" =~ [\ ] ]] && echo "sentence2 has more than one word"
For a robust, portable sh solution, see #JoSo's functions using set -f.
(Simple bash-only solution for answering (only) the "Is there at least 1 whitespace?" question; note: will also match leading and trailing whitespace, unlike the awk solution below:
[[ $v =~ [[:space:]] ]] && echo "\$v has at least 1 whitespace char."
)
Here's a robust awk-based bash solution (less efficient due to invocation of an external utility, but probably won't matter in many real-world scenarios):
# Functions - pass in a quoted variable reference as the only argument.
# Takes advantage of `awk` splitting each input line into individual tokens by
# whitespace; `NF` represents the number of tokens.
# `-v RS=$'\3'` ensures that even multiline input is treated as a single input
# string.
countTokens() { awk -v RS=$'\3' '{print NF}' <<<"$1"; }
hasMultipleTokens() { awk -v RS=$'\3' '{if(NF>1) ec=0; else ec=1; exit ec}' <<<"$1"; }
# Example: Note the use of glob `*` to demonstrate that it is not
# accidentally expanded.
v='I am *'
echo "\$v has $(countTokens "$v") token(s)."
if hasMultipleTokens "$v"; then
echo "\$v has multiple tokens."
else
echo "\$v has just 1 token."
fi
Not sure if this is exactly what you meant but:
$# = Number of arguments passed to the bash script
Otherwise you might be looking for something like man wc