I have something like that:
...
args=$*
echo $args
...
result is
unusable1 unusable2 useful useful ... useful unusable3
I need remove all "unusable" args. They always at first, second and last position.
After some investigation i find ${*:3} bash syntax. It help remove first two.
...
args=${*:3}
echo $args
...
result is
useful useful ... useful unusable3
But I can't find how to remove last word using same nice syntax.
You can use a function/script like this to print all but last arguments:
func() {
echo "${#:1:$#-1}";
}
func aa bb cc dd ee
aa bb cc dd
func foo bar baz hello how are you
foo bar baz hello how are
args=${*:3} flattens your argument list. You don't want to do that. Consider following the pattern given below instead:
# this next line sets "$#" for testing purposes; you don't need it in real life
set -- \
"first argument" \
"second argument" \
"third argument" \
"fourth argument" \
"fifth argument"
# trim the first two
args=( "${#:2}" )
# trim the last one
args=( "${args[#]:1:$(( ${#args[#]} - 2 ))}" )
# demonstrate the output content
printf '<%s>\n' "${args[#]}"
Running the above yields the following output:
<third argument>
<fourth argument>
...and, by doing so, demonstrates that it's correctly keeping arguments together, even when they contain spaces or wildcard characters.
For a shell completion script, you might also consider:
printf '%q ' "${args[#]}"
...which quotes content in such a way as to be eval'able by the shell.
Using your syntax, you can use this :
args=${*:3:$#-3}
Explanation :
${*:offset:length}
offset is 3 to begin to the third argument and length is the number of arguments minus 3 (two first and last one).
You could use awk:
args="unusable1 unusable2 useful useful ... useful unusable3"
args=$(awk '{$1=$2=$NF="";print}' <<< "$args")
echo "$args"
Output:
useful useful ... useful
The command sets the the first, second and the last ($NF) position to an empty string. NF holds the number of fields in awk. Therefore $NF is the last column.
bash doesn't really provide such filtering methods for arrays. The best option is to just use a loop and filter one at a time.
for arg; do # Implicitly iterate over $#
[[ $arg =~ unusable ]] && continue
args+=( "$arg" )
done
Try with this:
args=$*
useful=${args#* }
useful=${useful#* }
useful=${useful% *}
Then you wlill get the interesting parameters in the $result variable.
Related
I have to write a simple bash script for my programming class. The idea is to use a for loop with $* (names of Files as Command Line Parameters). The task is to reverse and print out the Command Line parameters while still using the for inFile in $*; do loop.
I have no idea how to do this.
#!/bin/bash
for inFile in $*;do
echo $inFile
done
I know this doesn't work it just prints out the command line parameters in order.
The idea to loop over $* to reverse command line arguments is broken,
when any command line argument contains a white space. For example when the command line arguments are foo and "bar baz", the output of the script in the question will be:
foo
bar
baz
When the correct output should be:
foo
bar baz
The exact wording of the task is important.
For example, if the task is to print the arguments in reverse, and it doesn't mention $*, then you can use a counting loop in reverse, and ${!i} to expand to the value of the numbered positional parameters:
# nice clean solution
for ((i = $#; i > 0; i--)); do
echo "${!i}"
done
Another example, if the task insists that you must use $* and accepts that the command line arguments will only have supported characters, then you could collect the parameters into an array, and then print the content of the array in reverse, again with a counting loop:
args=()
# not recommended, unsafe due to shell expansion of $*
for arg in $*; do
args+=("$arg")
done
for ((i = ${#args[#]} - 1; i >= 0; i--)); do
echo "${args[i]}"
done
If you are not allowed to use arrays, then you can prepend values to a string, and then iterate over that string:
# dirtiest solution, with unsafe expansions and unquoted variables, not recommended
args=
for arg in $*; do
args="$arg $args"
done
for arg in $args; do
echo "$arg"
done
I have the following problem: (Its about dates)
The user will set the following variable.
variable1=33_2016
now I Somehow want to to automatically set a second variable which sets the "33" +1
that I get
variable2=34_2016
Thanks for any advice.
My first choice would be to break the first variable apart with read, then put the (updated) pieces back together.
IFS=_ read f1 f2 <<< "$variable1"
# Option 1
variable2=$((f1 + 1))_$f2
# Option 2
printf -v variable2 '%s_%s" "$((f1 + 1))" "$f2"
You can also use parameter expansion to do the parsing:
f1=${variable%_*}
f2=${variable#*_}
You can also use a regular expression, which is more readable for parsing but much longer to put the pieces back together (BASH_REMATCH could use a shorter synonym).
[[ $variable1 =~ (.*)_(.*) ]] &&
f1=$((${BASH_REMATCH[1]}+1)) f2=${BASH_REMATCH[2]}
The first and third options also allow the possibility of working with an array:
# With read -a
IFS=_ read -a f <<< "$variable1"
variable2=$(IFS=_; echo "${f[*]}")
# With regular expression
[[ $variable1 =~ (.*)_(.*) ]]
variable2=$(IFS=_; echo "${BASH_REMATCH[*]:1:2}")
You can use awk:
awk 'BEGIN{FS=OFS="_"}{$1+=1}1' <<< "${variable1}"
While this needs an external process to spawn (a bit slower) it's easier to read/write. Decide for yourself what is more important for you here.
To store the return value in a variable, use command substitution:
variable2=$(awk 'BEGIN{FS=OFS="_"}{$1+=1}1' <<< "${variable1}")
You can do somewhat the same thing with parameter expansion with substring substitution, e.g.
$ v1=33_2016
$ v2=${v1/${v1%_*}/$((${v1%_*}+1))}
$ echo $v2
34_2016
It's six to one, a half-dozen to another.
Or anything in shell script to implement the same thing?
I was doing an assignment that requires us to write a Bourne shell script that shows the last argument of a bunch, e.g.:
lastarg arg1 arg2 arg3 ..... argN
which would show:
argN
I was not sure if there's any equivalencies to hasNext in Java as it's easy to implement.
Sorry if I was rude and unclear.
#!/bin/bash
all=($#)
# to make things short:
# you can use what's in a variable as a variable name
last=$(( $# )) # get number of arguments
echo ${!last} # use that to get the last argument. notice the !
# while the number of arguments is not 0
# put what is in argument $1 into next
# move all arguments to the left
# $1=foo $2=bar $4=moo
# shift
# $1=bar $2=moo
while [ $# -ne 0 ]; do
next=$1
shift
echo $next
done
# but the problem was the last argument...
# all=($#): put all arguments into an array
# ${all[n]}: get argument number n
# $(( 1+2 )): do math
# ${#all[#]}: get the count of element in an array
echo -e "all:\t ${all[#]}"
echo -e "second:\t ${all[1]}"
echo -e "fifth:\t ${all[4]}"
echo -e "# of elements:\t ${#all[#]}"
echo -e "last element:\t ${all[ (( ${#all[#]} -1 )) ]}"
ok, last edit (omg :p)
$ sh unix-java-hasnext.sh one two three seventyfour sixtyeight
sixtyeight
one
two
three
seventyfour
sixtyeight
all: one two three seventyfour sixtyeight
second: two
fifth: sixtyeight
# of elements: 5
last element: sixtyeight
POSIX-based shell languages don't implement iterators.
The only things you have are for V in words ; do ... ; done or implementing the loop with while and manual stuff to update and test the loop variable.
This is not place to make wild guesses, still: Bash provide shift operator, for loop and more.
(If this is for argument processing you have getopt library. More info in Using getopts in bash shell script to get long and short command line options )
$1 is the first argument.
$# is all of them.
How can I find the last argument passed to a shell
script?
This is Bash-only:
echo "${#: -1}"
This is a bit of a hack:
for last; do true; done
echo $last
This one is also pretty portable (again, should work with bash, ksh and sh) and it doesn't shift the arguments, which could be nice.
It uses the fact that for implicitly loops over the arguments if you don't tell it what to loop over, and the fact that for loop variables aren't scoped: they keep the last value they were set to.
$ set quick brown fox jumps
$ echo ${*: -1:1} # last argument
jumps
$ echo ${*: -1} # or simply
jumps
$ echo ${*: -2:1} # next to last
fox
The space is necessary so that it doesn't get interpreted as a default value.
Note that this is bash-only.
The simplest answer for bash 3.0 or greater is
_last=${!#} # *indirect reference* to the $# variable
# or
_last=$BASH_ARGV # official built-in (but takes more typing :)
That's it.
$ cat lastarg
#!/bin/bash
# echo the last arg given:
_last=${!#}
echo $_last
_last=$BASH_ARGV
echo $_last
for x; do
echo $x
done
Output is:
$ lastarg 1 2 3 4 "5 6 7"
5 6 7
5 6 7
1
2
3
4
5 6 7
The following will work for you.
# is for array of arguments.
: means at
$# is the length of the array of arguments.
So the result is the last element:
${#:$#}
Example:
function afunction{
echo ${#:$#}
}
afunction -d -o local 50
#Outputs 50
Note that this is bash-only.
Use indexing combined with length of:
echo ${#:${##}}
Note that this is bash-only.
Found this when looking to separate the last argument from all the previous one(s).
Whilst some of the answers do get the last argument, they're not much help if you need all the other args as well. This works much better:
heads=${#:1:$#-1}
tail=${#:$#}
Note that this is bash-only.
This works in all POSIX-compatible shells:
eval last=\${$#}
Source: http://www.faqs.org/faqs/unix-faq/faq/part2/section-12.html
Here is mine solution:
pretty portable (all POSIX sh, bash, ksh, zsh) should work
does not shift original arguments (shifts a copy).
does not use evil eval
does not iterate through the whole list
does not use external tools
Code:
ntharg() {
shift $1
printf '%s\n' "$1"
}
LAST_ARG=`ntharg $# "$#"`
From oldest to newer solutions:
The most portable solution, even older sh (works with spaces and glob characters) (no loop, faster):
eval printf "'%s\n'" "\"\${$#}\""
Since version 2.01 of bash
$ set -- The quick brown fox jumps over the lazy dog
$ printf '%s\n' "${!#} ${#:(-1)} ${#: -1} ${#:~0} ${!#}"
dog dog dog dog dog
For ksh, zsh and bash:
$ printf '%s\n' "${#: -1} ${#:~0}" # the space beetwen `:`
# and `-1` is a must.
dog dog
And for "next to last":
$ printf '%s\n' "${#:~1:1}"
lazy
Using printf to workaround any issues with arguments that start with a dash (like -n).
For all shells and for older sh (works with spaces and glob characters) is:
$ set -- The quick brown fox jumps over the lazy dog "the * last argument"
$ eval printf "'%s\n'" "\"\${$#}\""
The last * argument
Or, if you want to set a last var:
$ eval last=\${$#}; printf '%s\n' "$last"
The last * argument
And for "next to last":
$ eval printf "'%s\n'" "\"\${$(($#-1))}\""
dog
If you are using Bash >= 3.0
echo ${BASH_ARGV[0]}
For bash, this comment suggested the very elegant:
echo "${#:$#}"
To silence shellcheck, use:
echo ${*:$#}
As a bonus, both also work in zsh.
shift `expr $# - 1`
echo "$1"
This shifts the arguments by the number of arguments minus 1, and returns the first (and only) remaining argument, which will be the last one.
I only tested in bash, but it should work in sh and ksh as well.
I found #AgileZebra's answer (plus #starfry's comment) the most useful, but it sets heads to a scalar. An array is probably more useful:
heads=( "${#: 1: $# - 1}" )
tail=${#:${##}}
Note that this is bash-only.
Edit: Removed unnecessary $(( )) according to #f-hauri's comment.
A solution using eval:
last=$(eval "echo \$$#")
echo $last
If you want to do it in a non-destructive way, one way is to pass all the arguments to a function and return the last one:
#!/bin/bash
last() {
if [[ $# -ne 0 ]] ; then
shift $(expr $# - 1)
echo "$1"
#else
#do something when no arguments
fi
}
lastvar=$(last "$#")
echo $lastvar
echo "$#"
pax> ./qq.sh 1 2 3 a b
b
1 2 3 a b
If you don't actually care about keeping the other arguments, you don't need it in a function but I have a hard time thinking of a situation where you would never want to keep the other arguments unless they've already been processed, in which case I'd use the process/shift/process/shift/... method of sequentially processing them.
I'm assuming here that you want to keep them because you haven't followed the sequential method. This method also handles the case where there's no arguments, returning "". You could easily adjust that behavior by inserting the commented-out else clause.
For tcsh:
set X = `echo $* | awk -F " " '{print $NF}'`
somecommand "$X"
I'm quite sure this would be a portable solution, except for the assignment.
After reading the answers above I wrote a Q&D shell script (should work on sh and bash) to run g++ on PGM.cpp to produce executable image PGM. It assumes that the last argument on the command line is the file name (.cpp is optional) and all other arguments are options.
#!/bin/sh
if [ $# -lt 1 ]
then
echo "Usage: `basename $0` [opt] pgm runs g++ to compile pgm[.cpp] into pgm"
exit 2
fi
OPT=
PGM=
# PGM is the last argument, all others are considered options
for F; do OPT="$OPT $PGM"; PGM=$F; done
DIR=`dirname $PGM`
PGM=`basename $PGM .cpp`
# put -o first so it can be overridden by -o specified in OPT
set -x
g++ -o $DIR/$PGM $OPT $DIR/$PGM.cpp
The following will set LAST to last argument without changing current environment:
LAST=$({
shift $(($#-1))
echo $1
})
echo $LAST
If other arguments are no longer needed and can be shifted it can be simplified to:
shift $(($#-1))
echo $1
For portability reasons following:
shift $(($#-1));
can be replaced with:
shift `expr $# - 1`
Replacing also $() with backquotes we get:
LAST=`{
shift \`expr $# - 1\`
echo $1
}`
echo $LAST
echo $argv[$#argv]
Now I just need to add some text because my answer was too short to post. I need to add more text to edit.
This is part of my copy function:
eval echo $(echo '$'"$#")
To use in scripts, do this:
a=$(eval echo $(echo '$'"$#"))
Explanation (most nested first):
$(echo '$'"$#") returns $[nr] where [nr] is the number of parameters. E.g. the string $123 (unexpanded).
echo $123 returns the value of 123rd parameter, when evaluated.
eval just expands $123 to the value of the parameter, e.g. last_arg. This is interpreted as a string and returned.
Works with Bash as of mid 2015.
To return the last argument of the most recently used command use the special parameter:
$_
In this instance it will work if it is used within the script before another command has been invoked.
#! /bin/sh
next=$1
while [ -n "${next}" ] ; do
last=$next
shift
next=$1
done
echo $last
Try the below script to find last argument
# cat arguments.sh
#!/bin/bash
if [ $# -eq 0 ]
then
echo "No Arguments supplied"
else
echo $* > .ags
sed -e 's/ /\n/g' .ags | tac | head -n1 > .ga
echo "Last Argument is: `cat .ga`"
fi
Output:
# ./arguments.sh
No Arguments supplied
# ./arguments.sh testing for the last argument value
Last Argument is: value
Thanks.
There is a much more concise way to do this. Arguments to a bash script can be brought into an array, which makes dealing with the elements much simpler. The script below will always print the last argument passed to a script.
argArray=( "$#" ) # Add all script arguments to argArray
arrayLength=${#argArray[#]} # Get the length of the array
lastArg=$((arrayLength - 1)) # Arrays are zero based, so last arg is -1
echo ${argArray[$lastArg]}
Sample output
$ ./lastarg.sh 1 2 buckle my shoe
shoe
Using parameter expansion (delete matched beginning):
args="$#"
last=${args##* }
It's also easy to get all before last:
prelast=${args% *}
$ echo "${*: -1}"
That will print the last argument
With GNU bash version >= 3.0:
num=$# # get number of arguments
echo "${!num}" # print last argument
Just use !$.
$ mkdir folder
$ cd !$ # will run: cd folder
I have a string containing many words with at least one space between each two. How can I split the string into individual words so I can loop through them?
The string is passed as an argument. E.g. ${2} == "cat cat file". How can I loop through it?
Also, how can I check if a string contains spaces?
I like the conversion to an array, to be able to access individual elements:
sentence="this is a story"
stringarray=($sentence)
now you can access individual elements directly (it starts with 0):
echo ${stringarray[0]}
or convert back to string in order to loop:
for i in "${stringarray[#]}"
do
:
# do whatever on $i
done
Of course looping through the string directly was answered before, but that answer had the the disadvantage to not keep track of the individual elements for later use:
for i in $sentence
do
:
# do whatever on $i
done
See also Bash Array Reference.
Did you try just passing the string variable to a for loop? Bash, for one, will split on whitespace automatically.
sentence="This is a sentence."
for word in $sentence
do
echo $word
done
This
is
a
sentence.
Probably the easiest and most secure way in BASH 3 and above is:
var="string to split"
read -ra arr <<<"$var"
(where arr is the array which takes the split parts of the string) or, if there might be newlines in the input and you want more than just the first line:
var="string to split"
read -ra arr -d '' <<<"$var"
(please note the space in -d ''; it cannot be omitted), but this might give you an unexpected newline from <<<"$var" (as this implicitly adds an LF at the end).
Example:
touch NOPE
var="* a *"
read -ra arr <<<"$var"
for a in "${arr[#]}"; do echo "[$a]"; done
Outputs the expected
[*]
[a]
[*]
as this solution (in contrast to all previous solutions here) is not prone to unexpected and often uncontrollable shell globbing.
Also this gives you the full power of IFS as you probably want:
Example:
IFS=: read -ra arr < <(grep "^$USER:" /etc/passwd)
for a in "${arr[#]}"; do echo "[$a]"; done
Outputs something like:
[tino]
[x]
[1000]
[1000]
[Valentin Hilbig]
[/home/tino]
[/bin/bash]
As you can see, spaces can be preserved this way, too:
IFS=: read -ra arr <<<' split : this '
for a in "${arr[#]}"; do echo "[$a]"; done
outputs
[ split ]
[ this ]
Please note that the handling of IFS in BASH is a subject on its own, so do your tests; some interesting topics on this:
unset IFS: Ignores runs of SPC, TAB, NL and on line starts and ends
IFS='': No field separation, just reads everything
IFS=' ': Runs of SPC (and SPC only)
Some last examples:
var=$'\n\nthis is\n\n\na test\n\n'
IFS=$'\n' read -ra arr -d '' <<<"$var"
i=0; for a in "${arr[#]}"; do let i++; echo "$i [$a]"; done
outputs
1 [this is]
2 [a test]
while
unset IFS
var=$'\n\nthis is\n\n\na test\n\n'
read -ra arr -d '' <<<"$var"
i=0; for a in "${arr[#]}"; do let i++; echo "$i [$a]"; done
outputs
1 [this]
2 [is]
3 [a]
4 [test]
BTW:
If you are not used to $'ANSI-ESCAPED-STRING' get used to it; it's a timesaver.
If you do not include -r (like in read -a arr <<<"$var") then read does backslash escapes. This is left as exercise for the reader.
For the second question:
To test for something in a string I usually stick to case, as this can check for multiple cases at once (note: case only executes the first match, if you need fallthrough use multiple case statements), and this need is quite often the case (pun intended):
case "$var" in
'') empty_var;; # variable is empty
*' '*) have_space "$var";; # have SPC
*[[:space:]]*) have_whitespace "$var";; # have whitespaces like TAB
*[^-+.,A-Za-z0-9]*) have_nonalnum "$var";; # non-alphanum-chars found
*[-+.,]*) have_punctuation "$var";; # some punctuation chars found
*) default_case "$var";; # if all above does not match
esac
So you can set the return value to check for SPC like this:
case "$var" in (*' '*) true;; (*) false;; esac
Why case? Because it usually is a bit more readable than regex sequences, and thanks to Shell metacharacters it handles 99% of all needs very well.
Just use the shells "set" built-in. For example,
set $text
After that, individual words in $text will be in $1, $2, $3, etc. For robustness, one usually does
set -- junk $text
shift
to handle the case where $text is empty or start with a dash. For example:
text="This is a test"
set -- junk $text
shift
for word; do
echo "[$word]"
done
This prints
[This]
[is]
[a]
[test]
$ echo "This is a sentence." | tr -s " " "\012"
This
is
a
sentence.
For checking for spaces, use grep:
$ echo "This is a sentence." | grep " " > /dev/null
$ echo $?
0
$ echo "Thisisasentence." | grep " " > /dev/null
$ echo $?
1
echo $WORDS | xargs -n1 echo
This outputs every word, you can process that list as you see fit afterwards.
(A) To split a sentence into its words (space separated) you can simply use the default IFS by using
array=( $string )
Example running the following snippet
#!/bin/bash
sentence="this is the \"sentence\" 'you' want to split"
words=( $sentence )
len="${#words[#]}"
echo "words counted: $len"
printf "%s\n" "${words[#]}" ## print array
will output
words counted: 8
this
is
the
"sentence"
'you'
want
to
split
As you can see you can use single or double quotes too without any problem
Notes:
-- this is basically the same of mob's answer, but in this way you store the array for any further needing. If you only need a single loop, you can use his answer, which is one line shorter :)
-- please refer to this question for alternate methods to split a string based on delimiter.
(B) To check for a character in a string you can also use a regular expression match.
Example to check for the presence of a space character you can use:
regex='\s{1,}'
if [[ "$sentence" =~ $regex ]]
then
echo "Space here!";
fi
For checking spaces just with bash:
[[ "$str" = "${str% *}" ]] && echo "no spaces" || echo "has spaces"
$ echo foo bar baz | sed 's/ /\n/g'
foo
bar
baz
For my use case, the best option was:
grep -oP '\w+' file
Basically this is a regular expression that matches contiguous non-whitespace characters. This means that any type and any amount of whitespace won't match. The -o parameter outputs each word matches on a different line.
Another take on this (using Perl):
$ echo foo bar baz | perl -nE 'say for split /\s/'
foo
bar
baz