How may I echo all but the last parameter in bash?
I have the following
#!/bin/bash
function f1 ()
{
echo "${@:1:-2}"
}
f1 1 2 3 4 5 6
I need to echo 1 2 3 4 5
man bash tells me that when I use @ I can't use a negative length.
I resorted to using a calculating ("${@:1:$((${#@}-1))}") which is seeming unorthodox to me.
How do I exclude the last parameter from outputting?
bash parameter bash-functions
edited Dec 10 '18 at 11:25
Kusalananda
128k16241399
asked Dec 10 '18 at 10:56
Bret JosephBret Joseph
758
add a comment |
I have the following
#!/bin/bash
function f1 ()
{
echo "${@:1:-2}"
}
f1 1 2 3 4 5 6
I need to echo 1 2 3 4 5
man bash tells me that when I use @ I can't use a negative length.
I resorted to using a calculating ("${@:1:$((${#@}-1))}") which is seeming unorthodox to me.
How do I exclude the last parameter from outputting?
bash parameter bash-functions
edited Dec 10 '18 at 11:25
Kusalananda
128k16241399
asked Dec 10 '18 at 10:56
Bret JosephBret Joseph
758
add a comment |
I have the following
#!/bin/bash
function f1 ()
{
echo "${@:1:-2}"
}
f1 1 2 3 4 5 6
I need to echo 1 2 3 4 5
man bash tells me that when I use @ I can't use a negative length.
I resorted to using a calculating ("${@:1:$((${#@}-1))}") which is seeming unorthodox to me.
How do I exclude the last parameter from outputting?
bash parameter bash-functions
edited Dec 10 '18 at 11:25
Kusalananda
128k16241399
asked Dec 10 '18 at 10:56
Bret JosephBret Joseph
758
I have the following
#!/bin/bash
function f1 ()
{
echo "${@:1:-2}"
}
f1 1 2 3 4 5 6
I need to echo 1 2 3 4 5
man bash tells me that when I use @ I can't use a negative length.
I resorted to using a calculating ("${@:1:$((${#@}-1))}") which is seeming unorthodox to me.
How do I exclude the last parameter from outputting?
bash parameter bash-functions
bash parameter bash-functions
edited Dec 10 '18 at 11:25
Kusalananda
128k16241399
asked Dec 10 '18 at 10:56
Bret JosephBret Joseph
758
edited Dec 10 '18 at 11:25
Kusalananda
128k16241399
asked Dec 10 '18 at 10:56
Bret JosephBret Joseph
758
edited Dec 10 '18 at 11:25
Kusalananda
128k16241399
edited Dec 10 '18 at 11:25
Kusalananda
128k16241399
edited Dec 10 '18 at 11:25
Kusalananda
128k16241399
128k16241399
asked Dec 10 '18 at 10:56
Bret JosephBret Joseph
758
asked Dec 10 '18 at 10:56
Bret JosephBret Joseph
758
asked Dec 10 '18 at 10:56
Bret JosephBret Joseph
758
758
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
echo "${@:1:$#-1}"
The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.
answered Dec 10 '18 at 11:19
KusalanandaKusalananda
128k16241399
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
votes
echo "${@:1:$#-1}"
The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.
answered Dec 10 '18 at 11:19
KusalanandaKusalananda
128k16241399
add a comment |
echo "${@:1:$#-1}"
The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.
answered Dec 10 '18 at 11:19
KusalanandaKusalananda
128k16241399
add a comment |
echo "${@:1:$#-1}"
The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.
answered Dec 10 '18 at 11:19
KusalanandaKusalananda
128k16241399
echo "${@:1:$#-1}"
The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.
answered Dec 10 '18 at 11:19
KusalanandaKusalananda
128k16241399
answered Dec 10 '18 at 11:19
KusalanandaKusalananda
128k16241399
answered Dec 10 '18 at 11:19
KusalanandaKusalananda
128k16241399
answered Dec 10 '18 at 11:19
KusalanandaKusalananda
128k16241399
128k16241399
add a comment |
add a comment |
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