How may I echo all but the last parameter in bash?
up vote
2
down vote
favorite
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 at 11:25
Kusalananda
120k16225367
asked Dec 10 at 10:56
Bret Joseph
365
New contributor
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
2
down vote
favorite
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 at 11:25
Kusalananda
120k16225367
asked Dec 10 at 10:56
Bret Joseph
365
New contributor
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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 at 11:25
Kusalananda
120k16225367
asked Dec 10 at 10:56
Bret Joseph
365
New contributor
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 at 11:25
Kusalananda
120k16225367
asked Dec 10 at 10:56
Bret Joseph
365
New contributor
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 10 at 11:25
Kusalananda
120k16225367
asked Dec 10 at 10:56
Bret Joseph
365
New contributor
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 10 at 11:25
Kusalananda
120k16225367
edited Dec 10 at 11:25
Kusalananda
120k16225367
edited Dec 10 at 11:25
Kusalananda
120k16225367
120k16225367
asked Dec 10 at 10:56
Bret Joseph
365
New contributor
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 10 at 10:56
Bret Joseph
365
asked Dec 10 at 10:56
Bret Joseph
365
365
New contributor
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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3
down vote
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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 $@.
answered Dec 10 at 11:19
Kusalananda
120k16225367
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
echo "${@: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 $@.
answered Dec 10 at 11:19
Kusalananda
120k16225367
add a comment |
up vote
3
down vote
accepted
echo "${@: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 $@.
answered Dec 10 at 11:19
Kusalananda
120k16225367
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
echo "${@: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 $@.
answered Dec 10 at 11:19
Kusalananda
120k16225367
echo "${@: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 at 11:19
Kusalananda
120k16225367
answered Dec 10 at 11:19
Kusalananda
120k16225367
answered Dec 10 at 11:19
Kusalananda
120k16225367
answered Dec 10 at 11:19
Kusalananda
120k16225367
120k16225367
add a comment |
add a comment |
Bret Joseph is a new contributor. Be nice, and check out our Code of Conduct.
Bret Joseph is a new contributor. Be nice, and check out our Code of Conduct.
Bret Joseph is a new contributor. Be nice, and check out our Code of Conduct.
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