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 at 11:25
Kusalananda
121k16228372
asked Dec 10 at 10:56
Bret 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 at 11:25
Kusalananda
121k16228372
asked Dec 10 at 10:56
Bret 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 at 11:25
Kusalananda
121k16228372
asked Dec 10 at 10:56
Bret 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 at 11:25
Kusalananda
121k16228372
asked Dec 10 at 10:56
Bret Joseph
758
edited Dec 10 at 11:25
Kusalananda
121k16228372
asked Dec 10 at 10:56
Bret Joseph
758
edited Dec 10 at 11:25
Kusalananda
121k16228372
edited Dec 10 at 11:25
Kusalananda
121k16228372
edited Dec 10 at 11:25
Kusalananda
121k16228372
121k16228372
asked Dec 10 at 10:56
Bret Joseph
758
asked Dec 10 at 10:56
Bret Joseph
758
asked Dec 10 at 10:56
Bret 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 $@.
answered Dec 10 at 11:19
Kusalananda
121k16228372
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "106"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f487085%2fhow-may-i-echo-all-but-the-last-parameter-in-bash%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
echo "${@: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 $@.
answered Dec 10 at 11:19
Kusalananda
121k16228372
add a comment |
echo "${@: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 $@.
answered Dec 10 at 11:19
Kusalananda
121k16228372
add a comment |
echo "${@: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 $@.
answered Dec 10 at 11:19
Kusalananda
121k16228372
echo "${@: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
121k16228372
answered Dec 10 at 11:19
Kusalananda
121k16228372
answered Dec 10 at 11:19
Kusalananda
121k16228372
answered Dec 10 at 11:19
Kusalananda
121k16228372
121k16228372
add a comment |
add a comment |
Thanks for contributing an answer to Unix & Linux Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f487085%2fhow-may-i-echo-all-but-the-last-parameter-in-bash%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
