How do we do inputrc command by a shell line code as gave failure here
$ bind -x '"\C-l":next-history'
bash: next-history: command not found
$
Please do sincere help
According to man bash:
-x keyseq:shell-command
Cause shell-command to be executed whenever keyseq is entered. [...]
So for your case when you press CTRL-L, Bash will try to execute a command named next-history.
To map to Readline's next-history you can write like this:
bind '"\C-l": next-history'
Related
In a bash script I want to get the name of the last command executed in terminal and store it in the variable for later use. I know that !:0 doesn't work in bash script, and I'm looking for some replacement of it.
For example:
#user enters pwd
> pwd
/home/paul
#I call my script and it show the last command
> ./last_command
pwd
this didn't help, it just prints empty line.
getting last executed command from script
Tell the shell to continuously append commands to the history file:
export PROMPT_COMMAND="history -a"
Put the following into your script:
#!/bin/bash
echo "Your command was:"
tail -n 1 ~/.bash_history
as far as I benefit the working one in my .bashrc;
export HISTCONTROL=ignoredups:erasedups
then do this, on console or in script respectively
history 2
cm=$(history 1)
As an example, I am trying to capture the raw commands that are output by the following script:
https://github.com/adampointer/go-deribit/blob/master/scripts/generate-models.sh
I have tried to following a previous answer:
BASH: echoing the last command run
but the output I am getting is as follows:
last command is gojson -forcefloats -name="${struct}" -tags=json,mapstructure -pkg=${p} >> models/${p}/${name%.*}_request.go
What I would like to do is capture the raw command, in other words have variables such as ${struct}, ${p} and ${p}/${name%.*} replaced by the actual values that were used.
How do I do this?
At the top of the script after the hashbang #!/usr/bin/env bash or #!/bin/bash (if there is any) add set -x
set -x Print commands and their arguments as they are executed
Run the script in debug mode which will trace all the commands in the script: https://stackoverflow.com/a/10107170/988525.
You can do that without editing the script by typing "bash generate-models.sh -x".
I have written the following code:
#!/bin/bash
#Simple array
array=(1 2 3 4 5)
echo ${array[*]}
And I am getting error:
array.sh: 3: array.sh: Syntax error: "(" unexpected
From what I came to know from Google, that this might be due to the fact that Ubuntu is now not taking "#!/bin/bash" by default... but then again I added the line but the error is still coming.
Also I have tried by executing bash array.sh but no luck! It prints blank.
My Ubuntu version is: Ubuntu 14.04
Given that script:
#!/bin/bash
#Simple array
array=(1 2 3 4 5)
echo ${array[*]}
and assuming:
It's in a file in your current directory named array.sh;
You've done chmod +x array.sh;
You have a sufficiently new version of bash installed in /bin/bash (you report that you have 4.3.8, which is certainly new enough); and
You execute it correctly
then that should work without any problem.
If you execute the script by typing
./array.sh
the system will pay attention to the #!/bin/bash line and execute the script using /bin/bash.
If you execute it by typing something like:
sh ./array.sh
then it will execute it using /bin/sh. On Ubuntu, /bin/sh is typically a symbolic link to /bin/dash, a Bourne-like shell that doesn't support arrays. That will give you exactly the error message that you report.
The shell used to execute a script is not affected by which shell you're currently using or by which shell is configured as your login shell in /etc/passwd or equivalent (unless you use the source or . command).
In your own answer, you say you fixed the problem by using chsh to change your default login shell to /bin/bash. That by itself should not have any effect. (And /bin/bash is the default login shell on Ubuntu anyway; had you changed it to something else previously?)
What must have happened is that you changed the command you use from sh ./array.sh to ./array.sh without realizing it.
Try running sh ./array.sh and see if you get the same error.
Instead of using sh to run the script,
try the following command:
bash ./array.sh
I solved the problem miraculously. In order to solve the issue, I found a link where it was described to be gone by using the following code. After executing them, the issue got resolved.
chsh -s /bin/bash adhikarisubir
grep ^adhikarisubir /etc/passwd
FYI, "adhikarisubir" is my username.
After executing these commands, bash array.sh produced the desired result.
i've created simple bash script that do the following
:
#!/usr/bin/env bash
cf ssh "$1"
When I run the command line from the CLI like cf ssh myapp its running as expected, but when I run the script like
. myscript.sh myapp
I got error: App not found
I dont understand what is the difference, I've provided the app name after I invoke the script , what could be missing here ?
update
when I run the script with the following its working, any idea why the "$1" is not working ...
#!/usr/bin/env bash
cf ssh myapp
When you do this:
. myscript.sh myapp
You don't run the script, but you source the file named in the first argument. Sourcing means reading the file, so it's as if the lines in the file were typed on the command line. In your case what happens is this:
myscript.sh is treates as the file to source and the myapp argument is ignored.
This line is treated as a comment and skipped.
#!/usr/bin/env bash
This line:
cf ssh "$1"
is read as it stands. "$1" takes the value of $1 in the calling shell. Possibly - most likely in your case - it's blank.
Now you should know why it works as expected when you source this version of your script:
#!/usr/bin/env bash
cf ssh myapp
There's no $1 to resolve, so everything goes smoothly.
To run the script and be able to pass arguments to it, you need to make the file executable and then execute it (as opposed to sourcing). You can execute the script for example this way:
./script.bash arg1 arg2
I am trying to associate a hotkey with opening vim with recent history browsing, thus I have wrote the following line
gnome-terminal -x "vim -c ':browse old'"
However this gives
Error: Failed to execute child process "vim -c ':browse old'" (No such file or directory)
What am I doing wrong?
Good news! The -x option of gnome-terminal makes it very easy to start a new terminal and run a new program in it. Just do:
gnome-terminal -x vim -c ':browse old'
The meaning of -x is that all subsequent arguments are passed to the program that you run, so no quoting is needed.